Problem:
Call the form f (left) non-degenerate if $f(\alpha,\beta)=0 \; \forall \beta \implies \alpha =0$. Prove that f is non-degenerate if and only if its associated linear operator $T$ is non-singular (i.e. invertible)
By Theorem 1, suppose V is an inner product space, then $L(V)\rightarrow SLF(V)$ (SLF means sesqui-linear form) , such that $T\mapsto f_T$ with $ f_T(v,w)=(Tv|w)$, establishes an isomorphism of vector spaces.
$"\Leftarrow ":f_T(v,w)=(Tv|w)=(v|T^{-1}w) \implies T^{-1}=T^*, f_T(v,w)=(Tv|w)=(v|T^{-1}w)=0 \; \forall w $. In particular, pick $w\in V$ such that $T^{-1}w=v$, then $f_T(v,w)=(v|v)=0\implies v=0$
$"\Rightarrow": \forall w\in V, (Tv|w)=0=(v|T^*w)\implies v=0 \implies Tv=0$
Claim: $T^*T=TT^*=I$
$f_{T^*T}(v,w)=(T^*Tv|w)=(v|T^*Tw)=(Tv|Tw)$
This is all I got so far. How may I proceed to prove the claim? Or is the claim wrong? In the case of the latter, how may I proceed to solve the problem?
Suppose that $f$ is non-degenerate. We want to show that $T_f = T$ is invertible. Note that therefore it is sufficient to show that $T_f \alpha = 0$ implies $\alpha = 0$. So let $T_f \alpha = 0$ for some $\alpha \in V$. Then we have
$$f(\alpha, \beta) = (T\alpha \vert \beta) = 0.$$
Since $f$ is non-degenerate, we conclude $\alpha = 0$.
On the other hand, suppose $f(\alpha, \beta) = 0$ for all $\beta$, then
$$f(\alpha, \beta) = (T\alpha \vert \beta) = 0$$
for all $\beta$. In particular, choosing $\beta = T\alpha$, we have $(T\alpha \vert T \alpha) = 0$, hence $T\alpha = 0$. But $T$ is invertible, so $\alpha = 0$. Therefore, $f$ is non-degenerate.