1. MATRIX AND DETERMINANTS 1.1. Field and Number Matrix. - Goal: Conceptually speaking, matrix is describing a linear transformation when elements are in R, but matrix itself is more than that. In this section, you are going to learn the most basic, strict, or even abstract concepts of number matrix, to start with, we should define every term: Number and Matrix. To define what is numbers, we define the terminology: Fields. 1.1.1. Fields.
Definition 1.1.1 We call a set F that equipped with two two-term-operations + and × a Field if this set satisfies the following axioms (1) Closed under addition and multiplication if a ∈ F and b ∈ F , then a + b ∈ F and a × b ∈ F (2) Commutativity of + For any two elements a, b ∈ F , we have a + b = b + a (2) Associativity of + For any three elements a, b, c ∈ F we have (a + b) + c = a + (b + c) (3) Existence of 0 There exists an element 0, such that for any elemnt a ∈ F ,a + 0 = a. (4) Existence of opposite For any element a ∈ F , There exists an element called −a, such that a + (−a) = 0 (5) Commutativity of × For any two elements a, b ∈ F , we have a × b = b × a (6) Associativity of × For any three elements a, b, c ∈ F we have (a × b) × c = a × (b × c) (7) Existence of 1 There exists an element 1, such that for any elemnt a ∈ F , a × 1 = a. (8) Existence of reciprocal For any non-zero lement a 6= 0 ∈ F , There exists an element called a−1, such that a × a−1 = 1 (9) Distribute property for any a, b, c ∈ F , we have a × (b + c) = a × b + a × c
As you see, there is nothing strage for us. The importance of this axioms is that not only real number or complex numbers could be numbers. There are more other type of useful numbers
A simple example of a field is the field of rational numbers, consisting of numbers which can be a written as fractions b , where a and b are integers, and b 6= 0. The opposite of such a fraction is a b simply − b , and the reciprocal (provided that a 0) is a .
Suppose F only has two elements, namely 0 and 1, they are zero element and unit element respec- tively. And we define 1 + 1 = 0 ,0 + 1 = 1,0 + 0 = 0,0 ∗ 0 = 0,0 ∗ 1 = 0, 1 ∗ 1 = 1. Then this is a field. This is the minium size field in the universe — field of two element.
It is clear that real numbers(all rational and irrational numbers) formed a field. Later on we will use complex numbers, which can be expressed as a + bi, where i is the mysterious element that saitisfying i2 = −1. We can verify that complex numbers also satisfies the axioms of field. In the book ,we denote real number field as R, and complex number field as C
In this book, the element in field are always called number or scalar. 1 In the study of linear algebra, you will not only encounter calculation, but also proofs. proofs are logic analysis of why some statement is right based on axioms. To show an example, we proof the following proposition.
Proposition 1.1
For any element a ∈ F , a × 0 = 0
Proof. We proof as following: Assume c = a × 0 We Claim that c = c + c, because: By definition of 0, we have 0 + 0 = 0. Thus c = a × 0 = a × (0 + 0) = a × 0 + a × 0 = c + c So we verified c = c + c By the existance of −c, we add −c on both side of equation Thus, (−c) + c = (−c) + c + c So 0 = c Thus we proved c = 0 This means a × 0 = 0 1.1.2. Matrices.
Definition 1.1.2 A Matrix M over a field F or, simply, a matrix M(when the field is clear) is a rectangular array of numbers aijin F , which are usually presented in the following form: a11 a12 ··· a1n a21 a22 ··· a2n . . .. . . . . . am1 am2 ··· amn
The following are matrices over Q (Remember Q is the field of rational numbers, i.e. numbers of a 1 2 the form b , where each a and b are integers. Like 2 , 7 ): 1 2 4 8 3 1 3 7 6 0.212121 ··· 0.5 1 2 3 , 5 5 1 , 8 , (5), 4 23 0.03 0.24 0.333 ··· 3 7 6 where in the last matrix the dots represents repeating digits, not the ignored elements.
The following are matrices over R (Remember R is the field of all rational and irrational numbers, i.e. numbers√ that can be expressed as finitely many integer digit and infinitely many decimal digits. 1 Like 3 , 2, π)
2 √ 4 2 4 5 π 3 2 − 1 3π π , , 10 3 cos 1 1 7 5 e 5 2 6 10
The following are matrices over C (Remember C is the field of all complex numbers, i.e. numbers that can be expressed as a+bi where a and b are real numbers and i is the element satisfies i2 = −1)
1 3 2 3 + i 1 3 1 + i 2i 5 , 6 2 8 , π + 1 i 5 1 3 + πi 5 + 2i 5 6 1 2 4 9 + 9i 1 5i
Because we know rational numbers are included in real numbers, real numbers are included in complex numbers, so a matrix over Q is a matrix over R, and a matrix over R is a matrix over C
a11 a12 ··· a1n a21 a22 ··· a2n A = For a matrix . . .. . over F, . . . . am1 am2 ··· amn the rows of Matrix are called row vectors of Matrix, always denoted as rk, where the subindex are arranged by order: 1st row vector r1 = a11 a12 ··· a1n 2nd row vector r2 = a21 a22 ··· a2n ··· m’th row vector rm = am1 am2 ··· amn The columns of Matrix are called column vectors of Matrix, always denoted as ck, where the subindex are arranged by order: a11 a12 a21 a22 c = c = 1st column vector 1 . 2nd column vector 2 . . . am1 am2 a1n a2n ··· c = n’th column vector n . . amn
√ 3 1 2 For the matrix over R: A = 1 5 8 . The row vectors are 1 3 5 2 √ 1 r1 = 3 1 2 , r2 = 1 5 8 , r3 = 3 5 2
3 The column vectors are √ 3 1 2 c1 = 1 , c2 = 5 , c3 = 8 1 3 5 2
The element aij, which located at i’th row and j’th column are called ij-entry of the Matrix. We frequently denote a matrix by simply writting A = (aij)1≤i≤m if the formula or rule of aij is explicitly given or clear. 1≤j≤n
1 3 5 Suppose (aij)1≤i≤3 = 2 1 9 , What is a12? 1≤j≤3 7 0 8 Answer: a12 is the element getting by the first row and second column, so a12 = 3
What is the matrix (i + j)1≤i≤3? 1≤j≤3 2 3 4 Answer: 3 4 5 4 5 6
2 2 What is the matrix (i + j )1≤i≤1? 1≤j≤2 Answer: 2 5
Besids the notation (aij)1≤i≤m where aij is bunch of numbers. People can also write a matrix as a row 1≤j≤n vector of column vectors, or column vector of row vectors. Explicitely, row vector of column vectors is like A = c1 c2 ··· cn , this only make sense when each the size of each column vector ci are r1 r2 A = equal. The column vector of row vectors is like . where each ri are row vectors, and only . rm make sense when size of each of them are equal.
r1 Suppose r1= 6 2 3 , r2= 4 9 , what is the matrix ? r2 Answer:This does not make sense because the size of r1 are not equal to the size of r2
r1 Suppose r1= 2 3 , r2= 1 5 , what is the matrix ? r2
4 2 3 Answer: 1 5
1 2 Suppose c = , c = , what is the matrix c c ? 1 0 2 9 1 2 1 2 Answer: 0 9
A matrix with m rows and n columns is called an m by n matrix, written as m × n. The pair of the number m × n is called the size of the matrix.
What is the size of 1 3 ? Answer: 1 × 2
Two matrices A and B over F are called equal, written A=B, if they have the same size and each of the corresponing elements are equal. Thus the equality of two m × n matrices is equivalent to a system of mn equalities, each of the equality corresponds to pair of elements.
x y 1 4 Solving the equation = z 5 2 5 Answer: as the 22-entry matches each other, this is possible for the equation make sense. By definition of equal of matrix, we have x = 1, y = 4, z = 2
A matrix of size 1 × n over F, namely, only consists of 1 row, are called row matrix or directly abuse of notation called row vector. A matrix of size m × 1 over F, namely, only consists of 1 column, are called column matrix or directly called column vector
Which of the following are row vectors? Which of them are column vectors?
4 1 3 6 2 4 , 2 , (5), 9 1 7 6 Answer: The 1st is a row vector, the 2nd is a column vector, the third is both row vector and column vector,the last one is neither.
For convenient, people always ignore 0 in the entry of matrices, thus when you see some number matrix has missing entry, that entry represents 0
0 0 0 1 means 0 0 1 0 0 0
5 2 0 2 0 6 3 means 6 0 3 2 0 2 0
1.2. Matrix Multiplication. - Goal:Our Matrix is defined over a field. We have + and × in field. So it should be inherit by the matrix. With these operation, we can do basic algebra with matrices. You will see in the future how matrix algebra simplify all the linear algebra problem. Definition 1.2.3 Suppose Aand Bare two matrices over F and have the same size m × n. The sum of the two matrices are defined to be a11 a12 ··· a1n b11 b12 ··· b1n a21 a22 ··· a2n b21 b22 ··· b2n + . . .. . . . .. . . . . . . . . . am1 am2 ··· amn bm1 bm2 ··· bmn a11 + b11 a12 + b12 ··· a1n + b1n a21 + b11 a22 + b12 ··· a2n + b2n = . . .. . . . . . am1 + bm1 am2 + bm2 ··· amn + bmn
1 7 1 3 5 Does + 2 2 make sense? if it is, please calculate. 7 1 0 1 4 9 Answer: This expression doesn’t make sense because the former is of size 2 × 3, and later 3 × 2, so their size are not equal.
