Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Jordan normal form wikipedia , lookup
Orthogonal matrix wikipedia , lookup
Matrix multiplication wikipedia , lookup
Vector space wikipedia , lookup
Matrix calculus wikipedia , lookup
Perron–Frobenius theorem wikipedia , lookup
Covariance and contravariance of vectors wikipedia , lookup
System of linear equations wikipedia , lookup
MATH 108A HW 6 SOLUTIONS RAHUL SHAH Problem 1. [§3.15] Solution. ‘⇒’ Let V be a finite dimensional vector space and let T ∈ L(V, W ). Assume that T is surjective. Let B = {v1 , . . . vn } be a basis for V . Notice that even though a priori W need not be finite dimensional, the fact that T is surjective implies that T (B) = {T (v1 ), . . . T (vn )} is a spanning set for W and thus W is finite dimensional. Since T (B) spans W (a finite dimensional vector space), we can reduce it to a basis, B 0 , for W . We will denote B 0 = {w1 , . . . wk } where for each wi ∈ B 0 , wi = T (vj ) for some vj ∈ B. Since for each wi ∈ B 0 , there exists at least one (there might be more, but that won’t matter) vj ∈ B such that T (vj ) = wi , we will define S(wi ) to be vj (i.e. vj is one of the elements of B such that T (vj ) = wi ). Since S has been defined on B 0 , a basis of W , S ∈ L(W, V ). Thus T ◦ S ∈ L(W, W ) = L(W ). To show that T ◦ S = IW , the identity linear map on W , it is enough to show that for each wi ∈ B 0 , S ◦ T (wi ) = wi . So let wi ∈ B 0 . T ◦ S(wi ) = T (vj ), where vj is an element in B such that T (vj ) = wi . Thus T ◦ S(wi ) = T (vj ) = wi . Thus T ◦ S = IW and we are done. ‘⇐’ Let V be a finite dimensional vector space and let T ∈ L(V, W ). Assume that there exists S ∈ L(W, V ) such that T ◦ S = IW , the identity map on W . Assume for contradiction that T is not surjective. Thus, there exists w ∈ W such that w ∈ / T (V ), the image of V . T ◦ S(w) = IW (w) = w. Thus w ∈ T (S(W )). However, since S(W ) ⊂ V (it is actually a linear subspace of V ), we know that T (S(W )) ⊂ T (V ). But then w ∈ T (S(W )) ⇒ w ∈ T (V ) which contradicts the assumption that w ∈ / T (V ) and thus it is a contradiction to assume that such a w exists. Thus ∀w ∈ W , there exists some v ∈ V such that T (v) = w and ergo, T is surjective. Problem 2. [§3.18] Solution. Let A, B and C be matrices such that (A · B) · C and A · (B · C) make sense. Thus if A = (ai,j )m,n , i.e. the i, j-th entry of A is given by ai,j , and A is an m × n matrix, then B = (bi,j )n,o and C = (ci,j )o,p . We will repeatedly use the representation of matrix multiplication given at the bottom of page 51 [Axler, §3]. Thus (A · B) = n X r=1 1 ! ai,r br,j . m,o 2 RAHUL SHAH We thus find that (A · B) · C o n X X = s=1 ai,r br,s ! cs,j r=1 o n X X = (2.1) ! s=1 m,p !! ai,r br,s cs,j r=1 m,p And similarly, (B · C) = o X ! bi,r cr,j r=1 . n,p Which thus gives us that A · (B · C) = n X ai,s s=1 = = (2.2) = o X bs,r cr,j r=1 n o X X s=1 r=1 o X n X r=1 s=1 o n X X s=1 ! m,p !! ai,s bs,r cr,j m,p !! ai,s bs,r cr,j m,p !! by switching r and s as indices. ai,r br,s cs,j r=1 m,p But the i, j-th entry of A · (B · C), given by Equation 2.2, equals the i, j-th entry of (A · B) · C, given by Equation 2.1 and thus we find that matrix multiplication is associative. Problem 3. [§3.20] Solution. Let B = (v1 , . . . vn ) be a basis for V . Thus dim V = n. For v ∈ V , define M(v) to be the matrix of v with respect to the basis B of V . This is well defined since for each v ∈ V , v can be uniquely represented as v = a1 v1 + . . . + an vn , where ai ∈ F. Thus M(v) is