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 $$A^{-1}=A$$.

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

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?

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

 

[gabbagabbahey]

He was asking how you compute/simplify

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

 

[Dustinsfl]

He was asking how you compute/simplify

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

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

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

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

 

[Dustinsfl]

Right, so

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

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

Ok I understand thanks.