1 3 5 1 7 2 Does + 1 make sense? if it is, please calculate. 7 1 0 2 4 9 1 3 5 1 7 2 1 + 1 3 + 7 5 + 2 2 10 7 Answer: + 1 = 1 = 5 7 1 0 2 4 9 7 + 2 1 + 4 0 + 9 9 4 9
The definition of matrix multiplication is somewhat complicated. To see how complecated it is, we give the strict definition: Definition 1.2.4 Suppose Aand Bare two matrices over F. Then we define the product of this two matrix to be
6 a11 a12 ··· a1n b11 b12 ··· b1p a21 a22 ··· a2n b21 b22 ··· b2p × . . .. . . . .. . . . . . . . . . am1 am2 ··· amn bn1 bn2 ··· bnp Pn Pn Pn j=1 a1jbj1 j=1 a1jbj2 ··· j=1 a1jbjp Pn Pn Pn j=1 a2jbj1 j=1 a2jbj2 ··· j=1 a2jbjp = . . .. . . . . . Pn Pn Pn j=1 amjbj1 j=1 amjbj2 ··· j=1 amjbjp
It might cost you whole life to understand this definition by above words. let’s analysis what is going on. Firstly, to define multiplication. the last number of the size of first matrix should match the first number of the size of last matrix. In other words, the number of the columns of A should be equal to the number of rows of B.
Suppose A is a 3 × 2 matrix, B is a 5 × 9 matrix, does AB make sense? Answer: No, because 2 6= 5
1 5 1 8 0 Suppose A = 8 1 , B = , Does AB make sense? 3 7 2 3 9 Answer: Yes, because the size of A is 3 × 2, and B is of size 2 × 3, the final result AB would be a ∗ ∗ ∗ 3 × 3 matrix.like ∗ ∗ ∗ ∗ ∗ ∗
1 5 1 Suppose A = 8 1 1, B = , Does AB make sense? 8 3 9 Answer: Yes, because the size of A is 3 × 2, and B is of size 2 × 1, the final result AB would be a ∗ 3 × 1 matrix. like ∗ ∗
If the matrix is of right size, suppose we have a11 a12 ··· a1j ··· a1n b11 b12 ··· b1k ··· b1p c11 c12 ··· c1k ··· c1p a21 a22 ··· a2j ··· a2n b21 b22 ··· b ··· b2p c21 c22 ··· c ··· c2p 2k 2k ...... ...... ...... ...... ...... ...... = ai1 ai2 ··· aij ··· ain bj1 bj2 ··· bjk ··· bjp ci1 ci2 ··· cik ··· cip ...... ...... ...... ...... ...... ...... am1 am2 ··· amj ··· amn bn1 bn2 ··· bnk ··· bnp cm1 cm2 ··· cmk ··· cmp 7 Then the element cik is given by multiplying ith row of A and kth column of B:
cik = ai1b1k + ai2b2k + ai3b3k + ··· + ainbnk
5 7 1 5 7 1 × 5 + 5 × 1 + 7 × 3 1 × 7 + 5 × 0 + 7 × 3 1 0 = = 2 8 1 2 × 5 + 8 × 0 + 1 × 3 2 × 7 + 8 × 0 + 1 × 3 8 3 66 28 26 17
1 5 2 1 2 0 0 3 2 8 2 1 − 2 4 6 = 6 −2 −5 0 2 3 8 1 2 −8 1 1
Definition 1.2.5
The zero matrix of size m × n is a m × n matrix with all entries equal to 0, denote as 0m×n or simply 0 if we konw the size from content.
0 0 0 is a zero matrix. 0 0 0
We call a matrix square matrix if it is of size n × n for some integer n. The multiplication of two square matrix make sense if and only if they are of the same size, and the result is still a square matrix of the same size. Now let’s concentrate on square matrix. diagnal of square matrix: We call the left-right, up-down diagnal as diagnal of matrix. Pay attention, we don’t call the other diagnal line the diagnal of matrix. When we say diagnal, we always assume we start from left up and end with right down.
∗ ∗ ∗ In , all the starts lies in the diagnal of the square matrix, but in ∗ , ∗ ∗ ∗ the stars are not lies in the diagnal of the matrix
8 Definition 1.2.6 Unit Matrix: Unit Matrix is a square matrix who has entry 1 in diagnals and 0 elsewhere. we always denote it as In, where n × n is the size of it. Unit Matrix looks like 1 1 1 1 .. . 1
Proposition 1.2
For any square matrix A of size n × n, we have InA = AIn = A
this means the unit matrix acts like multiplicative identity. Definition 1.2.7 Suppose A is a square matrix of size n × n, if there exists a matrix B of the same size, such that BA = In, then we call B the inverse of A, and we call the matrix A invertible matrix we always denote such B as A−1
Proposition 1.3
If a matrix is invertible, then the inverse of a matrix unique, and is commute with the original matrix, −1 −1 that is AA = A A = In
The proof needs the fact that if A have left inverse, then A should also have right inverse. We will prove this proposition after we learned the rank of the matrix. Proposition 1.4
The product of two invertible matrix is invertible, and the inverse is given by: (AB)−1 = B−1A−1
Proof. We calculate B−1A−1AB = B−1(A−1A)B = B−1IB = B−1B = I, so indeed, we can write(B−1A−1)(AB) = −1 −1 −1 I and by the uniqueness of inverse, this implies (AB) = B A Proposition 1.5
Suppose A is an invertible matrix (1) (AT )−1 = (A−1)T (2) (A−1)−1 = A
Proposition 1.6
If the following notation make sense, then (1) (A + B) + C = A + (B + C) (2) A + 0 = 0 + A (Here 0 means 0 matrix)
9 (3) A + (−A) = (−A) + A = 0 (4) A + B = B + A (5) k(A + B) = kA + kB (Here k ∈ F is a scalar in feild) (6) (k + k0)A = kA + k0A (Here k and k’ are two scalars in feild) (7) (kk0)A = k(k0A) (8) 1A = A
Proposition 1.7
If the following notation of matrix make sense, then (1) A(B + C) = AB + AC (2) (B + C)A = BA + CA (3) (AB)C = A(BC) (4) k(AB) = (kA)B = A(kB)
With distributive law, we can apply our algebra skills of numbers to matrices, this is what we called Matrix Algebra. Keep in mind that A and B are not necessarily commute.
Suppose A, B are all 2 × 2 matrices, expand (A + B)(A − B), Is this equal to A2 − B2? When are they equal? Answer: (A + B)(A − B) = A(A − B) + B(A − B) = AA − AB + BA − BB = A2 − B2 + (BA − AB). But A2 − B2 + (BA − AB) = A2 − B2 if and only if BA − AB = 0, that is AB = BA. So (A + B)(A − B) = A2 − B2 if and only if A and B commute.
This example shows when two matrices commute each other, most of the laws for numbers applies to A and B. But tipically are not. Commutativety is a very important problem in linear algebra. It is not exaggeration to say that the uncertainty of the future is just because of the existance of non-commute matrices. Definition 1.2.8
T Suppose A = (aij)1≤i≤m, then we define the transpose of A to be A = (aji)1≤j≤n , to be precise, 1≤j≤n 1≤i≤m T a11 a12 ··· a1n a11 a21 ··· am1 a21 a22 ··· a2n a12 a22 ··· am2 = . . .. . . . .. . . . . . . . . . am1 am2 ··· amn a1n a2n ··· amn
T 2 0 2 3 1 = 3 6 0 6 5 1 5 1 5 T 1 6 = 6 7 5 7
10 1 T 3 = 1 3 5 5
Proposition 1.8
Suppose the following notation of matrix make sense, then (1) (AB)T = BT AT (2) (A + B)T = AT + BT (3) (λA)T = λAT (4) (AT )T = A
We will prove this proposition after we study the block matrix multiplication.
1.2.1. Some special square matrix. - Diagnal Matrix: Diagnal Matrix is a square matrix who only has entry in diagnals and 0 elsewhere. we always denote it as diag(a1, a2, ··· , an), where n × n is the size of it, and each ai is the entry of it. Diagnal Matrix looks like
a1 a2 a3 diag(a , a , ··· , a ) = 1 2 n a4 .. . an
Scalar Matrix: Scalar Matrix is a diagnal matrix with the diagnal entry all the same. It could be viewed as a scalar multiply the unit matrix Unit matrix is scalar matrix, every scalar matrix is diagnal matrix. Zero square matrix is scalar matrix, is diagnal matrix.
−1 0 0 1 Diagnal Matrix over Q: 0 2 0 0 0 5 1 − 2 0 0 1 Scalar Matrix over Q: 0 − 2 0 1 0 0 − 2
Now we discuss what will happen when we multiplying the diagnal matrix. When left multiplying diagnal matrix on a given matrix, the corresponding entries of diagnal matrix multiplied on each row of the given matrix. When right multiplying diagnal matrix, the corresponding entries multiplied on each column of the given matrix. To be precise, look at the following example:
11 2 2 1 3 4 2 6 5 10 20 15 = 50 100 75 6 100 200 120 600 1200 720 This answer is got by applying the matrix multiplication rule. the final result the first row multiplied by 2, 2nd row by 5, 3rd row by 6.
2 10 300 2 4 50 1800 6 70 800 5 = 12 350 4800 1 20 1200 6 2 100 7200 This answer is got by applying the matrix multiplication rule. the final result the first column multiplied by 2, 2nd column by 5, 3rd column by 6
Proposition 1.9
The product of diagnal matrices is diagnal matrix. If the diagnal elements are all non-zero, then the diagnal matrix is invertible, and the inverse is still a diagnal matrix.
Upper triangular matrix We call a matrix A to be upper triangular, if all the non-zero entries lies on or above the diagnal. In other words, all entris below the diagnal is 0. The upper triangular matrices looks like:
a11 a12 ··· a1n a22 ··· a2n .. . . . ann
Proposition 1.10
The product of two upper triangular matrices is still a upper triangular matrix, and the entries of the diagnals of the product is the product of their corresponding entries in diagnals. if all the entries on the diagnal are non-zero, then upper triangular matrix is invertible, and the inverse is still a upper triangular matrix.
