I was given two problems in homework
Let $V$ be a finite dimensional inner product space and $T$ be a linear operator on it then,
Prove the following
(1) $ \operatorname{range}$$ ({T^{\dagger}})^{\perp} = \ker(T)$$ $
(2) $ \operatorname{range}$$ ({T^{\dagger}}) = \ker(T)^{\perp}$$ $
The first one was proved using following method, from the definition of adjoint
Let $x$ ${\in}\ $ ${\ker}(V)$ and $y \ {\in} \ V$, then we have $$ 0 \ = \ <Tx,y>\ = \ <x,T^{\dagger}y>$$
So,as the dot product is zero, we can say that, $$ 0 = <x,T^{\dagger}y>$$
From above identity, we say that, $ \operatorname{range}$$({T^{\dagger}})^{\perp} = \ker(T)$$ $
I am stuck at problem 2, by similar method, I am not getting to the answer. What can be other method?
Hint: You can take the orthogonal complement on both sides. All you have to prove is that $U^{\perp\perp}=U$ for any subspace $U$ in a finite dimensional space.
Note that only the inclusion $(\mathrm{ran} T^\dagger)^\perp \supseteq \ker T$ is rigorously proved here, but the other inclusion also holds by the same equality, since if $x\perp\mathrm{ran} T^\dagger$ then for all $y$, we have $0=\langle x, T^\dagger y\rangle=\langle Tx, y\rangle$, so $Tx$ must be $0$.