$\displaystyle A$ is a square matrix such that $\displaystyle A^2$=0. Prove that for any real number $\displaystyle \alpha$ the matrix $\displaystyle I+\alpha A$ is invertible

Attempt: if $\displaystyle A^2$=0 then $\displaystyle A$ is a singular matrix. A singular matrix times $\displaystyle \alpha$ is still a singular matrix. so the idea is to prove that $\displaystyle I+A$ 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 = $\displaystyle \left(\begin{array}{cc}1&1\\2&2\end{array}\right)$ D = $\displaystyle \left(\begin{array}{cc}-1&0\\0&0\end{array}\right)$ C and D are both singular matrices.

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

however,

2) D + $\displaystyle \left(\begin{array}{cc}1&0\\0&1\end{array}\right)$ = $\displaystyle \left(\begin{array}{cc}0&0\\0&1\end{array}\right)$ = 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 $\displaystyle A^2=0$

Thanks!