Let $T$ be an invertible linear operator on a finite-dimensional inner product space. I just want a hint as to how I should prove that $T^{*}$ is also invertible and $( T^{-1} )^{*} = ( T^{*} )^{-1}$.
$$ \circ \circ \circ ~ Answer ~ from ~ Below ~ \circ \circ \circ $$
$$
\langle(T^{-1})^*(T^*(v))\mid w\rangle\overset{1}{=}
\langle T^*(v)\mid T^{-1}(w)\rangle\overset{2}{=}
\langle v\mid T(T^{-1}(w))\rangle\overset{3}{=}
\langle v\mid w\rangle
$$
Could somebody explain steps $1$ through $3$, please?
Actually, I think @egreg is using this property:
$\langle Ax,y \rangle = \langle x,A^*y \rangle$
Best Answer
The transpose $T^*$ of an endomorphism $T$ is characterized by the property that $$ \langle v\mid T(w)\rangle=\langle T^*(v)\mid w\rangle $$ (where I denote by $\langle x\mid y\rangle$ the inner product of the two vectors $x,y\in V$). The other thing to note is that if $S_1$ and $S_2$ are endomorphisms of $V$, then $S_1=S_2$ if and only if $\langle S_1(v)\mid w\rangle=\langle S_2(v)\mid w\rangle$, for all $v,w\in V$.
By definition, for all $v,w\in V$, $$ \langle(T^{-1})^*(T^*(v))\mid w\rangle= \langle T^*(v)\mid T^{-1}(w)\rangle= \langle v\mid T(T^{-1}(w))\rangle= \langle v\mid w\rangle $$ and therefore $(T^{-1})^*\circ T^*$ is the identity.
Similarly, $$ \langle T^*((T^{-1})^*(v))\mid w\rangle= \langle (T^{-1})^*(v)\mid T(w)\rangle= \langle v\mid T^{-1}(T(w))\rangle= \langle v\mid w\rangle $$ so also $T^*\circ (T^{-1})^*$ is the identity.
Therefore $$ (T^{-1})^*=(T^*)^{-1}. $$ as we wanted to prove.
Note that no hypothesis of finite dimensionality is needed, but just the existence of the transpose endomorphism, which is true for Hilbert spaces.