Vector Spaces §4.6 Rank of a Matrix

Home , Row and column spaces

Preview Row and Column Spaces Solutions of Linear Systems Summary Equivalent Conditions Exercises and Solutions Homework

Satya Mandal, KU

October 25

Satya Mandal, KU Vector Spaces §4.6 Rank of a Matrix Preview Row and Column Spaces Solutions of Linear Systems Summary Equivalent Conditions Exercises and Solutions Homework Goals

Suppose A is a m × n matrix ◮ The row space and column space of A will be defined, as a vector space. ◮ rank(A) will be defined. ◮ Then Null space N(A) of A will be defined, as a vector space. ◮ It will be proved rank(A)+ dim(N(A)) = n.

Satya Mandal, KU Vector Spaces §4.6 Rank of a Matrix Preview Row and Column Spaces Solutions of Linear Systems Summary Equivalent Conditions Exercises and Solutions Homework Row and Column vectors

Suppose A is a m × n matrix. ◮ The n−tuples corrospeding to the rows of A are called row vectors of A. So, the row vectors of A are in Rn. ◮ The m−tuples corrospeding to the columns of A are called columns vectors of A. So, the column vectors of A are in Rm.

Satya Mandal, KU Vector Spaces §4.6 Rank of a Matrix Preview Row and Column Spaces Solutions of Linear Systems Summary Equivalent Conditions Exercises and Solutions Homework Row and Column vectors Suppose row vectors of A a11 a12 ··· a1n (a11, a12, ··· , a1n)  a21 a22 ··· a2n  A = Then, (a21, a22, ··· , a2n) ··· ··· ··· ···   ···  am1 am2 ··· amn    (am1, a12, ··· , amn) The column vectors of A are a11 a12 a1n  a21   a22   a2n  , , ··· ··· ··· ···        am1   am2   amn        Satya Mandal, KU Vector Spaces §4.6 Rank of a Matrix Preview Row and Column Spaces Solutions of Linear Systems Summary Equivalent Conditions Exercises and Solutions Homework Row and Column Space

Suppose A is a m × n matrix, as above. ◮ The row space of A is deﬁned to be the subsapce of Rn spanned by row vectors of A. ◮ The column space of A is deﬁned to be the subsapce of Rn spanned by column vectors of A.

Satya Mandal, KU Vector Spaces §4.6 Rank of a Matrix Preview Row and Column Spaces Solutions of Linear Systems Summary Equivalent Conditions Exercises and Solutions Homework Continued So, ◮ row − space(A)=

(a11, a12, ··· , a1n)  (a21, a22, ··· , a2n)  n span ⊆ R . ···    (am1, a12, ··· , amn)    ◮ column − space(A)=

a11 a12 a1n  a21   a22   a2n  m span , , ··· ⊆ R . ··· ··· ···        am1   am2   amn        Satya Mandal, KU Vector Spaces §4.6 Rank of a Matrix Preview Row and Column Spaces Solutions of Linear Systems Summary Equivalent Conditions Exercises and Solutions Homework Some Properties

Remark. The properties of row-spaces and column-spaces would be analogous. Theorem 4.13 Suppose A, B are two m × n matrices. ◮ If A and B are row equivalent then, rowSpace(A)= rowSpace(B). ◮ If A and B are column equivalent then, columnSpace(A)= columnSpace(B).

Satya Mandal, KU Vector Spaces §4.6 Rank of a Matrix Preview Row and Column Spaces Solutions of Linear Systems Summary Equivalent Conditions Exercises and Solutions Homework

Proof.We will prove only the ﬁrst statement. It is enough to prove it when, B is obtained from A by application of one elementary row operations (to be explained in class). In this case, we have: ◮ All three elementary operations (switching two rows, multiplication of a row by a c =6 0 and addition of a scalar multiple of a row to another) lead to a linear combination of rows of A. So, row − space(B) ⊆ row − space(A). ◮ Since the process is reversible, row − space(A) ⊆ row − space(B). ◮ So, row − space(B)= row − space(A). ◮ The proof is complete.

