# OC of Adjoint Operator’s Image is subset of Kernel

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 $$T^{\star}(y)=x$$

$$z \in (ImT^{\star})^{\perp}$$ $$\Rightarrow$$ $$\forall x \in ImT^{\star}$$ $$=0==$$

But I cannot obtain $$T(z)=0$$

Could you please explain it in the easiest way?

The key here lies in realizing that, by the definition of $$(\text{Im}(T^\ast))^\bot$$,

$$z \in (\text{Im}(T^\ast))^\bot \Longrightarrow \forall y \in K, \; \langle z, T^\ast(y) \rangle = 0, \tag 1$$

$$\forall y \in K, \; \langle T(z), y \rangle = \langle z, T^\ast(y) \rangle = 0 \Longrightarrow T(z) = 0 \Longrightarrow z \in \ker T, \tag 2$$
$$(\text{Im}(T^\ast))^\bot \subset \ker T. \tag 3$$
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 $$y \in K$$.
• Thanks a lot for answer. I have used $\exists y$ instead of $\forall y$ but I couldn’t catch why for all? – user519955 Dec 19 '18 at 19:24
• @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! – Robert Lewis Dec 19 '18 at 19:27
• @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! – Robert Lewis Dec 19 '18 at 19:39