Linear Algebra II Lecture 22 Xi Chen University of Alberta March 4, 24
Outline Characteristic Polynomial, Eigenvalue, Eigenvector and
Eigenvalue, Eigenvector and Let T : V V be a linear endomorphism. We call λ an eigenvalue of T and v an eigenvector of T corresponding to λ if T (v) = λv. All eigenvectors v of T corresponding to λ lie in the eigenspace {v : T (v) = λv} = {v : (λi T )v} = K (λi T ) of T corresponding to λ. If dim V <, we call det(xi [T ] B,B ) = x n + a x n +... + a n the characteristic polynomial of T, which is independent of the choice of B. T (v) = λv f (T )(v) = f (λ)v for all polynomials f (x). (Cayley-Hamilton) f (T ) = for the characteristic polynomial f (x) of T.
Characteristic Polynomial, Eigenvalue, Eigenvector and We call T : V V diagonalizable if there exists an ordered basis B of V such that [T ] B,B is diagonal. Let V be a vector space of dimension n. Then a linear transformation T : V V is diagonalizable if and only if T has n linearly independent eigenvectors v, v 2,..., v n. In addition, λ λ 2 [T ] B,B =... λn for B = {v, v 2,..., v n } if v, v 2,..., v n are linearly independent eigenvectors of T corresponding to λ, λ 2,..., λ n.
Eigenvalues, Eigenvectors and Let λ, λ 2,..., λ m be all the eigenvalues of T : V V. Then T is diagonalizable if and only if V = K (λ I T ) + K (λ 2 I T ) +... + K (λ m I T ) or equivalently, dim V = dim K (λ I T )+dim K (λ 2 I T )+...+dim K (λ m I T ) or equivalently, (m ) dim V = rank(λ I T )+rank(λ 2 I T )+...+rank(λ m I T ). T is diagonalizable f (T ) is diagonalizable for all polynomials f (x).
Let T : R 3 R 3 be a linear transformation with two eigenvalues and 2. If dim K (I T ) = and dim K (2I T ) =, is T diagonalizable? No since dim K (I T ) + dim K (2I T ) < 3. There exists an ordered basis B such that [T ] B,B = or 2 2. 2
Let T : R 3 R 3 be a linear transformation with two eigenvalues and 2. If dim K (I T ) = and dim K (2I T ) = 2, is T diagonalizable? Yes since dim K (I T ) + dim K (2I T ) = 3. Let B = {v, v 2, v 3 } with {v } a basis of K (I T ) and {v 2, v 3 } a basis of K (2I T ). Then [T ] B,B = 2. 2
Let T : R 3 R 3 be the linear transformation T (x, y, z) = (x + y, y + z, 2z) Find the characteristic polynomial, eigenvalues and eigenvectors of T and find a basis B such that [T ] B,B is diagonal if such B exists. Solution. Let C be the standard basis. Then [T ] C,C =. 2 So the characteristic polynomial of T is det(xi [T ] C,C ) = (x ) 2 (x 2).
Solution (CONT). T has eigenvalues and 2 with eigenvectors {v : T (v) = v} = K (I T ) = x y z : x y z = = Span {v : T (v) = 2v} = K (2I T ) = x y z : x y z = = Span T is not diagonalizable since Span{(,, ), (,, )} R 3.
Let T : R 3 R 3 be the linear transformation T (x, y, z) = (x, y + z, 2z) Find the characteristic polynomial, eigenvalues and eigenvectors of T and find a basis B such that [T ] B,B is diagonal if such B exists. Solution. Let C be the standard basis. Then [T ] C,C =. 2 So the characteristic polynomial of T is det(xi [T ] C,C ) = (x ) 2 (x 2).
Solution (CONT). T has eigenvalues and 2 with eigenvectors {v : T (v) = v} = K (I T ) = x y z : x y z = = Span, {v : T (v) = 2v} = K (2I T ) = x y z : x y z = = Span T is diagonalizable since Span{(,, ), (,, ), (,, )} = R 3.
Solution (CONT). Let B = {(,, ), (,, ), (,, )}. Then [T ] B,B = = P B C [T ] C,CP B C 2 = 2 and 2 2 have the same characteristic polynomials and but they are not similar.
Let V be a vector space of finite dimension and T : V V be a linear transformation satisfying T 2 = I. Show that T is diagonalizable. Proof. Since T 2 = I, I T 2 = (I T )(I + T ) = R(I T ) K (I + T ) rank(i T ) dim K (I + T ). By Rank Theorem, rank(i T ) = dim V dim K (I T ). Therefore, dim K (I + T ) + dim K (I T ) dim V So T is diagonalizable. dim K (I + T ) + dim K (I T ) = dim V
Let T : R 2 R 2 be the linear transformation given by T (x, y) = (3x + y, 4x + 3y). Find T n and T 24 (, ). Solution. We try to diagonalize T. Let C be the standard basis. Then [ ] 3 [T ] C,C = and det(xi [T ] 4 3 C,C ) = x 2 6x + 5. So T has eigenvalues and 5 with eigenvectors K (I T ) = Span{(, 2)} and K (5I T ) = Span{(, 2)}. Let B = {(, 2), (, 2)}. Then [ [T ] B,B = 5 ]
Solution (CONT). So Therefore, [T ] C,C = P B C [T ] B,B P B C [ ] [ ] [ = 2 2 5 2 2 [ ] [ [T n ] C,C = 2 2 = [ ] [ 4 2 2 5 n ] [ 2 2 ] n [ 5 2 2 ] = 4 ] ] [ 2(5 n + ) (5 n ) 4(5 n ) 2(5 n + ) ]
Solution (CONT). So T n (x, y) = 4 (2(5n + )x + (5 n )y, 4(5 n )x + 2(5 n + )y) and T 24 (, ) = 4 (3 524 +, 6 5 24 2). Solution 2. By Cayley-Hamilton, f (T ) = for the characteristic polynomial of T. Therefore, T 2 6T + 5I =. By long division, T n = (T 2 6T + 5I)g(T ) + at + bi T n = at + bi. It suffices to find a and b.
Solution 2 (CONT). Suppose that x n (x 2 6x + 5)g(x) + ax + b = (x )(x 5)g(x) + ax + b. Let x =. Then a + b =. Let x = 5. Then 5a + b = 5 n. Solve the system of linear equations: { a + b = 5a + b = 5 n a = 4 (5n ) b = 4 (5 5n ) T n = at + bi T n (x, y) = at (x, y) + b(x, y) = ((3a + b)x + ay, 4ax + (3a + b)y).