1 3 5 2 5 1 2 14 42 We compute 6 7 3 2 = 18 61 9 7 63 This example reflects the fact that the product of two upper triangular matrices is upper triangular, and the diagnal entries correspond to product of each entries in diagnal.2 = 1×2, 18 = 6×3, 63 = 9 × 7
We will prove this result after studying the blockwise computation of matrices. Lower triangular matrix We call a square matrix A to be lower triangular, if all the non-zero entries lies on or below the diagnal. In other words, all entris above the diagnal is 0. The upper triangular matrices looks like: 12 a11 a21 a22 . . .. . . . an1 an2 ··· ann Proposition 1.11
The product of two lower triangular matrices is still a lower triangular matrix, and the entries of the diagnals of the product is the product of their corresponding entries in diagnals. If all the entries in the diagnal are non-zero, then the lower triangular matrix is invertible, and the inverse is a lower triangular matrix.
Symmetric matrix :we call a square matrix A to be symmetric, if it saitisfies AT = A. The entry of symmetric matrix is symmetric with respect to diagnal. See the following examples
The following are symmetric matrices 1 4 6 1 3 , 4 7 5 3 1 6 5 0
1.3. Block Matrix Multiplicaiton. Matrix multiplication is not only valid if we do it numberwise, it’s also make sense blockwise. Definition 1.3.9
A partition of n is an ordered numbers P = (n1, n2, ··· , nk), such that n1 + n2 + ··· + nk = n . We call each summand nk as k’th part of partition Two partition P1 = (a1, a2, ··· , ak) and P2 = (b1, b2, ··· , bk) are said to be the same if each correspond part are the same, that is ai = bi
Let (2, 3, 1) be a partition of 6, this partition separate object like this: oo|ooo|o
Definition 1.3.10
A block matrix is a matrix with partition Pr on its rows and Pc on its columns. These partition separate matrix into blocks. We denote the ij-block the block located in the i’th part row and j’th part column
Suppose we have a partition Pr = (2, 1) and Pc = (1, 2), and a 3 × 3 matrix A. then we can separate A by row partition Pr and column partition Pc into a block matrix: 1 2 3 1 2 3 A = 1 7 9 → 1 7 9 0 2 5 0 2 5 A A and then by this partition we can denote A as a block matrix A = 11 12 A21 A22
13 1 2 3 where The 11-block is A = , 12-block is A = , 21-block is A = 0, 11 1 12 7 9 21 22-block is A22 = 2 5
Proposition 1.12
Block Matrix Multiplication: Suppose A is a block matrix, with row partition Pr and column partition Pc. and B is a block matrix, with row partition Qr and column partition Qc. If Pc = Qr, then the product AB is precisely the same with the same rule applied to blocks, to be precise, A11 A12 ··· A1n B11 B12 ··· B1p A21 A22 ··· A2n B21 B22 ··· B2p × . . .. . . . .. . . . . . . . . . Am1 Am2 ··· Amn Bn1 Bn2 ··· Bnp equals to Pn Pn Pn j=1 A1jBj1 j=1 A1jBj2 ··· j=1 A1jBjp Pn Pn Pn A2jBj1 A2jBj2 ··· A2jBjp j=1 j=1 j=1 . . .. . . . . . Pn Pn Pn j=1 AmjBj1 j=1 AmjBj2 ··· j=1 AmjBjp and the row partition of the product is given by Pr, the column partition is given by Qc
We illustrate this property by an example.
1 5 2 1 2 0 By the normal method of matrix multiplication, we can compute: 8 1 2 2 0 1 = 1 3 4 0 3 5 11 8 15 10 22 11 . Now we use another computation method: 7 14 23 We separate matrices as follows:
1 5 2 1 2 0 8 1 2 2 0 1 1 3 4 0 3 5 And apply our general method of multiply matrices, first we compute each entries:
1 5 1 2 11 + 0 = 8 1 2 2 10 1 5 2 0 2 8 15 + 3 5 = 8 1 0 1 2 22 11 1 1 3 + 4 × 0 = 7 2
14 2 0 1 3 + 4 × 3 5 = 14 23 0 1 11 8 15 Thus, the final result is 10 22 11 7 14 23 It doesn’t matter which partition we use to calculate the product.
Proposition 1.13 A11 A12 ··· A1n A21 A22 ··· A2n A = Transpose of Block Matrix Suppose . . .. . is a block matrix, then . . . . Am1 Am2 ··· Amn T T T A11 A21 ··· An1 T T T A12 A22 ··· An2 AT = . . .. . . . . . T T T A1m A2m ··· Anm
This proposition means before we transpose the position of each block, we should transpose each block itself first.
3 1 2 T 3 2 7 2 5 6 = 1 5 1 7 1 6 2 6 6
To illustrate an application of block matrix multiplication, we give the proof of (AB)T = BT AT Proposition 1.14
(AB)T = BT AT
Proof. Separate A and B into Block Matrix, r1 r2 A = B = c c ··· c we assume . , 1 2 p . rm r1c1 r1c2 ··· r1cp r2c1 r2c2 ··· r2cp AB = We see that . . .. . . . . . rmc1 rmc2 ··· rmcp For column vector c row vector r, we should have rc = cT rT
15 a1 a2 c = r = b b ··· b Assume . , 1 2 n . an T T We compute rc = b1a1 + ··· + bnan, c r = a1b1 + ··· + anbn.
because number multiplication is commute,
Then we have b1a1 + ··· + bnan = a1b1 + ··· + anbn
Then we have rc = cT rT Because a transpose of a number is this number itself, so (rc)T = rc = cT rT
T T T T T T c1 r1 c1 r2 ··· c1 rm T T T T T T c2 r1 c2 r2 ··· c2 rm (AB)T = Thus, . . .. . . . . . T T T T T T cp r1 cp r2 ··· cp rm On the other hand, T c1 T c2 BT = AT = rT rT ··· rT . , 1 2 m . T cp T T T T T T c1 r1 c1 r2 ··· c1 rm T T T T T T c2 r1 c2 r2 ··· c2 rm BT AT = we see that . . .. . . . . . T T T T T T cp r1 cp r2 ··· cp rm T T T So we conclude (AB) = B A
1.4. The row space and column space of a matrix. A very important view of matrix multiplication is by viewing matrix as row vectors of column vectors, or column vectors of row vectors. We can use block matrix to strict our word. The trivial partition of n is (n), which means no partition at all. The singleton partition of n is (1, 1, 1, ··· , 1), which means separate an n-element set in to singletons. Thus, row vector of column vector is a block matrix with trivial partition of rows and singleton partition on columns, and column vector of row vector is a block matrix with trivial partition of columns and singleton partition on rows. Now we use this idea to discuss more about matrix multiplication.
Definition 1.4.11
Suppose c1, c2, ··· , cn are column vectors over F. a linear combination of column vectors c1, c2, ··· , cn with coefficient a1, a2, ··· , an means the column vector formed by c1a1 + c2a2 + ··· + cnan. We also define linear combination of row vectors in the same manner. The lienar combination of row vectors r1, r2, ··· , rn means a1r1 + a2r2 + ··· + anrn.
16 1 1 2 Over field Q. Suppose c1 = 3 , c2 = 1 . Then 4 is a linear combination of c1 and 5 1 6 2 1 1 c2, with the assignments of coefficients a1 = 1, a2 = 1, which means 4 = 3 + 1 6 5 1
1 1 2 Over field Q. Suppose c1 = 3 , c2 = 1 . Then 7 is not a linear combination of 5 1 6 c1 and c2, because we can’t find any coefficient that can combine such a vector. (To prove, show that the entry of any linear combination of c1 and c2 must be a arithmetic sequence)
1 0 1 3 Over field . Suppose c = , c = , c = . Then is a linear Q 1 0 2 1 3 1 4 combination of c1,c2,c3, could with coefficient a1 = 3, a2 = 4, a3 = 0, also could with coefficient a1 = 0, a2 = 1, a3 = 3. There are infinitely many choice of coefficient. In this case, we always 3 say that to represent , these three vectors are redundant, which means we can only choose 4 3 two of them, such as c and c and represents our uniquely as 3c + 4c . 1 2 4 1 2
Proposition 1.15
Suppose Am×n, and Bn×p are two matrices. Let Cm×p = AB. Then each column of C is linear combination of columns of A,with coefficient given by the corresponding rows of B. each row of C is linear combination of rows of B, the coefficeint of i’th row of C comes from i’th row A. The coefficient of j’th column of C comes from j’th column of B.
a11 a12 ··· a1n b11 b12 ··· b1p a21 a22 ··· a2n b21 b22 ··· b2p Proof. A = B = Suppose . . .. . , . . .. . . . . . . . . . am1 am2 ··· amn bn1 bn2 ··· bnp To prove, write A in to a block matrix, a11 a12 ··· a1n a21 a22 ··· b2n A = that is . . .. . . . . . am1 aa2 ··· bmn To simplify notation, denote as A = (c1, c2, ··· , cn) which ci are column vectors. Now using block matrix multiplication, 17 b11 b12 ··· b1p b21 b22 ··· b2p (c , c , ··· , c ) = (d , d , ··· , d ); d = c b +c b +···+ 1 2 n . . .. . 1 2 n where 1 1 11 2 21 . . . . bn1 bn2 ··· bnp cnbn1, d2 = c1b12 + c2b22 + ··· + cnbn2,... This means the column vectors of product is linear combinations of column vectors of A. For the rest of the proof, we left it to the reader.
