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5. Solutions to Exercise 5 Exercise 5.1. Suppose V Is Finite-Dimensional. Then Each Operator on V Has at Most Dim V Distinct E

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5. Solutions to Exercise 5 Exercise 5.1. Suppose V Is Finite-Dimensional. Then Each Operator on V Has at Most Dim V Distinct E 24 5. Solutions to Exercise 5 Exercise 5.1. Suppose V is finite-dimensional. Then each operator on V has at most dim V distinct eigenvalues. Solution 5.1. Let T 2 L(V ). Suppose λ1; : : : ; λm are distinct eigenvalues of T . Let v1; : : : ; vm be corresponding eigenvectors. Then fv1; : : : ; vmg is linearly independent. Thus m ≤ dim V . Exercise 5.2. (1) Suppose S; T 2 L(V ) are such that ST = TS. Prove that Nul(S) is invariant under T . Solution 5.2. For any v 2 Nul(S), S(v) = 0. Since ST = TS, S(T (v)) = T (S(v)) = T (0) = 0. Then T (v) 2 Nul(S). Then Nul(S) is invariant under T . (2) Suppose S; T 2 L(V ) are such that ST = TS. Prove that im(S) is invariant under T . Solution 5.2. For any v 2 im(S), there exists w 2 V such that S(w) = v. Since ST = TS, T (v) = T (S(w)) = S(T (w)). Then T (v) 2 im(S). Then im(S) is invariant under T . Exercise 5.3. See the proof of Theorem 5.4.1. Let v 2 V . Let f[v1];:::; [vk]g be a basis V / Span(v). Please show that fv; v1; : : : ; vkg is linearly independent using the definition of linearly independence. Solution 5.3. Set up the equation in V : cv + c1v1 + ::: + ckvk = 0: (**) Then c1v1 + ::: + ckvk = −cv 2 Span(v). Then [c1v1 + ::: + ckvk] = 0 2 V / Span(v). So c1[v1] + ::: + ck[vk] = [c1v1 + ::: + ckvk] = 0: Since f[v1];:::; [vk]g is a basis of V / Span(v), c1 = ::: = ck = 0. Then in Equation (**), we have cv = 0. Therefore c = 0. Then fv; v1; : : : ; vkg is linearly independent. □ Exercise 5.4. Suppose V is a finite-dimensional complex vector space and T 2 L(V ). Prove that T has an invariant subspace of dimension k for each k = 1;:::; dim V . 25 h i Proof. Let n = dim V . By Theorem 5.4.1, there is a basis B = fv1; : : : ; vng such that T B B is upper-triangular. Then T (v1) 2 Span(v1); T (v2) 2 Span(v1; v2); :::::: T (vn) 2 Span(v1; : : : ; vn): Then Wk = Span(v1; : : : ; vk) is the desired T -invariant subspace of dimension k for any k = 1; : : : ; n. □ Exercise 5.5. Suppose W is a complex vector space and T 2 L(W ) has no eigenvalues. Prove that every subspace of W invariant under T is either f0g or infinite-dimensional. Proof. Use contradiction. Assume that there is a finite-dimensional non-zero subspace U of W invariant under T . Then by Theorem 5.4.2, T jU has an eigenvector in U. Then there exists w 2 U, λ 2 C such that w =6 0 and T (w) = λw. Since w 2 U ⊂ W , w is an eigenvector of T in W . This is a contradiction. So the assumption is wrong. Then every subspace of W invariant under T is either f0g or infinite-dimensional. □ Exercise 5.6. Let T 2 L(C3) which is defined by the matrix 2 3 − 62 2 07 6 7 6 7 A = 40 3 05: 1 6 2 Find a basis C3 to write T as an upper-triangular matrix. 8 2 3 2 3 2 39 > > <> 617 607 607=> 3 6 7 6 7 6 7 Solution 5.6. Let V = C . Let S = e1 = 607; e2 = 617; e3 = 607 be the standard basis. > 4 5 4 5 4 5> :> ;> 0 0 1 Then 2 3 − h i 62 2 07 6 7 T = 60 3 07: S S 4 5 1 6 2 26 (1) Find an eigenvector of T on V : det(A − λI) = 0. Then (2 − λ)2(3 − λ) = 0. So 2 and 3 are eigenvalues. We use 2 as our current eigenvalue here. Solve the equation (A − 2I)X = 0 for X 2 C3. 2 3 − − 62 2 2 0 7 6 7 6 − 7 4 0 3 2 0 5X = 0: 1 6 2 − 2 02 31 2 3 B607C 607 B6 7C 6 7 2 B6 7C 6 7 Solutions are X Span @405A. We take v1 = 405 as our first eigenvector. 1 1 nh i h io (2) Consider the quotient space V / Span(v1). Consider the basis e1 ; e2 where 2 3 2 3 h i 617 h i 607 6 7 6 7 = 6 7 + (v ); = 6 7 + (v ): e1 405 Span 1 e2 415 Span 1 0 0 2 3 2 3 h i 1 h i 0 ' C2 7! 4 5 7! 4 5 V / Span(v1) by e1 , e2 . 0 1 (3) Since 22 33 ([ ]) h i 66277 h i 66 77 T = = 66 77 = 2 ; e1 T (e1) 44055 e1 1 22 33 − ([ ]) h i 66 277 h i h i 66 77 T = = 66 77 = −2 + 3 : e2 T (e2) 44 3 55 e1 e2 6 2 3 nh i h io − 42 25 Therefore under the basis e1 ; e2 , the matrix of T is B = . 0 3 (4) Find an eigenvector of T on V / Span(v1): det(B − λI) = 0. Then (2 − λ)(3 − λ) = 0. So 2 and 3 are eigenvalues. We use 2 as our current eigenvalue here. Solve the equation 27 2 (B − 2I)X = 0 for X 2 C ' V / Span(v1). 2 3 2 − 2 −2 4 5X = 0: 0 3 − 2 02 31 2 3 1 1 @4 5A 4 5 2 Solutions are X 2 Span . We take v2 = . By the isomorphism between C and 0 0 2 3 2 3 617 1 6 7 4 5 6 7 V / Span(v1), is corresponding to 405 + Span(v1). We take a representative from the 0 0 2 3 617 6 7 6 7 class to be v2. For example we can pick v2 = 405. 0 nh io (5) Consider the quotient space V / Span(v1; v2). Consider the basis e2 where 2 3 h i 607 6 7 = 6 7 + (v ; v ): e2 415 Span 1 2 0 h i ' C 7! V / Span(v1; v2) by e2 1. (6) Since 22 33 − ([ ]) h i 66 277 h i 66 77 T = = 66 77 = 3 : e2 T (e2) 44 3 55 e2 6 28 h i Then e is an eigenvector of T in V / Span(v1; v2). The vector is corresponding to the 2 32 607 6 7 6 7 class 415 + Span(v1; v2). We take a representative from the class to be v3. For example we 0 2 3 607 6 7 6 7 can pick v3 = 415. 0 8 2 3 2 3 2 39 > > <> 607 617 607=> 6 7 6 7 6 7 (7) We now have a basis B = v1 = 607; v2 = 607; v3 = 617 . The change-of-basis PS B = > 4 5 4 5 4 5> :> ;> 1 0 0 2 3 60 1 07 6 7 6 7 40 0 15. Then the matrix of T is 1 0 0 2 3 h i h i 62 1 6 7 −1 6 7 T = PS B T PS B = 60 2 −27: B B S S 4 5 0 0 3.
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