well defined as (a1 , . . . an )T ∈ Mat(n, 1, F). For each vi ∈ B, we define T 0 (vi ) = ei ∈ Mat(n, 1, F), where ei is the n×1 matrix that has 0 in all entries except for the i, 1-th entry, which is 1. Since T 0 is defined on the basis B of V , T 0 ∈ L(V, Mat(n, 1, F)). We will show that T (v) = T 0 (v) and thus T (v) is equal to and thus is a linear transformation from V to Mat(n, 1, F). For an arbitrary v ∈ V , v = a1 v1 + . . . + an vn . Thus T (v) = M(v) = (a1 , . . . an )T . However, T 0 (v) = T 0 (a1 v1 + . . . + an vn ) = T 0 (a1 v1 ) + . . . + T (an vn ) = a1 T (v1 ) + . . . + an + T (vn ) = a1 e1 + . . . + an en = (a1 , a2 , . . . an )T = M(v) = T (v) MATH 108A HW 6 SOLUTIONS 3 For each ei , an element in the standard basis of Mat(n, 1, F), define S(ei ) = vi ∈ B. Notice that since we have defined S on a basis for Mat(n, 1, F), S ∈ L(Mat(n, 1, F), V ). To show that T ◦ S = IMat(n,1,F) , we note that T ◦ S(ei ) = T (vi ) = ei = IMat(n,1,F) (ei ). Similarly, S ◦ T (vi ) = S(ei ) = vi = IV . Thus T is invertible (and by [§3.13], it is onto Mat(n, 1, F)). Problem 4. [§3.22] Solution. ‘⇒’ Suppose that V is a finite dimensional vector space and S, T ∈ L(V ). Assume that S ◦ T is invertible. By Theorem 3.21, S ◦ T is surjective. Thus S is surjective (if it was not, then there exists v ∈ V such that v ∈ / S(V ) ⊃ S(T (V )) = S ◦ T (V ) which contradicts the assumption that S ◦ T is surjective). Again by Theorem 3.21, we have that since S ◦ T is invertible, it is injective. Thus T is injective (if it was not, then ∃v1 , v2 ∈ V such that T (v1 ) = T (v2 ) and thus S(T (v1 )) = S(T (v2 )) ⇒ S ◦ T (v1 ) = S ◦ T (v2 ), which contradicts the assumption that S ◦ T is injective). Since S, T ∈ L(V ) such that S is surjective and T is injective, by Theorem 3.21, we have that S, T are both invertible. ‘⇐’ Suppose that V is a finite dimensional vector space and S, T ∈ L(V ). Assume that both S, T are invertible. Thus by Theorem 3.21, both S and T are injective. Thus the composition, S ◦ T , is injective and thus by Theorem 3.21, S ◦ T is invertible. Problem 5. [§3.23] Solution. ‘⇒’ Suppose that V is a finite dimensional vector space and S, T ∈ L(V ). Assume that S ◦ T = I. Since I is invertible, so is S ◦ T . By [§3.22], we then find that both S and T are invertible and thus T is surjective by Theorem 3.21. Thus for each wα ∈ V , there exists vα ∈ V such that T (vα ) = wα . Since S ◦ T = I, S ◦ T (vα ) = vα ⇒ S(wα ) = vα . Now for an arbitrary wβ ∈ V , T ◦ S(wβ ) = T (S(wβ )) = T (vβ ) = vβ = I(vβ ). Thus T ◦ S = I. ‘⇐’ By symmetry of T and S (i.e. T and S are interchangable), the result follows. Problem 6. [§4.3] Solution. Assume that for p, q ∈ P(F), p 6= 0 there exist s1 , r1 , s2 , r2 ∈ P(F) such that q = s1 p + r1 and q = s2 p + r2 such that deg r1 < deg p and deg r2 < deg p. We thus find that s1 p + r1 = q = s2 p + r2 ⇒ (s1 − s2 )p = r2 − r1 . Thus (6.3) deg((s1 − s2 )p) = deg(r2 − r1 ). 4 RAHUL SHAH But deg(r2 − r1 ) ≤ max{deg r1 , deg r2 } < deg p. Since for all a(x), b(x) ∈ P(F), deg(a(x)b(x)) = deg a(x) + deg b(x), and the degree is always non-negative, we thus find that if neither p nor s1 − s2 has degree −∞, i.e. is the zero polynomial, we have that deg((s1 − s2 )p) = deg(s1 − s2 ) + deg p ≥ deg p > deg(r1 − r2 ), which is a contradiction, because deg(r2 − r1 ) = deg(p(s1 − s2 )) by Equation 6.3. We have assumed that p 6= 0 and thus s1 − s2 = 0 and thus s1 = s2 . We hence have that s1 p + r1 = s1 p + r2 ⇒ r1 = r2 . Thus the division algorithm gives us a unique solution.