1 3 6 1 2 0 Problem 1.1. What is the second column of the matrix product 2 7 8 0 0 2 ? 1 9 0 3 4 3 1 3 6 Answer: The second column of product is linear combination of columns of 2 7 8 , and 1 9 0 1 2 0 2 the coefficient comes from the second column of 0 0 2 , that is 0 . Thus we compute: 3 4 3 4 1 6 26 2 × 2 + 8 × 4 = 36 1 0 2
Problem 1.2. Suppose P is a 3 × 3 matrix. And 1 0 1 3 b 2 0 1 0 P = 4 0 1 1 0 1 a 3 c What is a, b, c? We have two method for this question. Solution 1. Since the columns of the product comes from the linear combination of columns from 1 0 first factor. Since any linear combination of 0 and 1 has the form 1 0 x y . x Therefore a = 3, b = 3, c = 2. Next we determine the middle row of P . Since each column of P serves as assignment of coefficients to produce corresponding columns of the product. Therefore, since 3 1 0 4 = 0 3 + 1 4 3 1 0
18 3 1 0 0 = 0 3 + 1 0 3 1 0 2 1 0 1 = 0 2 + 1 1 2 1 0 Then the middle row of P must be 4 0 1
Next we define the row space and column space of a matrix A Definition 1.4.12 Suppose A is an m × n matrix. The set of all possible linear combinations of columns of A is called as the column space of A, denoted by Im(A).
Definition 1.4.13 Suppose A is an m × n matrix. The set of all possible linear combinations of rows of A is called as the row space of A, denoted by coIm(A).
1 2 2 Problem 1.3. For the matrix 3 9 0 , determine (by observation) whether the following 1 2 2 object is an element of ImA. 1 4 1 1 2 6 ; 6 ; 2 ; ; 1 2 2 3 4 1 3 4 4 1 2 4 Solution: Since 6 = 3 2 + 0 1, Therefore 6 is a linear combination of 4 1 2 4 4 columns of A, so 6 ∈ Im(A). 4 1 1 2 1 Since 6 = 3 (−1) + 9 1. So 6 is a linear combination of columns of A, 1 2 2 1 1 so 6 ∈ Im(A). 1 1 The column vector 2 can not be written as a linear combination of columns of A by any 3
19 1 means. So 2 ∈/ Im(A) 3 1 2 Since is a 4 × 1 matrix! it does not make sense at all! All elements in Im(A) should be 3 4 1 2 3 × 1 matrix. So ∈/ Im(A). 3 4 Since 1 2 2 is a 1 × 3 matrix, it also does not make sense. 1 2 2 ∈/ Im(A). But, 1 2 2 ∈ coIm(A)
Definition 1.4.14
Let A be an m × n matrix, let c1, ··· , cn be columns of A. v ∈ Im(A) is an element in the column space of A. By definition, there exists coefficients a1, ··· , an such that v = c1a1+···+cnan. We call a1, ··· , an as an assignments of coefficients for columns of A to produce v by linear combination.
!!
By block matrix multiplication. The expression v = c1a1 + ··· + cnan can also be written as a1 . v = A . an Therefore, the column vector form is a standard way to represent an assignment of coefficients for column vectors. In the definition1.4, we will write the assignment as a1 . . an
Definition 1.4.15
Let A be an m × n matrix, let r1, ··· , rn be rows of A. v ∈ coIm(A) is an element in the row space of A. By definition, there exists coefficients a1, ··· , an such that v = a1r1 + ··· + anrn. We call a1, ··· , an as an assignments of coefficients for rows of A to produce v by linear combination.
!!
By block matrix multiplication. The expression v = a1r1 + ··· + anrn can also be written as v = a1 ··· an A
20 Therefore, the row vector form is a standard way to represent an assignment of coefficients for row vectors. In the definition1.4, we will write the assignment as a1 ··· an
1 2 2 Problem 1.4. For the matrix 3 9 0 , and the following column vectors, determine (by 1 2 2 observation) the assignments of coefficients on column of A to produce them. 4 1 6 ; 6 ; 4 1 Solution: 4 1 2 4 By observation, 6 = 3 2 + 0 1, Therefore 6 is a linear combination of 4 1 2 4 2 columns of A, with the assignment of coefficients 0 . In other words 1 4 2 6 = A 0 . 4 1 4 However, the answer for this problem is not unique. With another observation. 6 = 4 2 2 4 2 4 9 3 + 0 3 . So 6 is a linear combination of columns of A, with the assignment 2 2 4 0 2 of coefficients 3 . In other words 4 3 4 0 6 = A 2/3 . 4 4/3 In fact, we have infinitely many answers for this problem. 1 We left the decomposition of 6 to the reader. 1
In the above example, one might discovered in some cases, we may have more than one assignment of coefficients to produce the same object. To describe the differences. We use the concept of linear relation
21 Definition 1.4.16 For a matrix A.A linear relation on columns of A is an assignment of coefficients on columns of A 0 . to produce . by linear combination. The set of all linear relations on columns of A is denoted 0 as ker(A). In any cases, the zero column matrix is always a linear relation on columns. We call this as trivial linear relation. A linear relation is called non-trivial if it is not trivial.
Definition 1.4.17 For a matrix A.A linear relation on rows of A is an assignment of coefficients on rows of A to produce 0 ··· 0 by linear combination.The set of all linear relations on rows of A is denoted as coker(A). In any cases, the zero row matrix is always a linear relation on rows. We call this as trivial linear relation. A linear relation is called non-trivial if it is not trivial.
The nature of linear relations is describing the difference of assignments of coefficients to produce the same vector. Proposition 1.16
Let A be an m × n matrix, let c1, ··· , cn be columns of A. v ∈ Im(A) is an element in the column a1 b1 . . space of A. Then . and . are two assignments of coefficients to produce v if and only an bn if their difference is a linear relation on columns of A.
Proof. Suppose a1 b1 . . A . = A . an bn Then a1 − b1 0 . . A . = . an − bn 0
We will demonstrate proposition1.4 by our example. Previously we discovered for the matrix 1 2 2 A = 3 9 0 1 2 2
22 2 0 4 There are two assignments of coefficients 0 and 2/3 to produce 6 . The differ- 1 4/3 4 2 2 ence of them is −2/3 . Then −2/3 is a linear relation since clear −1/3 −1/3 1 2 2 0 −2 −1 3 2 + 9 + 0 = 0 1 2 3 2 3 0 or we can write 2 0 A −2/3 = 0 −1/3 0
In other way around, if we discovered a linear relation. Then we can use this linear relation to 2 produce infinitely many assignments of coefficients to produce the same thing. Since 0 is an 1 4 −6 assignment of coefficients for columns of A to produce 6 . And 2 is a linear relation 4 1 of A. Then all those assignments: 2 −6 8 0 − 2 = −2 1 1 0 2 −6 −4 0 + 2 = 2 1 1 2 2 −6 −58 0 + 2 (10) = 20 1 1 11 4 are all assignments of coefficients on columns of A to produce 6 4
Problem 1.5. For the following given matrix 1 0 1 0 1 0 0 1 0 (1) Find a non-trivial linear relation on columns. (2) Find a non-trivial linear relation on rows.
23 1 (3) Find an assignment of coefficients for linear combination of columns to produce 1 . 1 (4) Find couple more assignment of coefficients for linear combination of columns to produce the same thing. Solution 2. We solve the first question. We observed that 1 0 1 0 0 1 + 1 0 + 0 (−1) = 0 0 1 0 0 1 Therefore 0 is a non-trivial linear relation on columns. In other words, we mean −1 1 0 1 1 0 0 1 0 0 = 0 0 1 0 −1 0 We solve the second question. We observe hat 0 1 0 1 + 1 0 1 0 + (−1) 0 1 0 = 0 0 0 Therefore 0 1 −1 is a non-trivial linear relation on rows. In other words, we mean 1 0 1 0 1 −1 0 1 0 = 0 0 0 0 1 0 We solve the third question. We observed that 1 0 1 1 0 1 + 1 1 + 0 0 = 1 0 1 0 1 1 1 Therefore 1 is an assignment of linear coefficient for columns to produce 1 We solve 0 1 the last question. Since we can add or subtract from the assignment of linear coefficient by any scalar multiple of a linear relation of columns without changing the linear combination. So we can write couple more 1 1 2 1 + 0 = 1 0 −1 −1 1 3 4 1 + 0 = 1 0 −3 −3
24 Problem 1.6. Find a non-trivial linear relation on rows of the following matrix 1 2 3 1 1 1 7 8 9 We will study the method of finding a linear relation on rows by observation. Step1: Pick up a row of the matrix. Say the last row 7 8 9 . Step2: Write the row into linear combination of other rows. Warninig: This is not always can be done, you need to select some tricky rows. 7 8 9 = 1 2 3 + 6 1 1 1
Step3: Then move all terms into one side. 1 2 3 + 6 1 1 1 + (−1) 7 8 9 = 0 0 0 So you find a linear relation. 1 2 3 1 6 −1 1 1 1 = 0 0 0 7 8 9
Suppose 1 1 1 1 2 5 1 4 7 P = a b c 1 2 3 1 0 2 What is a,b c? What is the sum of three rows of P . Solution 3. We discovered a linear relation on rows. That is −2 1 1 1 − 1 4 7 + 3 1 2 3 = 0 0 0 Therefore 1 1 1 −2 −1 3 1 4 7 = 0 0 0 1 2 3 So right multiplying P on both hands, we have 1 1 1 −2 −1 3 1 4 7 P = 0 0 0 P = 0 0 0 1 2 3 That is 1 2 5 −2 −1 3 a b c = 0 0 0 1 0 2 So −2 1 2 5 − a b c + 3 1 0 2 = 0 0 0 , therefore a b c = 3 1 0 2 − 2 1 2 5 = 1 −4 −4
25 Now we answer the second question. The sum of three rows of P means the first row add the second row add the third row of P . Which can be written as 1 1 1 P But we discovered that the first row of the following product is exactly means the sum of three rows of P 1 1 1 1 2 5 1 4 7 P = 1 −4 −4 1 2 3 1 0 2 So 1 1 1 P = 1 2 5
1.5. Elementary Matrices, Row and Column Transformations. We have already studied in the last section that we can view the matrix multiplication as linear combi- nation of column vectors of the first matrix, or row vectors of the second matrix. And the coefficient of matrix multiplication is exactly given by other matrix. This shows that to understand matrix multiplication, we have to study the linear combination of row and column vectors. In this section, we will study the most basic linear combination of rows and columns, row and column transformation.