Satya Mandal, KU Vector Spaces §4.6 Rank of a Matrix Preview Row and Column Spaces Solutions of Linear Systems Summary Equivalent Conditions Exercises and Solutions Homework Basis of Row Space

Theorem 4.14 (Basis of Row Space) Suppose A, B are two m × n matrices. If B is in row-echelon form then the non-zero rows of B forms a basis of row − space(A) (which is same as row − space(B)). Proof.Follows from Theorem 4.14. Reading Assignment: §4.6 Example 1-4.

Satya Mandal, KU Vector Spaces §4.6 Rank of a Matrix Preview Row and Column Spaces Solutions of Linear Systems Summary Equivalent Conditions Exercises and Solutions Homework Rank of a Matrix

Theorem 4.15 Let A be m × n matrix. Then, the row space of A and the column space of A, have same dimension. Proof. Skip or look at the textbook.

Deﬁnition. Let A be m × n matrix. Then the rank of A (written as rank(A)) is deﬁned to be the dimension of the row space of A (equivalently, dimension of the column space of A).

rank(A) := dim(rowSpace(A))

Satya Mandal, KU Vector Spaces §4.6 Rank of a Matrix Preview Row and Column Spaces Solutions of Linear Systems Summary Equivalent Conditions Exercises and Solutions Homework Null Space of a Matrix Definitions Let A be m × n matrix. ◮ The Null Space N(A) is defined to be the set of all the solutions of the homogeneous system of equations Ax = 0. N(A)= {x ∈ Rn : Ax = 0}. The null space N(A) is also called the solution space of Ax = 0. ◮ Theorem. N(A) is a subspace of Rn. ◮ Then nullity of A (written, Nullity(A)) is defined to be the dimension of the null space of A. Notationally, Nullity(A) = dim(N(A)).

Satya Mandal, KU Vector Spaces §4.6 Rank of a Matrix Preview Row and Column Spaces Solutions of Linear Systems Summary Equivalent Conditions Exercises and Solutions Homework Continued

Proof.Only the second statement need a proof. We only need to show that N(A) is closed under addition and scalar multiplication. Suppose x, y ∈ N(A) and c is a scalar. Then

A(x + y)= Ax + Ay = 0 + 0 = 0

So, x + y ∈ N(A). So, N(A) is closed under addition. Also,

A(cx)= cA(x)= c0 = 0

So, cx ∈ N(A). So, N(A) is closed under scalar multipliction. Therefore N(A) is a subsapce of Rn. The proof is complete.

Satya Mandal, KU Vector Spaces §4.6 Rank of a Matrix Preview Row and Column Spaces Solutions of Linear Systems Summary Equivalent Conditions Exercises and Solutions Homework Theorem 4.17

Theorem 4.17 Let A be m × n matrix. ◮ Then rank(A)+ Nullity(A)= n.

◮ This means, if rank of A is r, then the dimension of the solution space of Ax = 0 is n − r. ◮ Note n is the number of variables in the system Ax = 0.

Satya Mandal, KU Vector Spaces §4.6 Rank of a Matrix Preview Row and Column Spaces Solutions of Linear Systems Summary Equivalent Conditions Exercises and Solutions Homework Solutions of Linear Systems

Let Ax = b be a system of linear m equations, in n variables. Remarks: ◮ Here A is a m × n matrix. ◮ For a homogeneous system Ax = 0, set of its solutions is the null space N(A), which is a subspace of Rn. ◮ If the system Ax = b is non-homogeneous (i.e. b =6 0), then the set of solutions is not a subspace of Rn. This is because the zero 0 is not a solution of such a system. ◮ However, the solutions of a system Ax = b is related to the solutions space of he corresponding homogeneous system Ax = 0, the null space N(A).

