Exercise #10 for Advanced Linear Algebra Instructor: Dr. Y-C. Roger Lin Exercise #10 for Advanced Linear Algebra Due: Monday, May 18, 2009 1. Let T be a linear operator on a vector space V of dimension n. Suppose that T has a nonzero eigenvalue, and the nullspace NT() of T has dimension n −1. Show that T is diagonalizable. =− −⋅ =− = Since dimNT ( ) n 1, we have dimNT ( 0 IV ) n 1. So t1 0 is an eigenvalue of T, =− and the geometric multiplicity of 0 is dn1 1. Thus the algebraic multiplicity of 0 is =− = =− = mn1 1 or mn1 . Since T has a nonzero eigenvalue, we have mn1 1. Thus md11. =− = = Let t2 be the nonzero eigenvalue of T. Since mn1 1, we have m2 1. So d2 1. Thus = md22. Hence T is diagonalizable. ∈ t 2. Let A MFnn× (). Show that A is diagonalizable if and only if its transpose A is also diagonalizable. Suppose A is diagonalizable. Then there are an invertible matrix Q and a diagonal matrix D such that DQAQ= −1 . Thus DQAQQAQQAQtttttttt==()−−−111 ()() = . Since D is diagonal, we see that Dt is diagonal. Hence At is diagonalizable. Conversely, suppose At is diagonalizable. Then (Att ) is diagonalizable. Hence A is diagonalizable. 4. Let A be an nn× matrix. Prove that 2 ≤ dim(span({In , AA, , ""})) n. = 2 => 21m− Let WIAAspan({n , , , ""}) and dim(Wmn ) . Then {IAAn , , , "", A} is a basis for W (Theorem 5.22, P.315). By Cayley-Hamilton theorem (P.317), there are scalars aaan−110, "", , such that −+nn n−1 +++= (1)A aAnn−110"" aAaIO, 21m− =≤ which contradicts that {IAAn , , , "", A} is a basis for W. Hence dim(Wmn ) . 5. Let T be a linear operator on a two-dimensional vector space V. Prove that either V is a = T-cyclic subspace of itself, or TcIV for some scalar c. ≠ ∈ Suppose TcIV for all scalar c. Let x V \{0}. Then {x , Tx( )} is linearly independent. Thus WxTx= span({ , ()}) is a subspace of V and dim(WV )== 2 dim( ). Hence VW= , that is, V is a T-cyclic subspace of itself. 17 Teaching Assistant: C-W. Hsu Exercise #10 for Advanced Linear Algebra Instructor: Dr. Y-C. Roger Lin 6. Let T be a linear operator on a finite-dimensional vector space V, and let W be a T- invariant subspace of V. Suppose that vv1, "", k are eigenvectors of T corresponding ++∈ ∈ to distinct eigenvalues. Prove that if vvW1 "" k , then vWi for all i. Hint: Use mathematical induction on k. = ∈ ∈ When k 1, if vW1 , then vW1 . Thus the statement holds. Suppose the statement holds = ++∈ ∈ = when kn, that is, if vvW1 "" n , then vWi for all in1, "", . When =+ ++∈ =+ kn1, assume vvW11"" n+ . For all in1, "", 1 , let vi be an eigvector of T λ ++∈ with eigenvalue i . Since W is a T-invariant subspace of V, we have Tv()11"" vn+ W λ ++∈ and nn++11()vvW"" 1 . Since T is a linear operator, we have =++= ++ =++λλ ∈ x Tv()()()11111111"" vnnnn++++ Tv "" Tv v "" v W, and =++λλ ∈ yvnnn+++11"" 1 vW 1 . Thus −=λλ − + + λλ − ∈ x yv()111nnnn++"" () 1 vW. λλ− λ = Note that ()ini+1 v is an eigvector of T with eigenvalue i for all in1, "", , and λλ−∈ these eigenvalues are distinct. By induction hypothesis, we have ()ini+1 vW for all = λλ ∈ = in1, "", . Since 11, "", n+ are distinct, we have vWi for all in1, "", . =+ + −+ +∈ Thus vvnn++11()()"" v 1 v 1 "" vW n. It means that the statement holds when kn=+1. Hence the statement holds for all k ∈N . 18 Teaching Assistant: C-W. Hsu .