1.5.1. Elementary Row transformation. We have three types of row transformation. row switching This transformation swiches two row of matrix.
1 4 8 3 3 5 r1↔r3 Switch the 1st and 3rd row of matrix 2 0 9 −−−−→ 2 0 9 3 3 5 1 4 8
row multiplying This transformation multiplies some row with a non-zero scalar λ
1 4 8 1 4 8 2×r2 Multiply the 2nd row of matrix by 2 : 2 0 9 −−−→ 4 0 18 3 3 5 3 3 5
row adding In this transformation, we multiply some row by a scalar, but add that into another row.
1 4 8 1 4 8 r3+2×r2 Add twice of the 2nd row to the 3rd row : 2 0 9 −−−−−→ 2 0 9 3 3 5 7 3 23
Caution: Write 2 × r instead of r × 2, the reason for that is simple, because scalar is 1 × 1 matrix. In this view, scalar can only appear in fromt of row vectors. Simillarly, we can define the column transformation in the same way.
1.5.2. column transformation. column switcing This transformation swiches two column of matrix.
26 1 4 8 8 4 1 c1↔c3 Switch the 1st and 3rd column of matrix 2 0 9 −−−−→ 9 0 2 3 3 5 5 3 3
column multiplying This transformation multiplies some column with a non-zero scalar λ
1 4 8 1 8 8 c2×2 Multiply the 2nd column of matrix by 2 : 2 0 9 −−−→ 2 0 9 3 3 5 3 6 5
column adding In this transformation, we multiply some column by a scalar, but add that into another column.
1 4 8 1 4 16 c3+c2×2 Add twice of the 2nd column to the 3rd column : 2 0 9 −−−−−→ 2 0 9 3 3 5 3 3 11
1.5.3. Realization of elementary transformation by matrix multiplication. In the view of last section, row transformation is equivalent to left multiplication, column transformation is equivalent to right multiplica- tion. In order to make it precise, we define the following Elementary Matrices
Definition 1.5.18 The Switching matrix is the matrix obtained by swapping ith and jth rows of unit matrix. Denote by Sij:
1 . .. i 0 1 .. Sij = . j 1 0 . .. 1
Proposition 1.17
Left multiplyting switching matrix Sij will switch ith and jth rows of the matrix.
27 1 0 0 0 0 1 is obtained by switching 2nd and 3rd row of unit matrix, left multiplying 0 1 0 1 0 0 0 0 1 will do the same thing to rows of other matrix. As we compute using definition of 0 1 0 matrix multiplication: 1 0 0 1 2 3 1 2 3 0 0 1 4 5 6 = 7 8 9 0 1 0 7 8 9 4 5 6 1 2 3 The result is exactly switch 2nd and 3rd row of 4 5 6 7 8 9
Proposition 1.18
Right multiplying switching matrix Sij will switch jth and ith columns of the matrix.
1 0 0 0 0 1 is obtained by switching 2nd and 3rd column of unit matrix, right multiplying 0 1 0 1 0 0 0 0 1 will do the same thing to columns of other matrix. As we compute using definition 0 1 0 of matrix multiplication: 1 2 3 1 0 0 1 3 2 4 5 6 0 0 1 = 4 6 5 7 8 9 0 1 0 7 9 8 1 2 3 The result is exactly switch 2nd and 3rd column of 4 5 6 7 8 9
Definition 1.5.19 The Multiplying matrix is the matrix obtained by multiplying ith row by non-zero scalar λ of unit matrix. Denote by Mi(λ): i 1 . .. Mi(λ) = i λ .. . 1
28 Proposition 1.19
Left multiplyting multiplying matrix Mi(λ) will multiplies i’th row of the matrix by λ.
1 0 0 0 3 0 is obtained by multiplying the 2nd row of unit matrix by 3, left multiplying 0 0 1 1 0 0 0 3 0 will do the same thing to rows of other matrix. As we compute using definition of 0 0 1 matrix multiplication: 1 0 0 1 2 3 1 2 3 0 3 0 4 5 6 = 12 15 18 0 0 1 7 8 9 7 8 9 1 2 3 The result is exactly multiplying the 2nd row of 4 5 6 by 3 7 8 9
Proposition 1.20
Right multiplyting multiplying matrix Mi(λ) will multiplies i’th column of the matrix by λ.
1 0 0 0 3 0 is obtained by multiplying the 2nd column of unit matrix by 3, right multi- 0 0 1 1 0 0 plying 0 3 0 will do the same thing to columns of other matrix. As we compute using 0 0 1 definition of matrix multiplication: 1 2 3 1 0 0 1 6 3 4 5 6 0 3 0 = 4 15 6 7 8 9 0 0 1 7 24 9 1 2 3 The result is exactly multiplying the 2nd column of 4 5 6 by 3 7 8 9
29 Definition 1.5.20 The Addition matrix is the matrix obtained by add jth row by scalar λ to the ith row of unit matrix. Denote by Aij(λ): j 1 . .. i 1 λ .. Aij(λ) = . 1 . .. 1
Proposition 1.21
Left multiplying addition matrix Mi(λ) will add λ times j’th row to the i’th row.
1 2 0 0 1 0 is obtained by adding twice of the 2nd row of unit matrix to 1st row, left mul- 0 0 1 1 2 0 tiplying 0 1 0 will do the same thing to rows of other matrix. As we compute using 0 0 1 definition of matrix multiplication: 1 2 0 1 2 3 9 12 15 0 1 0 4 5 6 = 4 5 6 0 0 1 7 8 9 7 8 9 1 2 3 The result is exactly adding twice of the 2nd row to the 1st row of 4 5 6 7 8 9
Proposition 1.22 right multiplying addition matrix Mi(λ) will add λ times i’th column to the j’th column.
30 1 2 0 0 1 0 is obtained by adding twice of the 1st column of unit matrix to 2nd column, 0 0 1 1 2 0 right multiplying 0 1 0 will do the same thing to columns of other matrix. As we 0 0 1 compute using definition of matrix multiplication: 1 2 3 1 2 0 1 4 3 4 5 6 0 1 0 = 4 13 6 7 8 9 0 0 1 7 22 9 1 2 3 The result is exactly adding twice of the 1st column to the 2nd column of 4 5 6 7 8 9
Caution: when Aij(m) serving at left, it is add m times j’th row to i’th row. But when it serving at right, it is add m times i’th column to j’th column. Keep in mind that the order is different. As previous example, we have seen there is an easy way to remember the operation. When left multi- plying, like AB, you think how can we get A by row transformation from unit matrix, and then to get the product, do the same row transformation to B. When right multiplying, like BA, you think how to get A by column transformation from unit matrix, and then to get the product, do the same column transformation to B. All the matrices we defined above Sij, Mi(λ), Aij(λ), are called Elementary matrices. Proposition 1.23
−1 −1 −1 −1 Elementary matrices are invertible, Indeed, Sij = Sij, Mi(λ) = Mi(λ ), Aij(λ) = Aij(−λ)
1.5.4. Row echelon form and column echelon form. Now we want to know by row transformation, how far could we go. We want to make our matrices looks cleaner and better. Now we begin with an example to see how neat we can do. We start by an example: 1 2 1 2 7 2 4 2 6 18 −1 −2 0 −1 −3 Firstly, we use the first row to kill the first entry of other two rows,
r2 − 2 × r1 1 2 1 2 7 1 2 1 2 7 r3 + 1 × r1 2 4 2 6 18 −−−−−−−−−−→ 2 4 −1 −2 0 −1 −3 1 1 4 Now, the first column is clear(with clear I mean that column only contains 1 element), funtunately so is the second column. Now in the third column, we have two 1. But we can’t use the first row to clean the second row, because if we do that. It would hit the 3-1 entry and 3-2 entry, we want to keep them to be 0. Because finally we want it to look like stairs. So we arrange each rows from longer to shorter. Thus, we might swap the last two rows 31 1 2 1 2 7 1 2 1 2 7 r2↔r1 2 4 −−−−→ 1 1 4 1 1 4 2 4
Now we can use the 2nd row to clean up the first row
1 2 1 2 7 1 2 1 3 r1−1×r2 1 1 4 −−−−−→ 1 1 4 2 4 2 4
1 Now in the last row, the leading number is 2, we can make it to be 1 by multiplying 2
1 2 1 3 1 1 2 1 3 2 ×r3 1 1 4 −−−→ 1 1 4 2 4 1 2
Now use the last row to clean up first two rows.
r1 − 1 × r3 1 2 1 3 1 2 1 r2 − 1 × r3 1 1 4 −−−−−−−−−−→ 1 2 1 2 1 2
1.5.5. Block Row and Column Transformations. If you separate Matrix into Blocks, can you do Row and Column transformations? The answer is yes. We define the following elementary block matrices. We have three types of row transformation. block row switching This transformation swiches two row of matrix.
1 2 3 7 8 9 br1↔br2 Switch the 1st and 2nd block row of matrix 4 5 6 −−−−−→ 1 2 3 7 8 9 4 5 6
block row multiplying This transformation left multiplies an invertible matrix to some block row λ
1 3 Left Multiply the 1st block row of matrix by (Note this is invertible): 1 2 1 3 ! ×br 1 2 3 1 2 1 13 17 21 4 5 6 −−−−−−−−−−→ 9 12 15 7 8 9 7 8 9
block row adding In this transformation, we left multiply some row by a matrix, but add that into another block row.
