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Linear Algebra (VII)

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Linear Algebra (VII) Linear Algebra (VII) Yijia Chen 1. Review Basis and Dimension. We fix a vector space V. Lemma 1.1. Let A, B ⊆ V be two finite sets of vectors in V, possibly empty. If A is linearly indepen- dent and can be represented by B. Then jAj 6 jBj. Theorem 1.2. Let S ⊆ V and A, B ⊆ S be both maximally linearly independent in S. Then jAj = jBj. Definition 1.3. Let e1,..., en 2 V. Assume that – e1,..., en are linearly independent, – and every v 2 V can be represented by e1,..., en. Equivalently, fe1,..., eng is maximally linearly independent in V. Then fe1,..., eng is a basis of V. Note that n = 0 is allowed, and in that case, it is easy to see that V = f0g. By Theorem 1.2: 0 0 Lemma 1.4. If fe1,..., eng and fe1,..., emg be both bases of V with pairwise distinct ei’s and with 0 pairwise distinct ei, then n = m. Definition 1.5. Let fe1,..., eng be a basis of V with pairwise distinct ei’s. Then the dimension of V, denoted by dim(V), is n. Equivalently, if rank(V) is defined, then dim(V) := rank(V). 1 Theorem 1.6. Assume dim(V) = n and let u1,..., un 2 V. (1) If u1,..., un are linearly independent, then fu1,..., ung is a basis. (2) If every v 2 V can be represented by u1,..., un, then fu1,..., ung is a basis. Steinitz exchange lemma. Theorem 1.7. Assume that dim(V) = n and v1,..., vm 2 V with 1 6 m 6 n are linearly indepen- dent. Furthermore, let fe1,..., eng be a basis of V. Then for some 1 6 i1 < i2 < ··· < in-m 6 n v1,..., vm, ei1 ,..., ein-m is a basis of V. 1 We do not assume beforehand that u1,..., un are pairwise distinct, although under the conditions in the theorem, they have to be, i.e., ui 6= uj for every 1 6 i < j 6 n. 1 Proof: We prove by induction on m and start with m = 1. Since fe1,..., eng is a basis, v1 can be represented by e1,..., en. Thus, there exist a1,..., an 2 R such that v1 = aiei. iX2[n] As v1 6= 0 (otherwise, v1 is linearly dependent), there is an i 2 [n] with ai 6= 0. It follows that 1 aj aj ei = v1 + - · ei + - · ei. ai ai ai i<j n j2X[i-1] X6 In other words, ei can be represented by fv1, e1,..., ei-1, ei+1,..., eng. Thus fe1,..., eng can be represented by fv1, e1,..., ei-1, ei+1,..., eng, and so does every vector in α, since fe1,..., eng is a basis. By Theorem 1.6 (2), fv1, e1,..., ei-1, ei+1,..., eng is a basis. Now assume that m > 1 and v1,..., vm 2 V are linearly independent. Of course v1,..., vm-1 are linearly independent too. By induction hypothesis on m - 1, there exist 1 6 i1 < i2 < ··· < in-m+1 6 n such that v1,..., vm-1, ei1 ,..., ein-m+1 is a basis of V. In particular, vmcan be represented by this basis, i.e., there exist a1,..., am-1, c1,..., cn-m+1 2 R such that vm = aivi + cjeij . i2X[m-1] j2[nX-m+1] Observe that j2[n-m+1] cjeij 6= 0, otherwise, v1,..., vm would be linearly dependent. Thus, for some j 2 [n - m + 1] we have c 6= 0, and thereby P j ai 1 c` eij = - · vi + · vm + - · ei` cj cj cj i2X[m-1] `2[n-Xm+1]nfjg Then it is easy to see that v1,..., vm, ei1 ,..., eij-1 , eij+1 ,..., ein-m+1 is a basis of V. 2 Remark 1.8. (i) The above proof is in fact essentially the same proof for Lemma 1.1. (ii) Again, we can drop the requirement 1 6 m, in particular, the case m = 0 holds trivially. 2. Back to the Textbook Matrix and matrix operations. Recall an m × n matrix has the form 2 3 a11 a12 ··· a1n 6 a21 a21 ··· a2n 7 A = 6 7 = a , 6 . .. 7 ij m×n 4 . 5 am1 a21 ··· amn T where each aij 2 R. The transpose matrix of A, denoted by A , is the n × m matrix 2 3 a11 a21 ··· am1 6a12 a22 ··· am2 7 6 7 . 6 . .. 7 4 . 