# Eigenvalues & Similar Matrices

1. Oct 21, 2007

### kingwinner

Q: Suppose the only eigenvalues of A are 1 and -1, and A is similar to a diagonal matrix. Prove that A^-1 = A

My Attempt:
Suppose the only eigenvalues of A are 1 and -1, and A is similar to a diagonal matrix.
=>A is invertible (since 0 is not an eigenvalue of A)
and there exists invertible P s.t.
(P^-1) A P = D is diagonal
=> A= P D (P^-1)
=> A^-1 = P (D^-1) (P^-1)

Now if I can prove that D = D^-1, then I am done. But I am stuck right here. The trouble is that the eignevalues of A can have any number of multiplicities, so D can be diag{1,1,-1,-1,-1}, diag{1,-1,-1,-1,-1,-1}, etc., there are infinite number of possible D's, how can I prove that D = D^-1 is always true for these infinite number of different settings?

Thanks a lot!

2. Oct 21, 2007

### Dick

The inverse of adiagonal matrix is formed by inverting the diagonal elements. 1^(-1)=1 and (-1)^(-1)=-1. What's the problem?

3. Oct 21, 2007

### JasonRox

Whenever you're stuck, write out an example and solve it. If you have done so, you would have noticed what Dick is saying.

4. Oct 21, 2007

### kingwinner

How can I prove this statement? I haven't learnt this before...

5. Oct 21, 2007

### Dick

Follow JasonRox's advice and just do it. Multiplying diagonal matrices just involves multiplying the diagonal elements. c_ij=a_ik*b_kj sum over k. If they are diagonal, c_ii=a_ii*b_ii.

6. Oct 22, 2007

### HallsofIvy

Staff Emeritus
You don't really need to calculate A-1 at all. Just showing that A2= A$\cdot$A= I is sufficient.