# Diagonalizable matrix A^-1=a

1. Apr 24, 2010

### Dustinsfl

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

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

How do I show they are equal?

2. Apr 24, 2010

### rock.freak667

Re: A^-1=a

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

3. Apr 24, 2010

### Dustinsfl

Re: A^-1=a

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

4. Apr 24, 2010

### gabbagabbahey

Re: A^-1=a

He was asking how you compute/simplify

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

5. Apr 24, 2010

### Dustinsfl

Re: A^-1=a

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

6. Apr 24, 2010

### gabbagabbahey

Re: A^-1=a

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

7. Apr 24, 2010

### Dustinsfl

Re: A^-1=a

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

8. Apr 24, 2010

### gabbagabbahey

Re: A^-1=a

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

9. Apr 24, 2010

### Dustinsfl

Re: A^-1=a

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

10. Apr 24, 2010

### gabbagabbahey

Re: A^-1=a

Right, so

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

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

11. Apr 24, 2010

### Dustinsfl

Re: A^-1=a

Ok I understand thanks.

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