If is an eigenvalue of an invertible matrix A, then is an eigenvalue of the matrix
.

Question

If  is an eigenvalue of an invertible  matrix A, then  is an eigenvalue of the matrix  
.

Quizplus · Accepted Answer

If λ is an eigenvalue of A, then det(A - λI) = 0, where I is the identity matrix of the same size as A. 
Since A is invertible, det(A) ≠ 0, which means det(A - λI) ≠ 0 for any λ. 
Now suppose λ = 11edaba5_20a5_e5a3_a975_19a637a8284f_TBMC1039_00 is an eigenvalue of A. Then det(A - λI) = 0, which means det(A - 11edaba5_20a5_e5a3_a975_19a637a8284f_TBMC1039_00*I) = 0. 
Since the determinant of a product of matrices is equal to the product of the determinants, we have 
det((A - 11edaba5_20a5_e5a3_a975_19a637a8284f_TBMC1039_00*I)(A^(-1))) = det(A - 11edaba5_20a5_e5a3_a975_19a637a8284f_TBMC1039_00*I) * det(A^(-1)) = 0 * (1/det(A)) = 0. 
This shows that the eigenvalues of the matrix (A - 11edaba5_20a5_e5a3_a975_19a637a8284f_TBMC1039_00*I)(A^(-1)) are the same as the eigenvalues of A, except for the addition of λ = 11edaba5_20a5_e5a3_a975_19a637a8284f_TBMC1039_00. 
Therefore, 11edaba5_20a6_0cb5_a975_a1cc73178ca7_TBMC1039_00 is an eigenvalue of the matrix 11edaba5_20a6_0cb6_a975_65a327bd0804_TBMC1039_00, since it is equal to λ + 1.

If Is an Eigenvalue of an Invertible Matrix