Show that if A is self-adjoint and $A^{2}x=0$, show that $Ax=0$. I feel like i'm overcomplicating this a bit.
Let $X$ be a finite-dimensional inner product space and $A$ be a linear transformation from $X$ to $X$. If A is self-adjoint and if $A^{2}x=0$, show that $Ax=0$.
Here's my thought:
$0=\left<A^{2}x,y\right>=\left<A(A^{*}x),y\right>=\left<A^{*}x,A^{*}y\right>=\left<Ax,A^{*}y\right>$
The first equality just came from the fact that $A^{2}x=0$. So the last equality has to be zero as well. I just don't know how to show that $Ax$ specifically must be zero, because as its written, I think $A^{*}y$ could also be zero while $Ax$ is non-zero. Thanks in advance.

It's useful to know that if $X$ and $Y$ are finite dimensional inner product spaces over $F$ (where $F$ is $\mathbb R$ or $\mathbb C$) and $A:X \to Y$ is a linear transformation, then $A$ and $A^* A$ have the same null space.  Here's a proof: Clearly $Ax = 0 \implies A^* Ax = 0$.  Conversely,
\begin{align*}
& A^* A x = 0 \\
\implies & \langle x, A^* A x \rangle = 0 \\
\implies & \langle Ax, Ax \rangle = 0 \\
\implies & \|Ax \|^2 = 0 \\
\implies & Ax = 0.
\end{align*}
