Eigenvalues & Similar Matrices

[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?


Can someone please help me?
Thanks a lot!

 

[Dick]

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

 

[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.

 

[kingwinner]

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

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

 

[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.

 

[HallsofIvy]

You don't really need to calculate A^-1 at all. Just showing that A²= A\cdotA= I is sufficient.