Let λk, k = 1, 2, ..., n, be the eigenvalues of an n × n invertible matrix A.(a) (2 pts) Show that λk 6 = 0, k = 1, 2, ..., n.(b) (8 pts) Show that the eigenvalues of A−1 are 1λk, k = 1, 2, ..., n

(a) The eigenvalues of a matrix are the roots of its characteristic polynomial, which is defined by det(A - λI) = 0, where A is the matrix, λ is a scalar, I is the identity matrix, and det denotes the determinant. If A is invertible, then its determinant is not zero. If λk were zero, then det(A - 0*I) = det(A) would be zero, which contradicts the assumption that A is invertible. Therefore, λk ≠ 0 for k = 1, 2, ..., n.

(b) The eigenvalues of the inverse of a matrix are the reciprocals of the eigenvalues of the original matrix. This can be shown as follows:

Let v be an eigenvector of A corresponding to the eigenvalue λk, so Av = λkv. Then A^-1Av = A^-1λkv, which simplifies to v = λkA^-1v. Dividing both sides by λk gives 1/λk * v = A^-1v. Therefore, 1/λk is an eigenvalue of A^-1. Since this is true for all k = 1, 2, ..., n, the eigenvalues of A^-1 are 1/λk, k = 1, 2, ..., n.