Let Ax = b be a system of linear m equations, in n variables. ◮ Fix a solution xp of Ax = b. Such a ”particular solution” will be called a ”particular solution”. ◮ Theorem 4.18 Any solution of Ax = b can be written as

x = xp + xh xh ∈ N(A).

That means, xh is any solution of the homogeneous system Ax = 0.

Satya Mandal, KU Vector Spaces §4.6 Rank of a Matrix Preview Row and Column Spaces Solutions of Linear Systems Summary Equivalent Conditions Exercises and Solutions Homework Consistent System

Theorem 4.19 (Consistent System) A system Ax = b is consistent, if b is in the column space of A. Proof.If the system has a solution x, then b is linear combination of the columns: b = x1(First − Column)+ x2(second − Column)+ · + xn(First − Column). The proof is complete.

Satya Mandal, KU Vector Spaces §4.6 Rank of a Matrix Preview Row and Column Spaces Solutions of Linear Systems Summary Equivalent Conditions Exercises and Solutions Homework A Summary

Let A be a square marix of order n. The following conditions are equivalent: 1. A is invertible. 2. Ax = b has a unique solution. 3. Ax = 0 has only trivial solution. 4. A is row equivalent to In. 5. |A| 6= 0. 6. rank(A)= n. 7. The n row vectors of A are linearly independent. 8. The n column vectors of A are linearly independent.

Satya Mandal, KU Vector Spaces §4.6 Rank of a Matrix Preview Row and Column Spaces Solutions of Linear Systems Summary Equivalent Conditions Exercises and Solutions Homework Continued

◮ This (the last frame) is an update of a list given in §3.3. ◮ Last three conditions were added to the old list. ◮ From deﬁnition of rank, it follows

(6) ⇐⇒ (7) ⇐⇒ (8).

Also, (4) ⇐⇒ (7). This completes the proof.

Suppose A is an m × n matrix. ◮ By Theorem 4.14, if B is in row-echelon form (or in reduced row-echelon form) and is row equivalent to A, then the non-zero rows of B forms a basis of row − space(A) Remark. In fact, if B is like a row-echelon matrix, but without ”leading 1”, same works. ◮ Look at Example 2. We work out one problem

4 20 31 Let A =  6 −5 −6  . Find basis of the row space of A. 2 −11 −16  

1. Find a basis of the row space of A. 2. Find the rank of A.

Using TI-84 (”ref”) a row echelon form is:

5 1 − 6 −1 3 B =  0 1 2  0 0 0   5 2 A basis of the row pace : S = 1, − , −1 , 0, 1, 6 3 So, rank(A) = dim(row − space)=2.

◮ Column space of A is the row space of AT (written as rows). ◮ Refer to Example 4. ◮ Let A be as in Exercise 7. Find basis of the column space of A. Solution. We have 4 6 2 T A =  20 −5 −11  31 −6 −16  

Satya Mandal, KU Vector Spaces §4.6 Rank of a Matrix Preview Row and Column Spaces Finding Basis for the row space Solutions of Linear Systems Finding basis of the column space n Summary Equivalent Conditions Finding basis of subspaces of R Exercises and Solutions Problem on Null Spaces Homework Continued To compute a row-echelon form of AT , I avoid TI becaue it gives non-terminating decimals. (a) Subtract 8 times first row from third, (b) multiply third row by -1, and switch with first row: 4 6 2 1 54 32  20 −5 −11  7→  20 −5 −11  −1 −54 −32 4 6 2     Subtract 20 times first row from second and subtract 4 times first row from third: 1 54 32  0 −1085 −651  0 −210 −126   Satya Mandal, KU Vector Spaces §4.6 Rank of a Matrix Preview Row and Column Spaces Finding Basis for the row space Solutions of Linear Systems Finding basis of the column space n Summary Equivalent Conditions Finding basis of subspaces of R Exercises and Solutions Problem on Null Spaces Homework Continued Divide both second row by -217 and third row by -42:

