Explanation $\ker(T^*)=(\operatorname{Im}(T))^\perp$ I have some misunderstandings about the claim :
$T: V\to V$ is an inner products space.
$\ker(T^*)=(\operatorname{Im}(T))^\perp$
I don't get why $(\operatorname{Im}(T))^\perp$ isn't $\ker (T)$ ? (I know that $\ker(T)=\ker(T)^* \iff TT^*=T^*T$, this is not that case)
I will be grateful for an explanation or example, thanks!
 A: I don't know if a proof is good enough explanation, but this is what works for me.
Lets take $ v \in \ker(T^*)$, we want to show that $ v \in \operatorname{Im}(T)^\perp$
so let's take $ v_2 \in \operatorname{Im}(T)$ and we want to show that $  \langle v,v_2 \rangle = 0$.
$ v_2 \in \operatorname{Im}(T)$ so there is $u \in V$ such that $ Tu = v_2$.
$$ \langle v,v_2 \rangle = \langle v, Tu \rangle = \langle T^*v, u \rangle = (v \in \ker T^*) =  \langle 0, u \rangle = 0$$
On the other hand, if we take $ v \in \operatorname{Im}(T)^\perp$ and we want to show that $ v \in \ker T^*$, so we need to show that $T^*v = 0$
one way of showing it is using the norm.
$$ \lVert T^*v \rVert = \langle T^*v, T^*v \rangle = \langle v , T(T^*v) \rangle = (T(T^*v) \in \operatorname{Im}(T), v \in \operatorname{Im}(T)^\perp) = 0 $$
$$ \Rightarrow T^*v = 0$$
It wasn't an example but I hope it helps.
A: Just compute.
\begin{align}
x \in \ker(T^*) &\Longleftrightarrow T^*(x) = 0
\\
&\Longleftrightarrow \forall y\quad \langle T^*(x),y\rangle = 0
\\
&\Longleftrightarrow \forall y\quad \langle x,T(y)\rangle = 0
\\
&\Longleftrightarrow x \perp \operatorname{im}(T)
\\
&\Longleftrightarrow x \in \big(\operatorname{im}(T)\big)^\perp
\end{align}
