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 $y \in K$.

  • 1
    $\begingroup$ Thanks a lot for answer. I have used $\exists y$ instead of $\forall y$ but I couldn’t catch why for all? $\endgroup$ – user519955 Dec 19 '18 at 19:24
  • $\begingroup$ @user519955: well, you clearly need for all $y$ to make it work; how about you read the last line of my answer and get back to me if you have more questions? Happy Holidays! $\endgroup$ – Robert Lewis Dec 19 '18 at 19:27
  • $\begingroup$ @user519955: one needs $\langle T(z), y \rangle = 0$ for every $y \in K$ to ensure $Tz = 0$. So I simply tried to state things in terms of $\forall y \in K$ and followed where it led. Cheers! $\endgroup$ – Robert Lewis Dec 19 '18 at 19:39
  • 1
    $\begingroup$ very very thanks. Happy holidays $\endgroup$ – user519955 Dec 19 '18 at 19:48

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