32 1 Add times the 2nd block row to the 1st block row(note: left multiply): 2 1 ! br + ×br 1 2 3 1 2 2 8 10 12 4 5 6 −−−−−−−−−−−→ 18 21 24 7 8 9 7 8 9
!! Note that when we do row transformation, we multiply everything on the left, when we do column transformation, we should multiply on the right
Simillarly, we can define the block column transformation in the same way.
1.5.6. block column transformation. column switching This transformation switches two column of ma- trix.
1 2 3 2 3 1 bc1↔bc2 Switch the 1st and 2nd block column of matrix 4 5 6 −−−−−→ 5 6 4 7 8 9 8 9 7
column multiplying This transformation right multiplies some block column with invertible matrix λ
1 3 ! 1 2 3 bc2× 1 3 1 2 Right Multiply the 2nd block column by : 4 5 6 −−−−−−−−−−→ 1 2 7 8 9 1 5 12 4 11 27 7 17 42
column adding In this transformation, we right multiplies some column by a matrix, but then add that into another column. The examples leave to the reader. Similarly, we also have the following elementary block matrices Suppose we fix a partition P, then the elementary block matrix is a square matrix with same partition P = (n1, n2, ··· , nk) on its columns and rows. So the diagonal block of block matrices are square
33 Definition 1.5.21 The Block Switching matrix is the matrix obtained by swapping ith and jth block of unit matrix. Denote by Sij:
I . .. i 0 I .. Sij = . j I 0 . .. I where each I is of the right size nk
Proposition 1.24
If the row partition of the block matrix A is the same as P, then left multiplying switching matrix Sij will switch i’th and j’th block rows of A.
Proposition 1.25
If the column partition of the block matrix A is the same as P, then right multiplying switching matrix Sij will switch i’th and j’th block column of A.
Definition 1.5.22 The Block Multiplying matrix is the matrix obtained by multiplying ith block row by invertible matrix Pni×ni of unit matrix. Denote by Mi(λ): i I . .. Mi(P ) = i P .. . I
Proposition 1.26
If the row partition of the block matrix A is the same as P, then left multiplying multiplication matrix Mi(P ) will left multiplies i’th block row of the matrix by P .
34 Proposition 1.27
If the column partition of the block matrix A is the same as P, then right multiplying multiplication matrix Mi(P ) will right multiplies i’th block column of the matrix by P .
Definition 1.5.23 The Block Addition matrix is the matrix obtained by add jth block row left multiplied by matrix
Mni×nj to the ith block row of unit matrix. Denote by Aij(M): j I . .. i I M .. Aij(M) = . I . .. I
Proposition 1.28
If the row partition of the block matrix A is the same as P, then left multiplying addition matrix Aij(M) will add M times j’th block row to the i’th block row.
Proposition 1.29
If the column partition of the block matrix A is the same as P, then right multiplying addition matrix Aij(M) will add i’th block column times M to the j’th block column.
Proposition 1.30
Elementary Block Matrices are invertible.
Proposition 1.31
Elementary Block row or column Transformation does not change the rank.
To show an example of the application of the block matrix, Suppose A and B are n × n square matrices, we shows an identity rank(I − AB) = rank(I − BA)
Proof. We consider the 2n × 2n Matrix blocked by partition (n, n) , that is IA BI. 35 IA BI bc −bc ×A I −−−−−−−→2 1 BI − BA br −B×br I −−−−−−−→2 1 I − BA
IA I From here we know rank = rank = n + rank(I − BA) BI I − BA P (Remember in the homework we proved rank = rank(P ) + rank(Q)) We can also do Q another transformation:
IA BI bc −bc ×B I − AB A −−−−−−−→1 2 I br −A×br I − AB −−−−−−−→1 2 I
IA I − AB From here we know rank = rank = rank(I − AB) + n BI I So, n+rank(I-BA)=n+rank(I-AB), thus, rank(I − AB) = rank(I − BA).
This method works for every matrices. The simplest matrix we can reach is the matrix looks like above. We call it Row Echlen Form Definition 1.5.24 A m × n matrix is called a Row Echelon Form if it saitiesfying (1) The first non-zero entry for every row vector is 1, we call this leading 1(or pivot) (2) If a row contains a leading 1, then each row below it contains a leading 1 further to the right (3) If a column contains a leading 1, then all other entries in that column are 0
Which of the following is Row Echelon Form? (1) 2 1 1 1 1 3 (2) 1 6 6 1 1 6 1
36 (3) 1 3 2 1 2 1 1 0 1 Answer: The first one is not a row echelon form because there is a row that leading non-zero entry is 2; The second one is not a echelon form, because if we circle the pivid: 1 6 6 1 1 6 1
In the second column, it contains an element of pivod, but there is other non-zero entry in the second column. The third one is a echelon form, and the pivod is
1 3 2 1 2 1 1 0 1
Now let’s summerize the steps we did in the first example and using the matrix multiplication to explicitly represent that.
r2 − 2 × r1 1 2 1 2 7 1 2 1 2 7 r3 + 1 × r1 2 4 2 6 18 −−−−−−−−−−→ 2 4 −1 −2 0 −1 −3 1 1 4
1 2 1 2 7 r2↔r3 −−−−→ 1 1 4 2 4
1 2 0 1 3 r1−1×r2 (1) −−−−−→ 1 1 4 2 4
1 1 2 1 3 2 ×r3 −−−→ 1 1 4 1 2
r1 − 1 × r3 1 2 1 r2 − 1 × r3 −−−−−−−−−−→ 1 2 1 2 37 1.6. The recorder. Assume A is an m×n matrix. P is an invertible m×m matrix. Consider the expression P −1A
Problem 1.7. Will the product P −1A changes if we apply the same row transformation to P and A simutaneously?
If we do row adding, Let Q be the matrix to represent row adding. Then A changes to QA, P changes to QP . Then (QP )−1(QA) = P −1Q−1QA = P −1A If we do row multiplying, Let Q be the matrix to represent row multiplying. Then A changes to QA, P changes to QP . Then (QP )−1(QA) = P −1Q−1QA = P −1A If we do row switching, Let Q be the matrix to represent row switching. Then A changes to QA, P changes to QP . Then (QP )−1(QA) = P −1Q−1QA = P −1A
Therefore we have the very important property of the expression P −1A Proposition 1.32
Assume A is an m × n matrix. P is an invertible m × m matrix. Then the product P −1A will not change under simultaneous row operation on P and A.
!! This is a natural result. Please think the case m = n = 1. Then P = p and A = a are numbers. The only possible row operation for 1 × 1 matrices are row multiplying. Then this proposition is p nothing more than saying the fraction a will not change if we multiply the same non-zero scalar on top and bottom.
Problem 1.8. How about the expression AP −1? (Now we modify the size of P equals n × n)
You will see this has something to do with the column. If we do column adding, Let Q be the matrix to represent column adding. Then A changes to AQ, P changes to PQ. Then (AQ)(PQ)−1 = AQQ−1P −1 = AP −1 If we do column multiplying, Let Q be the matrix to represent column multiplying. Then A changes to QA, P changes to QP . Then (AQ)(PQ)−1 = AQQ−1P −1 = AP −1
38 If we do column switching, Let Q be the matrix to represent column switching. Then A changes to QA, P changes to QP . Then (AQ)(PQ)−1 = AQQ−1P −1 = AP −1
Therefore we have the very important property of the expression AP −1 Proposition 1.33
Assume A is an m × n matrix. P is an invertible n × n matrix. Then the product AP −1 will not change under simultaneous column operation on P and A.
Therefore, sometime, it is useful to write row or column operations in equalities. The advantage of doing −1 so is no information losses in this process. Since for any m × n matrix A, we have A = (Im) A = −1 A(In) . Then we can do the row or column transformation without the product changes. Here the identity −1 −1 matrix (Im) on the left is called as row operation recorder, and the identity matrix (In) on the right is called as column operation recorder. We rewrite the example in equation(1) in the way of putting a row transformation recorder as following
1 −1 1 2 1 2 7 1 2 4 2 6 18 1 −1 −2 0 −1 −3 1 −1 1 2 1 2 7 = −2 1 2 4 1 1 1 1 4
1 −1 1 2 1 2 7 = 1 1 1 1 4 −2 1 2 4
−1 −1 1 2 0 1 3 = 1 1 1 1 4 −2 1 2 4
−1 −1 1 2 1 3 = 1 1 1 1 4 1 −1 2 1 2
1 −1 1 − 2 −1 1 2 1 1 = 2 − 2 1 1 2 1 −1 2 1 2
Proposition 1.34
For every matrix Am×n, there exists a invertible matrix Pm×m, such that PA is a row echelon form. In this case, PA is called reduced row echelon form of A, denoted as rref(A)
39 !! Whenever you see a statement saying that there exists an invertible matrix P such that PM = N, it is equavalent to say N can be obtained by row transformation from M, sometimes prove the later sentense is more clear. Whenever you see a statement saying that there exists an invertible matrix P such that MP = N, it is equavalent to say N can be obtained by column transformation from M.
1.7. The rank of a matrix. We will define abstractly the rank for the matrix, and explain the geometric intuition in the next section. Definition 1.7.25
For matrix Am×n, the rank of the matrix A is defined to be the number of non-zero rows of rref(A). Or equivalently, the rank is the number of leading ones of the reduced row echelon form of A.
