Is the following argument correct? In addition can someone please provide some insight concerning the second claim in the given proposition.
Proposition. Let $V$ be finite-dimensional inner product space, and let $T$ be a linear operator on $V$. Prove that if $T$ is
invertible then $T^*$ is invertible and $(T^*)^{-1} = (T^{-1})^*$.
Proof. Assume $T$ is invertible and $T^*w_0 = 0$ where $w_0\in V$ then by definition of adjoint $\langle Tv,w_0\rangle = 0,\forall v\in V$, but from hypothesis there exists a $v_0\in V$ such that $Tv_0 = w_0$ consequently $\langle Tv_0,w_0\rangle = \langle w_0,w_0\rangle = 0$ implying $w_0 = 0$.
Thus $\operatorname{null}T^* = \{0\}$ equivalently $T^*$ is injective and by theorem $\textbf{2.5}$ invertible.
$\blacksquare$
Note: $\textbf{2.5}$ is the result that given an operator $T$ on a finite-dimensional vector space $V$ invertibility,injectivity and surjectivity are all equivalent.
Best Answer
Your proof is correct.
Now, let $u\in V$. Since $T$ is surjective, $u=Tw$ for some vector $w$. If $v\in V$,$$\bigl\langle(T^{-1})^*v,Tw\bigr\rangle=\bigl\langle v,T^{-1}Tw\bigr\rangle=\langle v,w\rangle$$and$$\bigl\langle(T^*)^{-1}v,Tw\bigr\rangle=\bigl\langle T^*(T^*)^{-1}v,w\bigr\rangle=\langle v,w\rangle.$$Since $(\forall u,v\in V):\bigl\langle(T^{-1})^*v,u\bigr\rangle=\bigl\langle(T^*)^{-1}v,u\bigr\rangle$, you have $(T^{-1})^*=(T^*)^{-1}$.