Let $V$ be an inner product space, and let $T$ be a linear operator on $V$. Prove that $R(T^{*})^{\bot}=N(T)$.
I would like to understand the following proof:
We prove that $ N(T^*) \subset N(T)^{\bot}$. Let $x \in N(T^{*})$, need to prove for all $y \in R(T), \langle x,y \rangle=0$. Let $y \in R(T)$, $y=T(z)$ for some $z \in V$. Then $$\langle x,y \rangle= \langle x,T(z) \rangle= \underbrace{\langle T^{*}(x)}_{0},z \rangle,$$ so $x \in R(T)^{\bot}$, hence $N(T^{*} \subset R(T)^{\bot}$.
Then we prove the other direction. Let $x \in R(T)^{\bot}$, want to show $T^{*}(x)=0$. We have $$|| T^{*}(x)||^2=\langle T^{*}(x), T^{*}(x) \rangle=\langle \underbrace{x}_{\in R(T)^{\bot}},\underbrace{T( T^{*}(x))}_{\in R(T)} \rangle,$$ so $T^{*}(x)=0$, hence $R(T)^{\bot}=N(T^{*})$.
I am lost in this whole proof. Can someone explain what's going on, why are they attempting to show those things? Much appreciated.
I too found the "proof" given in the text of the question obscure and confusing, so much so that I can't really say more; hence my own work is presented below; it is not difficult.
To avoid technical difficulties I assume $T$ is a bounded operator on $V$.
These things being said, let
$x \in N(T); \tag 1$
then
$Tx = 0, \tag 2$
whence
$\forall y \in V, \: \langle y, Tx \rangle = 0; \tag 3$
thus
$\forall y \in V, \; \langle T^\dagger y, x \rangle = 0, \tag 4$
that is,
$x \in R(T^\dagger)^\bot, \tag 5$
and thus
$N(T) \subset R(T^\dagger)^\bot; \tag 6$
also, in the event that (5) binds, we have
$\forall y \in V, \; \langle T^\dagger y, x \rangle = 0, \tag 7$
or
$\forall y \in V, \; \langle y, Tx \rangle = 0, \tag 8$
which immediately yields
$Tx = 0 \Longrightarrow x \in N(T), \tag 9$
and hence
$R(T^\dagger)^\bot \subset N(T); \tag{10}$
finally, (6) and (10) together imply
$R(T^\dagger)^\bot = N(T), \tag{11}$
$OE\Delta$.