Proposition 1.35
The rank of A is at most the number of non-zero rows of A
1 2 1 Problem 1.9. Find the rank of A = 2 5 1 0 0 1 Solution 4. 1 −1 1 2 1 1 2 5 1 1 0 0 1
1 −1 1 2 1 r2−2r1 ===== −2 1 0 1 −1 1 0 0 1
1 −1 1 2 1 r2+r3 (2) ===== −2 1 1 0 1 0 1 0 0 1
1 −1 −1 1 2 0 r1−r3 ===== −2 1 1 0 1 0 1 0 0 1
3 −2 −3 −1 1 r1−2r2 ===== −2 1 1 1 1 1
40 The reduced rwo echelon form of A has three leading one, so the rank of A is 3, reff(A) = I3
Definition 1.7.26 We call a n × n square matrix A of full rank, if the rank of A is equal to n, the size of A.
Proposition 1.36
If A is an n × n reduced row echelon form, with n leading 1. Then A = I
1.8. The inverse of a square matrix.
1 2 1 Problem 1.10. Find the inverse of the matrix A = 2 5 1 0 0 1 Solution 5. The steps of problem1.9 is also suitable for finding the inverse. Since by calculation 1 2 1 3 −2 −3 −1 2 5 1 = −2 1 1 0 0 1 1 So 1 2 1 −1 3 −2 −3 2 5 1 = −2 1 1 0 0 1 1
Proposition 1.37
For square matrix A, A is invertible if and only if A is full rank.
1.9. Determinant. Introduction:If you think row vector of a matrix as coordinate of vectors in space, then the geometric meaning of the rank of the matrix is the dimension of the parallelpepid spanned by them. But we are not only care about the dimension, sometime we want the volumn of that. Then we will find that determinant is the right value to work out. Historically speaking, determinant determines when does a system of equation with n unknowns and n equalities have unique solution. In our matrix language, determinant determines when a matrix is invertible. You will find out that determinant gives you a fast way to determine when a matrix is invertible and what is the inverse. In 2 × 2 and 3 × 3 it is fast, but in 4 × 4 or more dimension cases, the way we learned before might be the most fast way. We denote the set of all m × n matrices over F by Mm×n Theorem 1.11
There exists function f defined on Mn×n. That is assign every n×n matrix over F a value in F. which saitisfying: (1) If we switch ith and jth rows of B, then the value becomes to opposite. namely
f(SijB) = −f(B)
41 (2) If we multiplied a row of B by λ, then the value is also multiplied by λ(here lambda could be 0). namely f(Mi(λ)B) = λf(B) (3) If we add any scalar times jth row to ith row of B, then the value doesn’t change. namely
f(Aij(λ)B) = f(B) Then this kind of function exists, and any two such function is differ by a scalar multiple.
As any such two functions differ by a scalar multiple, this means whenever we defines the value of unit matrix, then the function is uniquely determined. This let us able to make following definition.
Definition 1.9.27 The determinant is a unique function on n × n matrices, which assign every n × n matrix M over F a value in F. denoted by |M| or det(M) which saitisfying: (1) If we switch ith and jth rows of a matrix, then the determinant becomes opposite. explicitly:
a11 a12 a13 ··· a1n a11 a12 a13 ··· a1n
··············· ···············
ai1 ai2 ai3 ··· ain aj1 aj2 aj3 ··· ajn
··············· = − ···············
aj1 aj2 aj3 ··· ajn ai1 ai2 ai3 ··· ain
··············· ···············
an1 an2 an3 ··· ann an1 an2 an3 ··· ann (2) If we multiplied a any row by λ, then the determinant will also multiplied by λ(here lambda could be 0). explicitly:
a11 a12 ··· a1n a11 a12 ··· a1n
············ ············
λai1 λai2 ··· λain = λ ai1 ai2 ··· ain
············ ············
an1 an2 ··· ann an1 an2 ··· ann (3) If we add any scalar times jth row to ith row of B, then the value doesn’t change. explicitly:
a11 a12 a13 ··· a1n a11 a12 a13 ··· a1n
··············· ···············
ai1 + λaj1 ai2 + λaj2 ai3 + λaj3 ··· ain + λajn ai1 ai2 ai3 ··· ain
··············· = ···············
aj1 aj2 aj3 ··· ajn aj1 aj2 aj3 ··· ajn
··············· ···············
an1 an2 an3 ··· ann an1 an2 an3 ··· ann (4) The determinant for unit matrix is 1
In other words, determinant is the unique function that saitisfying det(SijB) = − det(B); det(Mi(λ)B) = λ det(B); det(Aij(λ)B) = det(B); det(In) = 1
This definition is basically what is determinant, and all other properties of determinant can all be deduced from its uniqueness
42
2 3 Compute the determinant of following matrices ; 4 5 We proceed as following:
2 3
4 5
2 3 = ······ r2 + (−2) × r1 −1
2 = ······ r1 + 3 × r2 −1
1 = 2 × −1
1 = 2 × (−1) × 1
= 2 × (−1) × 1
= −2
Proposition 1.38
If one row is a linear combination of other rows, then the determinant is 0
This is a skill when you can see that, but don’t worry if you can’t. you can calculate all determinant just by row transformation.(also column transformation is valid, we will talk about it in the next few propositions)
2 3 Calculate 4 6 Some one can see that second line is twice of the first line, so the determinant should be 0. But one can also calculate:
2 3
4 6
2 3 = ······ r2 + (−2) × r1
2 3 = 0 × * can be anything ∗ ∗ = 0 anyway, if one row of determinant is 0, the determinant is just 0
43
2 3 4
Calculate 1 1 1
7 8 9 Some one can see that 3rd row is the first line plus five times the second row, so the determinant should be 0. But one can also calculate:
2 3 4
1 1 1
7 8 9
2 3 4
= 1 1 1 ······ r3 + (−1) × r1
5 5 5
2 3 4
= 1 1 1 ······ r3 + (−5) × r2
= 0 because the last row of determinant is 0 row
Proof. The first property is clearly from the axiom.
Proposition 1.39
The determinant for diagonal matrix is the product of all diagonal elements
44 Proof.
a11
a22
a33
.. .
ann
1
a22
a33 = a11 .. .
ann
1
1
a33 = a11a12 .. .
ann
= ···
1
1
1 = a11a22a33 ··· ann .. .
1
Proposition 1.40
The determinant of the upper tirangular matrix is the product of diagonal elements, that is
a11 a12 a13 ··· a1n
a22 a23 ··· a2n
ai3 ··· ain = a11a22a33 ··· ann
······
ann
Proof. We can transfer upper triangular matrix to diagonal matrix by only row addition, with their diagonal elements kept. But row addition does not change the determinant, so it is the same with the determinant of diagonal matrix
45
1 4 5
Calculate 2 6
9 Answer: 1 × 2 × 9 = 18
Proposition 1.41
The matrix M is invertible if and only if det(M) 6= 0
Proof. If det(M) = 0, then for any elementary row transformation of M, the determinant is 0. in other words, for any invertible matrix P, det(PM) = 0, we choose a P such that PM=reff(M). so by this particular choice, we know det(reff(M)) = 0, because reff(M) is upper triangular, so determinant should be the product of diagonal. that means there are 0 in diagonal. so M is not of full rank.
If det(M) 6= 0 then for any elementary row transformation of M, the determinant does not equal to 0 because each time only differ by a non-zero scalar multiple. That is, for any invertible matrix P, det(PM) 6= 0. so choose P such that PM=reff(M), so det(reff(M)) 6= 0, so all element of reff(M) are non zero, that means reff(M) is an unit matrix. so M is of full rank, so M is invertible. Proposition 1.42
det(BA) = det(B) det(A)
Proof. If one of A or B is not invertible, then both side of the equation is 0, so true. If A is invertible, then we det(BA) det((Sij B)A) det(Sij (BA)) det(BA)) consider the function f(B) = det(A) Let’s check f(SijB) = det(A) = det(A) = − det(A) = −f(B). And we can also use the same method to check f(Mi(λ)B) = λf(B) and f(Aij(λ)B) = f(B) det(BA) and f(In) = 1. So f(B)=det(B). Then det(A) = det(B), that is det(BA) = det(B) det(A)
Proposition 1.43
The determinant also saitisfies column transformation property, that is det(BSij) = − det(B); det(BMi(λ)) = λ det(B); det(BAij(λ)) = det(B);
Proof. By previous proposition, determinant is multiplicative, so det(BSij) = det(B) det(Sij) = − det(B); 46 det(BMi(λ)) = det(B) det(Mi(λ)) = λ det(B); det(BAij(λ)) = det(B) det(Aij(λ)) = det(B); This means when we calculate determinant we can both use colomn and row transformation. Proposition 1.44
det(A) = det(AT )
T T T T Proof. Consider the function f(B) = det(B ), we have f(SijB) = det((SijB) ) = det(B Sij) = T T T det(B ) det(Sij) = − det(B ) = −f(B) and we can check f(Mi(λ)B) = λf(B); f(Aij(λ)B) = f(B); T f(In) = 1 by the same process, that is f(B) = det(B). this means det(B) = det(B )
Corollary 1.1
The determinant of lower triangular matrix is the product of all entries in diagonal.
