Let $T \in B(H,K)$ when $H,K$ are Hilbert spaces and $T^{\star}$ is adjoint of $T$
Show that
$(ImT^{\star})^{\perp} \subseteq KerT$
( $ImT$ means Image of $T$ and $KerT$ means kernel of $T$)
My attempt
$\forall x \in ImT^{\star}$ $\exists y \in K$ such that $T^{\star}(y)=x$
$z \in (ImT^{\star})^{\perp} $ $\Rightarrow$ $\forall x \in ImT^{\star}$ $<z,x>=0=<z,T^{\star}(y)>=<T(z),y>$
But I cannot obtain $T(z)=0$
Could you please explain it in the easiest way?
Very thanks in advance
The key here lies in realizing that, by the definition of $(\text{Im}(T^\ast))^\bot$,
$z \in (\text{Im}(T^\ast))^\bot \Longrightarrow \forall y \in K, \; \langle z, T^\ast(y) \rangle = 0, \tag 1$
which immediatly leads to
$\forall y \in K, \; \langle T(z), y \rangle = \langle z, T^\ast(y) \rangle = 0 \Longrightarrow T(z) = 0 \Longrightarrow z \in \ker T, \tag 2$
and thus,
$(\text{Im}(T^\ast))^\bot \subset \ker T. \tag 3$
Our OP user519955's attempt breaks down insofar as it doesn't properly emphasize the nearly self-evident fact $\forall y \in K, T^\ast(y) \in \text{Im}(T^\ast)$, the essential idea here being for all $y \in K$.