1 54 32 1 54 32  0 5 3  7→  0 5 3  0 5 3 0 0 0     It is in row-echelon-like form. So a basis for row space of AT is S = {(1, 54, 32), (0, 5, 3)}. A basis for column space of A is

1 0  54  ,  5  .  32 3        Satya Mandal, KU Vector Spaces §4.6 Rank of a Matrix Preview Row and Column Spaces Finding Basis for the row space Solutions of Linear Systems Finding basis of the column space n Summary Equivalent Conditions Finding basis of subspaces of R Exercises and Solutions Problem on Null Spaces Homework Exercise 12

4 0 23 1  2 −120 1  Let A = 5 2 21 −1    4 0 22 1     2 −200 1    1. Find the rank of A. 2. Find a basis of the row space of A 3. Find a basis of the column space of A. (Skip)

Proof.First, we reduce A to a reduced row echelon form. Substract ﬁrst row of A from third row, the switch ﬁrst row and third row: 4023 1 1 2 0 −2 −2  2 −12 0 1   2 −12 0 1  1 2 0 −2 −2 7→ 4023 1      4022 1   4022 1       2 −20 0 1   2 −20 0 1     

Subtract 2 times first row from second and fifth, and 4 times first row from third and fourth; then subtract fith row from second: 1 2 0 −2 −2 1 2 0 −2 −2  0 −52 4 5   0128 0  0 −82 11 9 7→ 0 −82 11 9      0 −82 10 9   0 −82 10 9       0 −6 0 −4 5   0 −6 0 −4 5     

Subtract fourth row from third; then add 8 time second row to fourth and 6 times second row to ﬁfth: 1 2 0 −2 −2 12 0 −2 −2  0128 0   012 8 0  0001 0 7→ 000 1 0      0 −82 10 9   0 0 18 74 9       0 −6 0 −4 5   0 0 12 44 5     

Satya Mandal, KU Vector Spaces §4.6 Rank of a Matrix Preview Row and Column Spaces Finding Basis for the row space Solutions of Linear Systems Finding basis of the column space n Summary Equivalent Conditions Finding basis of subspaces of R Exercises and Solutions Problem on Null Spaces Homework Switch third and fourth row, then divide third row by 18; then subtract 12 times third row from ﬁfth: 12 0 −2 −2 1 2 0 −2 −2  012 8 0   012 8 0  00 1 37 .5 7→ 0 0 1 37 .5  9   9   000 1 0   000 1 0       0 0 12 44 5   0 0 0 32 −1      Subtract 32 times 4throw from 5th, then multiply 5throw by -1: 1 2 0 −2 −2 1 2 0 −2 −2  012 8 0   012 8 0  0 0 1 37 .5 7→ 0 0 1 37 .5  9   9   000 1 0   000 1 0       000 0 −1   000 0 1      Satya Mandal, KU Vector Spaces §4.6 Rank of a Matrix Preview Row and Column Spaces Finding Basis for the row space Solutions of Linear Systems Finding basis of the column space n Summary Equivalent Conditions Finding basis of subspaces of R Exercises and Solutions Problem on Null Spaces Homework

The above in row-echelon form and a basis of the row space is given by: (1, 2, 0, −2, −2),  (0, 1, 2, 8, 0),  37 S =  (0, 0, 1, , .5),   9   (0, 0, 0, 1, 0),   (0, 0, 0, 0, 1)    So, rank(A) = dim(row − space)=5. 

Give a subspace spaned by a few vector in Rn ◮ Suppose V is subspace of Rn, spaned by a few vectors. To ﬁnd a basis of V do the following: ◮ Form a matrix A with these vectors. ◮ Then, row space of A is V . ◮ A basis of the row space would be a basis of V . ◮ Refer to Example 3

Let S = {(1, 2, 2), (−1, 0, 0), (1, 1, 1)}. Find a basis of span(S). Solution. Form the matrix A. 1 22 1 2 2  −1 0 0  A row − echelon B =  0 1 1  1 11 0 0 0     So, a basis is {(1, 2, 2), (0, 1, 1)}.