1 2 5 3 Without multiplying two matrix, compute det( 7 8 1 8 ) 9 9 2 5 Answer: 1 2 5 3 det( 7 8 1 8 ) 9 9 2 5
1 2 5 3 = det( 7 8 ) det( 1 8 ) 9 9 2 5
= 63 × 120
= 7560
Next, we give some calculation method for second order determinant and third order determinant. Before we proceed, we illustrate that there is one more important property of determinant that worth using: Proposition 1.45
a11 a12 ··· a1n a11 a12 ··· a1n a11 a12 ··· a1n
············ ············ ············
bi1 + ai1 bi1 + ai2 ··· bi1 + ain = ai1 ai1 ··· ai1 + bi1 bi1 ··· bi1
············ ············ ············
an1 an2 ··· ann an1 an2 ··· ann an1 an2 ··· ann
Proof. This property could also proved by uniqueness of determinant. For A = (aij)1≤i≤nconsider the 1≤j≤n function 47
a11 a12 ··· a1n a11 a12 ··· a1n
············ ············
f(A) = bi1 + ai1 bi1 + ai2 ··· bi1 + ain − bi1 bi1 ··· bi1
············ ············
an1 an2 ··· ann an1 an2 ··· ann As we checked the three properties of f, we should proceed slightly take care of b. We omit proof here, The reader could do that, a good exercise. Proposition 1.46
a b = ac − bd c d
a b a b a c d Proof. = + = − = ad − bc c d c d c d c d b
Proposition 1.47
a b c
d e f = aei + bfg + cdh − ceg − bdi − afh
g h i
Proof. We use the same method as we did before, until we can get an upper triangular, lower triangular or block upper or lower triangular matrix.
a b c
d e f
g h i
a b c
= d e f + d e f + d e f
g h i g h i g h i
a a b b c c
= e + d f + f + d e + e + d f
g h i g h i g h i g h i g h i g h i
a a b b c c
= e − d f + f − e d − e + f d
g h i g i h h i g h g i i h g i g h
= aei − afh + bfg − bdi − ceg + cdh
= aei + bfg + cdh − ceg − bdi − afh
48
2 5 Calculate the determinant of 2 7 Answer: 2 × 7 − 5 × 2 = 4
2 5 6
Calculate the determinant of 1 0 2
3 2 0 Answer: 2×0×0+5×2×3+6×1×2−6×0×3−5×1×0−2×2×2 = 0+30+12−0−0−8 = 34
2
Calculate the determinant of 1 3 2
9 Answer: I want you to practice property of row adding
2
1 3 2
9
2 2
= 3 2 + 1 0
9 9
= 2 × 3 × 9 + 0
= 54
1.10. Laplacian Expansion, Cofactor, Adjugate, Inverse Matrix Formula. In this section, We generalize the method that we used to compute the formula of 2 by 2 and 3 by 3 determinants to n by n matirx, the goal is not to write down the explicite formula, but to make use of it to analyses a method to find the inverse.
Definition 1.10.28 The block diagonal matrix is the block matrix with the same partition on rows and columns, such that all non-zero blocks lies on the diagonal cells. Like: A1 .. . An
49 The following could be viewed as block diagonal matrix with approperate partition 1 1 2 5 = 2 5 6 7 6 7 The above matrix could be seen as a block diagonal matrix with partition (1, 2)
2 3 2 3 4 6 4 6 2 4 7 2 4 7 = 1 2 0 1 2 0 3 5 6 3 5 6 6 6 The above matrix could be seen as block diagonal matrix with partition (1, 3, 2) 1 3 2 4 5 6 5 7 The above matrix is not a block diagonal matrix, as the partition on its columns and rows are not the same
Proposition 1.48
The determinant of block diagonal matrix is the product of the determinant of all diagonal blocks.
Compute the following determinant:
1 2
3 2
2 3
1 5 Answer: The determinant is just the product of the determinant of diagonal matrices. That is
50
1 2
3 2
2 3
1 5
1 2 2 3 = 3 2 1 5
= (−4) × 7
= −28
Definition 1.10.29 The block upper triangular matrix is the block matrix with the same partition on rows and columns, such that all non-zero blocks lies on or above the diagonal cells. Like: A11 ··· A1n .. . . . Ann
Definition 1.10.30 The block lower triangular matrix is the block matrix with the same partition on rows and columns, such that all non-zero blocks lies on or below the diagonal cells. Like: A11 . .. . . An1 ··· An
Proposition 1.49
The determinant of block upper triangular matrix or block lower triangular matrix is the product of determinant of blocks of diagonals.
Definition 1.10.31 Let M be a n × n square matrix over F , and 1 ≤ i ≤ n, 1 ≤ j ≤ n, then the cofactor of M at i,j is defined to be the determinant of (n − 1) × (n − 1) matrix which is obtained by deleting the ith row and jth column. If we multiply this determinant by (−1)i+j, then the value is called the algebraic cofactor, denoted by Mij, where the M is the notation for original matrix.
51 1 2 2 Let M= 8 2 8 , What is M12? −1 2 0 Answer: M12 is got by computing the determinant of cancelling the 1st row and 2nd column, and times (−1)i+j i.e.
1+2 8 8 M12 = (−1) = −8 −1 0
Now we trying to use our method to find the determinant for n × n matirx inductively, and look what we could find.
Trying to simplify the determinant of 5×5 matrix into the computation of determinant 4×4 matrix
1 2 2 1 1
2 4 5 4 4
4 6 7 8 6
5 7 1 1 2
6 1 0 0 7 We use our standard method:
52
1 2 2 1 1
2 4 5 4 4
4 6 7 8 6
5 7 1 1 2
6 1 0 0 7
1 2 2
2 4 5 4 4 2 4 5 4 4 2 4 5 4 4
= 4 6 7 8 6 + 4 6 7 8 6 + 4 6 7 8 6
5 7 1 1 2 5 7 1 1 2 5 7 1 1 2
6 1 0 0 7 6 1 0 0 7 6 1 0 0 7
1 1
2 4 5 4 4 2 4 5 4 4
+ 4 6 7 8 6 + 4 6 7 8 6
5 7 1 1 2 5 7 1 1 2
6 1 0 0 7 6 1 0 0 7
1 2 2
2 4 5 4 4 4 2 5 4 4 5 2 4 4 4
= 4 6 7 8 6 − 6 4 7 8 6 + 7 4 6 8 6
5 7 1 1 2 7 5 1 1 2 1 5 7 1 2
6 1 0 0 7 1 6 0 0 7 0 6 1 0 7
1 1
4 2 4 5 4 4 2 4 5 4
− 8 4 6 7 6 + 6 4 6 7 8
1 5 7 1 2 2 5 7 1 1
0 6 1 0 7 7 6 1 0 0
4 5 4 4 2 5 4 4 2 4 4 4
6 7 8 6 4 7 8 6 4 6 8 6 = 1 × − 2 × + 2 × 7 1 1 2 5 1 1 2 5 7 1 2
1 0 0 7 6 0 0 7 6 1 0 7
2 4 5 4 2 4 5 4
4 6 7 6 4 6 7 8 −1 × + 1 × 5 7 1 2 5 7 1 1
6 1 0 7 6 1 0 0
As you can see, in the expansion, we expanded out the all the entry of the first row, and left with the determinant of numbers that in the different row and column with the entry, with the sign alternating, that exactly the first row times the cofactor corresponds to first row. To conclude, we now have the Laplacian Expansion:
53 Definition 1.10.32 Suppose A is a square matrix, then we can expand A by any row or any column. namely Row expansion: det(A) = ai1Ai1 + ai2Ai2 + ··· + ainAin Column expansion: det(A) = a1iA1i + a2iA2i + ··· + aniAni (Please keep in mind to change the sign, as the term appeared on the formula means algebraic cofac- tors)
Expand the following matrix by the second row
6 8 9
2 4 1
1 2 5
8 9 6 9 6 8 = −2 × + 4 × − 1 × 2 5 1 5 1 2
Expand the following matrix by the second column
6 8 9
2 4 1
1 2 5
6 9 2 1 2 6 = −4 × + 8 × − 2 × 1 5 1 5 1 9
As you can see, the method we used above can inductively give the formula, but what is more than that is the following observation: What if I change the coefficient of each corfactor by the entry of other row? Then in the previous example it would be the laplacian expansion of the determinant:
In the expansion of
6 8 9
2 4 1
1 2 5
8 9 6 9 6 8 = −2 × + 4 × − 1 × 2 5 1 5 1 2
Now if we change the coefficient by the corresponding entries of other lines, that is
54
8 9 6 9 6 8 −1 × + 2 × − 5 × 2 5 1 5 1 2 What the value would be? as you can see, the line that we expanded only determined the coefficient in front of each algebraic cofacter, not appear in any of the cofacter . if we change the coefficient, the only change is the line that we expand by.
6 8 9 8 9 6 9 6 8 −1 × + 2 × − 5 × = −1 −2 −5 = 0 2 5 1 5 1 2 1 2 5 we get the last equation because the second row is scalar multiple of the third row
Proposition 1.50
The othogonal property of cofactor: let A = mataijnn be a matirx, r = k Al1 Al2 a a ··· a cl = k1 k2 kn be any row of A, and . be the cofactor of some row. then . Aln l rkc = 0 if k 6= l l rkc = detA if k = l
At this point, we can make the propersition: Now we construct a matrix, namely, Adjugate Matrix of A, it is constructed by putting Aij in the j’th row and i’th column(remember the order). Then with the adjugate matrix, we have the following property:
Proposition 1.51
∗ ∗ let A = (aij)1≤i≤n be a matirx, we define A = (Aji)1≤j≤n to be the adjunct matrix, then A A = 1≤j≤n 1≤i≤n det(A)In
Then if the determinant is not 0, simply divide the adjugate matrix by determinant, we can get the inverse
Proposition 1.52
∗ let A = (aij)1≤i≤n be a matirx, we define A = (Aji)1≤j≤n to be the adjunct matrix, then 1≤j≤n 1≤i≤n −1 A = (Aji)1≤j≤n 1≤i≤n
2 4 6 Find the inverse of 1 3 4 6 7 2 Answer:
55 2 4 6 1 3 4 6 7 2
3 4 4 6 4 6 − 7 2 7 2 3 4 1 1 4 2 6 2 6 = − − 6 2 6 2 1 4 2 4 6 1 3 4 1 3 2 4 2 4 − 6 7 2 6 7 6 7 1 3
−22 34 −2 1 = − 64 22 −32 −2 −11 10 2
56