Question

Question 3: Eigenvalue Theory 1 (a) Let A e Cnxn, and let (Ai, an), (Ak,Xk) be eigenpairs where all λί are distinct. Show that the corresponding eigenvectors r1,. .. Tk are linearly independent. (b) Let A, B e C"xn be similar. Show that A and B have the same char- acteristic polynomial, same eigenvalues including algebraic and geometric (c) Do A and B fro (b) share the same singular values? Justify.

Answer 1

) A elementof C^n times n , let (lambda_1, x_1) , lambda_2, x_2) , ....,(lambda_k , x_k) be eigenpais lambda_i are distinct Now, Ax_1 = lambda_1 x_1 Ax_2 = lambda_2 x_2 A x_k = lambda_k x_k } rightarrow (1) let us consider the relation c_1 x_1 + c_2 x_2 + ... + c_k x_k = 0 c_i , i = 1 to k are scalars Now, c_1 A x_1 + c_2 A x_2 + ... + c_k A x_k = 0 = > c_1 lambda_1 x_1 + c_2 lambda_2 x_2 + .... + c_k lambda_k x_k = 0 [from (1)] again c_1 lambda_1^2 x_1 + c_2 lambda^2_2 x_2 + ... + c_k lambda_k^2 x_k = 0 c_1 A^k - 1 x_1 + c_2 A^2 x_2 + .... + c_k A^k - 1 x_k = 0 = > c_1 lambda_1^k - 1 x_1 + c_2 lambda_2^k - 1 x_2 + ...+ c_k lambda_k^k - 1 x_k = 0 (c_1 x_1, c_2 x_2 .... c_k x_k) (1 1 ... 1 lambda_1 lambda_2 ... lambda_k ..... lambda_1^k - 1 lambda_2^k - 1 ... lambda_k^k - 1) = (0 0 .... 0) rightarrow (2) let B = (1 1 ... 1 lambda_1 lambda_2 ... lambda_k ..... lambda_1^k - 1 lambda_2^k - 1 ... lambda_k^k - 1), since all lambda_i are distinct B is non singular.