Find a basis for, and dimension of, the solution space of

3 −6 21 Ax = 0 where A =  −2 4 −14  1 −2 7   Solution. The matrix A can be reduce the row echelon form B, by elementary row operations, where

1 −2 7 B =  0 0 0  0 0 0  

◮ So, rank(A)= rank(B)=1. ◮ The solution space of Ax = 0 is the null space N(A). ◮ We have rank(A) + dim(N(A)) = 3 (which is the number of columns). So, dim N(A)=2. ◮ Now, homogeneous systmem, corresponding to B, is x − 2y +7z =0. So, x =2y − 7z. The solution space can be written as x 2s − 7t  y  =  s  where s, t ∈ R  z t      Satya Mandal, KU Vector Spaces §4.6 Rank of a Matrix Preview Row and Column Spaces Finding Basis for the row space Solutions of Linear Systems Finding basis of the column space n Summary Equivalent Conditions Finding basis of subspaces of R Exercises and Solutions Problem on Null Spaces Homework Continued

◮ We can write the solutions as x 2s − 7t 2 7  y  =  s  = s  1  + t  0  z t 0 1         ◮ So, a basis of the solution space is

2 7  1  ,  0   0 1       

Satya Mandal, KU Vector Spaces §4.6 Rank of a Matrix Preview Row and Column Spaces Finding Basis for the row space Solutions of Linear Systems Finding basis of the column space n Summary Equivalent Conditions Finding basis of subspaces of R Exercises and Solutions Problem on Null Spaces Homework Exercise 32 Find a basis for, and dimension of, the solution space of 1 4 21  2 −1 1 1  Ax = 0 where A = 4 2 11    0 4 20    Solution. Use TI-84, The matrix A can be reduce the row echelon form B, by elementary row operations, where 1 .5 .25 .25  0 1 .5 0  B = 1 00 1  3   00 0 1    Satya Mandal, KU Vector Spaces §4.6 Rank of a Matrix Preview Row and Column Spaces Finding Basis for the row space Solutions of Linear Systems Finding basis of the column space n Summary Equivalent Conditions Finding basis of subspaces of R Exercises and Solutions Problem on Null Spaces Homework Continued

◮ So, rank(A)= rank(B)=4. ◮ The solution space of Ax = 0 is the null space N(A). ◮ We have rank(A) + dim(N(A)) = 4 (which is the number of columns). So, dim N(A)=0. So, N(A)= {0} ◮ the solutions space N(A)= {0} of the homogeneous system has ”empty” basis. ◮ Remark. The textbook answers would have said the basis is {0}, which is incorrect way to say this.

Find a basis and the dimension of the solution space of the homogeneous system:

4x −y +2z =0 2x +3y −z =0 6x +2y +z =0

Solution. The coeﬃcient matrix is 1 1 3 −1 2 1 3 6 A =  3 3 −1  . A row−echelon : B =  0 1 −.75  6 2 1 00 0    

◮ So, rank(A)= rank(B)=2. ◮ Since rank(A)+dim N(A) = 3 (which is the number of columns or variables), we have dim N(A)=1. ◮ The homogeneous system corresponding to B is

1 1 x + 3 y + 6 z =0 y −.75z =0

◮ We can write the solutions as 5 5 x − 12 t − 12  y  =  .75t  = t  .75  z t 1       ◮ So, a basis of the solution space is

5 − 12  .75   1     

Satya Mandal, KU Vector Spaces §4.6 Rank of a Matrix Preview Row and Column Spaces Solutions of Linear Systems Summary Equivalent Conditions Exercises and Solutions Homework Homework

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Satya Mandal, KU Vector Spaces §4.6 Rank of a Matrix