Htykuut

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 $U^{perpperp}=U$ 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 $xperpmathrm{ran} T^dagger$ then for all $y$, we have $0=langle x, T^dagger yrangle=langle Tx, yrangle$, so $Tx$ must be $0$.