# If A^2 = 0, then A is not an invertible matrix

1. Feb 14, 2017

### Mr Davis 97

1. The problem statement, all variables and given/known data
Suppose that $A^2 = 0$. Show that $A$ is not an invertible matrix

2. Relevant equations

3. The attempt at a solution
We can do a proof by contradiction. Assume that $A^2 = 0$ and that $A$ is invertible. This would imply that $A=0$, which is to say that A is not invertible, since $0$ has no inverse. This is a contraction, so it must be the case that if $A^2 = 0$, then $A$ is not invertible.

Is this the way I should be doing this problem?

2. Feb 14, 2017

### Staff: Mentor

I first thought - and this might well have been intended - that you should show, that there is a non-trivial element in the kernel of $A$, namely the entire image of $A$, but I like your solution better.

3. Feb 15, 2017

### Math_QED

Looks good to me as well. Note that you can prove this directly by using determinants, but I suspect you are not allowed to use determinants at this stage.

4. Feb 15, 2017

### ehild

There are nonzero matrices so as A2=0. You should prove that they are not invertible.
For example, the square of the following matrix is zero.
\begin{pmatrix}

0 & 1 \\
0 & 0

\end{pmatrix}

Last edited: Feb 15, 2017
5. Feb 16, 2017

### willem2

If A^2 = 0 and A is invertible, this implies A^(-1) A^2 = A^(-1) 0 = 0. No need to bother with non-invertible A's here.