Matrix inversion problem

**ahmedzoro10** · January 14th, 2012, 12:43 PM

I've A^-1 = - A What is A
I've figured that A^2= -I but what's then?

**alexmahone** · January 14th, 2012, 12:55 PM

Originally Posted by ahmedzoro10

I've A^-1 = - A What is A
I've figured that A^2= -I but what's then?

$\mathbf{A}=\pm i\mathbf{I}$ , where $i=\sqrt{-1}$ .

**Opalg** · January 14th, 2012, 01:05 PM

Originally Posted by alexmahone

$\mathbf{A}=\pm i\mathbf{I}$ , where $i=\sqrt{-1}$ .

Not necessarily. There can be many matrices whose square is –I. For 2x2 matrices, you can for example take $A = \begin{bmatrix}0&1\\ -1&0 \end{bmatrix}.$

**ahmedzoro10** · January 14th, 2012, 01:12 PM

How did you get this one?

**alexmahone** · January 14th, 2012, 01:29 PM

Originally Posted by ahmedzoro10

How did you get this one?

Just note that $\sqrt{\mathbf{I}}=\mathbf{I}$ (there are others as well) and $\sqrt{-1}=\pm i$ .

**alexmahone** · January 14th, 2012, 01:32 PM

Originally Posted by Opalg

Not necessarily. There can be many matrices whose square is –I. For 2x2 matrices, you can for example take $A = \begin{bmatrix}0&1\\ -1&0 \end{bmatrix}.$

So how many solutions does a matrix equation like $\mathbf{A}^2=-\mathbf{I}$ have? (where $\mathbf{A}$ and $\mathbf{I}$ are n by n matrices.)

**HallsofIvy** · January 15th, 2012, 11:27 AM

Let $A= \begin{bmatrix}a & b \\ c & d\end{bmatrix}$ . Then $A^2= \begin{bmatrix}a^2+ bd & ab+ bd \\ ac+ cd & bc+ d^2\end{bmatrix}= \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$ . So we must have $a^2+ bd= 1$ , $ab+ bd= 0$ , $ac+ cd= 0$ , and $bc+ d^2= 1$ . You might try things like taking b (or c) equal to 0 or not equal to 0 and see what you get.

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