# Prove this matrix is invertible

• Oct 17th 2010, 04:43 AM
jayshizwiz
Prove this matrix is invertible
$A$ is a square matrix such that $A^2$=0. Prove that for any real number $\alpha$ the matrix $I+\alpha A$ is invertible

Attempt: if $A^2$=0 then $A$ is a singular matrix. A singular matrix times $\alpha$ is still a singular matrix. so the idea is to prove that $I+A$ is an invertible matrix when A is singular.

I'm not sure where to go from here but this is something interesting I found:

C = $\left(\begin{array}{cc}1&1\\2&2\end{array}\right)$ D = $\left(\begin{array}{cc}-1&0\\0&0\end{array}\right)$ C and D are both singular matrices.

1) C + $\left(\begin{array}{cc}1&0\\0&1\end{array}\right)$ = $\left(\begin{array}{cc}2&1\\2&3\end{array}\right)$ = invertible matrix

however,

2) D + $\left(\begin{array}{cc}1&0\\0&1\end{array}\right)$ = $\left(\begin{array}{cc}0&0\\0&1\end{array}\right)$ = singluar matrix

So I believe the idea is to figure out why D is not part of the matrix we are looking for, even though it is singular, and it probably has to do with the fact that $A^2=0$

Thanks!
• Oct 17th 2010, 04:48 AM
HappyJoe
Here's an idea:

Try computing

$(I+\alpha A)(I-\alpha A).$
• Oct 17th 2010, 06:00 AM
jayshizwiz
Quote:

Originally Posted by HappyJoe
Here's an idea:

Try computing

$(I+\alpha A)(I-\alpha A).$

Genius (:

Is this a common type of question. How'd you think to do that?
• Oct 17th 2010, 06:10 AM
HappyJoe
I'm not sure if it's common. I haven't seen it before, but it's a fun result.

As for my approach, I was first inspired by the condition $A^2=0$ to try and square $I+\alpha A$ to see what that'd get me. In the result of $(I+\alpha A)(I+\alpha A)$, the term $\alpha^2A^2$ of course vanishes, but the cross-term $2\alpha A$ doesn't, which is pretty useless. Then I just remembered that to get the cross-term out of the way, you can change the sign on $\alpha A$ in one of the factors. That's it.
• Oct 17th 2010, 06:42 AM
jayshizwiz
Thanks Joe! Stay happy (: