Let $V$ be a finite dimensional inner product space and $T:V\to V.$
- Prove Ker$T=($Im($T^*))^\bot$ and $($Ker$T^*)^\bot$ = Im$T$.
- Deduce dimIm$T$ = dimIm$T^*$.
I have written a proof but I'm very unsure about some of the steps and would love another set of eyes:
Let's prove this by using two-sided containment.
First notice, if $KerT =\{0\}$ we have $ImT=ImT^* = V$ and therefore since $V = ImT^* \oplus (ImT^*)^\bot$, we have $(ImT^*)^\bot=\{0\}.$
Thus meaning we can assume we have $0 \neq v \in KerT$.
Using inner product we can receive 0 = $<T(v),v>$ = $<v,T^*(v)>$ thus implementing since $v \neq 0$ that $T^*(v) = 0$ and so $v \in KerT^*$.
We have $V = KerT^* \oplus (KerT^*)^\bot$ thus this means $v \in KerT^* \to v \notin (KerT^*)^\bot$.
This means we have $v \in (ImT^*)^\bot$ and therfore we have $KerT \subseteq (ImT^*)^\bot$.
For the other side, consider $0 \neq v \in (ImT^*)^\bot$, (which exists from the same reasons as the previous containment).
Since $V = ImT^* \oplus (ImT^*)^\bot$ we have $v \notin ImT^*$ and since $v \neq 0$ we have $v \in KerT^*$.
This implements we have $(ImT^*)^\bot \subseteq KerT$ and so we have $KerT = (ImT^*)^\bot$.
Then we have $ImT = (ImT^*)^* = (((ImT^*)^\bot)^\bot)^* = ((KerT)^\bot)^* = (KerT^*)^\bot$.
By dimension theoreom we have: $dimV = dimImT + dimKerT = dimImT^* + dimKerT^*$
Since $<T(v),v>$ = $<v,T^*(v)>$ = 0 $\to$ $T(v) = T^*(v) = 0$,
$\forall v \in KerT$ or $v \in KerT^*$ that satsifies $v \neq 0$
then we have $kerT = KerT^*$ thus implamenting $dimKerT = dimKerT^*$.
So we have $dimImT = dimImT^*$, as needed.