# How do you show that for an nxn matrix A the eigenvalue being x, has an inverse and its eigenvalue is 1/x?

Basically we have a matrix A with and inverse A^-1. Matrix A has an eigenvalue x and A^-1 has eigenvalue of 1/x. Show that this is true.

### 2 Answers

- 1 decade agoBest Answer
By definition of the eigen value A X = x X

Multiply both sides by A^-1

A^-1 A X = A^-1 x X

Since A^-1 A = E (unit matrix)

LHS = E X = X

RHS = x A^-1 X

Or X = x A^-1 X

Divide both side by the scalar x

A^-1 X = (1/x) X

The last equation indicates that 1/x is the eigen value of A^-1 corresponding to eigen vector X.

- baldinoLv 43 years ago
The matrix M given by making use of three/2 -a million/2 -a million/2 3/2 is a 2x2 symmetric matrix with eigenvalues a million and a couple of, the situation a million corresponds to (a million,a million) and a couple of corresponds to (-a million,a million). there is in all probability a extra effective elegant technique than the brute rigidity recommendations-set that I employed, yet this is what I did: In M=PDP^-a million, the columns of P are the eigenvectors of M, so I placed D = a million 0 0 2 P = a million x a million y which makes P^-a million = (a million/(y-x)) y -x -a million a million and prolonged each and every of the above to type M=PDP^-a million, which gave M = PDP^-a million = (a million/(y-x)) y-2x x -y -x+2y and because that i needed M to be symmetric, I had to take x = -y. So I chosen x = -a million and y = a million. putting this lower back indoors the equation above with x and y, I have been given the respond for M.