5 a1n a2n ··· amn 2 Definition 2.1 (Matrix Addition). Let A = aij m×n and B = bij m×n be two m × n-matrices. Then 2 3 a11 + b11 ··· a1n + b1n A + B := a + b = 6 . .. 7 . ij ij m×n 4 . 5 am1 + bm1 ··· amn + bmn Definition 2.2. The zero m × n-matrix is 20 ··· 03 6 . .. 7 0m×n = 4 . 5 . 0 ··· 0 When m, n are clear from the context, we write 0 instead of 0m×n. Lemma 2.3. (i) A + B = B + A. (ii) (A + B) + C = A + (B + C). (iii) (A + B)T = AT + BT . Definition 2.4 (Scalar Multiplication). Let A = aij m×n be an m × n-matrix and k 2 R. Then 2 3 k · a11 ··· k · a1n k · A := k · a = 6 . .. 7 . ij m×n 4 . 5 k · am1 ··· k · amn Lemma 2.5. Let A and B be two m × n-matrices and k, ` 2 N. (i) k · (` · A) = (k · `) · A. (ii) (k + `) · A = k · A + ` · A. (iii) k · (A + B) = k · A + k · B. (iv) (k · A)T = k · AT . Definition 2.6. For every A = aij m×n we define -A := -1 · A = - aij m×n. Lemma 2.7. A + (-A) = 0. 2.1. Matrix multiplication. Definition 2.8. Let m, n, r > 1, A be an m × r-matrix, and B an r × n-matrix. Then C := AB = cij m×n is an m × n-matrix where each cij := ai`b`j. `X2[r] 3 Although matrix multiplication seems arbitrary at first sight, we have seen that it could be understood in the context of substituting the variables in one system of linear equations by another system of linear equations. Three matrices, A, B, and C, correspond to the coefficients of the three systems. Remark 2.9. Compared to most multiplications we have encountered before, matrix multiplica- tion is not commutative. – AB and BA can have different size, or even one of them is not defined. For instance, A = 213 1 0 4 and B = 415. Then AB is a 1 × 1-matrix, while BA is 3 × 3. 0 0 0 – Even if they have the same size, AB and BA can be different matrices. Let A := and 0 1 0 1 0 0 0 1 B := . Then AB = and BA = . 0 0 0 0 0 0 This is hardly a surprise by viewing AB as substituting variables xi’s with yi’s and BA the other way around for two systems of linear equations. Definition 2.10. For n > 1 we define 21 0 ··· 03 60 1 ··· 07 I = 6 7 . m 6 . .. 7 4 . 5 0 0 ··· 1 2 In is called the (n × n) identity matrix. Lemma 2.11. (i) (AB)C = A(BC). (ii) C(A + B) = CA + CB = and (A + B)C = AC + BC. (iii) k · (AB) = (k · A)B = A(k · B). (iv) Assume that A is an m × n-matrix. Then ImA = AIn = A. 3. Block Matrix For mostly computational reasons, sometimes we need to partition a matrix into several submatri- ces, or more precisely, block matrices. The following is an example. 2 1 2 -1 0 3 A A A = 6 2 5 0 -2 7 = 11 12 4 5 A A 3 1 -1 3 21 22 where 1 2 -1 0 A = , A = , 11 2 5 0 12 -2 A21 = 3 1 -1 , A22 = 3 . 2 Note that this is not the identity in the matrix space Mn×n(R), which is the scalar 1 2 R. 4 In general, for some p, q 2 N with 1 6 p 6 m and 1 6 q 6 n we break m rows of A into p blocks, and similarly break n columns of A into q blocks: n1 columns nq columns 2 3 m1 rows A11 A12 ··· A1q 6A A ··· A 7 6 21 22 2q 7 6 7 6 . .. 7 4 . 5 mp rows A A ··· A r1 p2 pq . Here, each Aij is an mi × nj-matrix for i 2 [p] and j 2 [q]. Remark 3.1. Note for fixed p and q, we might still have different Aij p×q, which actually de- pends on the choices of mi’s and nj’s. Example 3.2. Let 2 1 1 0 0 0 3 2 3 6 -1 1 0 0 0 7 A1 0 0 6 7 A = 6 0 0 1 0 0 7 = 4 0 A2 0 5 , 6 7 4 0 0 1 1 0 5 0 0 A3 0 0 0 0 1 where 1 1 1 0 A = , A = and A = 1 . 1 -1 1 2 1 1 3 The matrix 2 3 A1 0 0 4 0 A2 0 5 0 0 A3 is a block diagonal matrix. 3.1. Operations on block matrices. Transpose. Assume 2 3 A11 A12 ··· A1q 6A21 A22 ··· A2q 7 A = 6 7 . 6 . .. 7 4 . 5 Ap1 Ap2 ··· Apq Then 2 T T T 3 A11 A21 ··· Ap1 T T 6A12 A22 ··· Ap2 7 AT = 6 7 . 6 . .. 7 4 . 5 T T T A1q A2q ··· Apq Addition.
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