Proving Equality of Inverse Diagonalizable Matrix with Eigenvalues of 1 and -1

[Dustinsfl]

Let A be a diagonalizable matrix whose eigenvalues are all either 1 or -1. Show that A^{-1}=A.

A=X\begin{bmatrix}<br /> \pm 1 &amp; \cdots &amp; 0 \\<br /> \vdots &amp; \ddots &amp; \vdots \\ <br /> 0 &amp; \cdots &amp; \pm 1<br /> \end{bmatrix}X^{-1}<br /> and A^{-1}=X\begin{bmatrix}<br /> \pm 1 &amp; \cdots &amp; 0 \\<br /> \vdots &amp; \ddots &amp; \vdots \\ <br /> 0 &amp; \cdots &amp; \pm 1<br /> \end{bmatrix}^{-1}X^{-1}<br />

How do I show they are equal?

 

[rock.freak667]

How would you compute the diagonal matrix raised to any power n?

 

[Dustinsfl]

How would you compute the diagonal matrix raised to any power n?

<br /> A^n=X\begin{bmatrix}<br /> \pm 1 &amp; \cdots &amp; 0 \\<br /> \vdots &amp; \ddots &amp; \vdots \\ <br /> 0 &amp; \cdots &amp; \pm 1<br /> \end{bmatrix}^nX^{-1}<br /> <br />

 

[gabbagabbahey]

He was asking how you compute/simplify

\begin{bmatrix}<br /> \lambda_1 &amp; \cdots &amp; 0 \\<br /> \vdots &amp; \ddots &amp; \vdots \\ <br /> 0 &amp; \cdots &amp; \lambda_m<br /> \end{bmatrix}^n

 

[Dustinsfl]

He was asking how you compute/simplify

\begin{bmatrix}<br /> \pm 1 &amp; \cdots &amp; 0 \\<br /> \vdots &amp; \ddots &amp; \vdots \\ <br /> 0 &amp; \cdots &amp; \pm 1<br /> \end{bmatrix}^n

Raise the diagonal terms by n since this just a diagonal matrix.

 

[gabbagabbahey]

Right, so do that with n=-1...what is (\pm 1)^{-1}?

 

[Dustinsfl]

Right, so do that with n=-1...what is (\pm 1)^{-1}?

I don't understand what you mean with your latex code

 

[gabbagabbahey]

I don't understand what you mean with your latex code

It should be displaying properly now, try refreshing your page.

 

[Dustinsfl]

(\pm 1)^{-1}=\pm 1

 

[gabbagabbahey]

Right, so

\begin{bmatrix}<br /> \pm 1 &amp; \cdots &amp; 0 \\<br /> \vdots &amp; \ddots &amp; \vdots \\ <br /> 0 &amp; \cdots &amp; \pm 1<br /> \end{bmatrix}^{-1}=\begin{bmatrix}<br /> \pm 1 &amp; \cdots &amp; 0 \\<br /> \vdots &amp; \ddots &amp; \vdots \\ <br /> 0 &amp; \cdots &amp; \pm 1<br /> \end{bmatrix}

...Plug that into your equation for A^{-1}

 

[Dustinsfl]

Right, so

\begin{bmatrix}<br /> \pm 1 &amp; \cdots &amp; 0 \\<br /> \vdots &amp; \ddots &amp; \vdots \\ <br /> 0 &amp; \cdots &amp; \pm 1<br /> \end{bmatrix}^{-1}=\begin{bmatrix}<br /> \pm 1 &amp; \cdots &amp; 0 \\<br /> \vdots &amp; \ddots &amp; \vdots \\ <br /> 0 &amp; \cdots &amp; \pm 1<br /> \end{bmatrix}

...Plug that into your equation for A^{-1}

Ok I understand thanks.