Show $\alpha$ is selfadjoint.

Question

Question: Let $V$ be an inner product space finitely generated over $\Bbb C$ and let $\alpha$ be an endomorphism of $V$ satisfying $\alpha \alpha^* = \alpha^2$. Show that $\alpha$ is selfadjoint.

**Edited after more thought:

I know that in order to be selfadjoint $\alpha^*=\alpha$. Becuase this inner product space is over $\Bbb C$ does it change the proof?

Or, can I show selfadjoint for $\Bbb R$ and then state the proposition that for an inner product space over $\Bbb C$ and an End$(V)$, If $\langle{\alpha(v), v}\rangle \in \Bbb R$ for all $v \in V$, then $\alpha$ is selfadjoint.

Answer 1

Let $v,\:w\in V$, and so we have, $$ \big\langle (\alpha^*)^2(v), w\big\rangle=\big\langle v,\alpha^2(w) \big\rangle=\big\langle v,\alpha\alpha^*(w)\big\rangle,$$ which implies that $\alpha^2=(\alpha^*)^2=\alpha\alpha^*$. Now we claim that $\alpha\alpha^*=\alpha^*\alpha$.

This is because $$\big\langle(\alpha^*\alpha-\alpha\alpha^*)(v),(\alpha^*\alpha-\alpha\alpha^*)(v)\big\rangle=$$ $$\langle\alpha^*\alpha(v),\alpha^*\alpha(v)\rangle+\langle\alpha\alpha^*(v),\alpha\alpha^*(v)\rangle-\langle\alpha^*\alpha(v),\alpha\alpha^*(v)\rangle-\langle\alpha\alpha^*(v),\alpha^*\alpha(v)\rangle=0$$ Thus, $$\alpha\alpha^*=\alpha^*\alpha.$$

Now, as $$\big{\langle}(\alpha-\alpha^*)(v),(\alpha-\alpha^*)(v)\big\rangle=$$ $$\langle\alpha(v),\alpha(v)\rangle+\langle\alpha^*(v),\alpha^*(v)\rangle-\langle\alpha^*(v),\alpha(v)\rangle-\langle\alpha(v),\alpha^*(v)\rangle=0,$$ we get, $\alpha=\alpha^*.$

Answer 2

Given $\alpha^*\alpha=\alpha^2$

\begin{align} \langle{\alpha^*\alpha(v),v}\rangle & =\langle{\alpha^2(v),v}\rangle \\ &=\langle{\alpha(v),\alpha(v)}\rangle \\ &=\langle{v,\alpha\alpha^*(v)}\rangle \end{align}

Therefore, $\alpha^*\alpha=\alpha^2=\alpha\alpha^*$ and $\alpha=\alpha^*$ over $\Bbb R$.

Now, since V is an inner product space over $\Bbb C$, $\alpha \in End(V)$ has an adjoint, and we hae shown that if $\langle{\alpha(v), v}\rangle \in \Bbb R$ for all $v \in V$, then we can say that $\alpha$ is selfadjoint. (Taken from Golan, The linear algebra a beginning graduate student ought to know, Proposition 17.2).

Check and constructive criticism greatly appreciated.