normal operator and prove $N(T)=N(T^{*})$

Question

Let T be a normal operator on a finite-dimensional inner product space V. Prove that $N(T)=N(T^{*})$ and $R(T)=R(T^{*})$.

Let V is an inner product space and T is a linear operator on V, $||T(x)||=||T^{*}(x)||$ for all $x \in V$. For $x \in V$, $||x||=\sqrt{ \langle x,x, \rangle}$

Thus $T(x)=0$ if and only if $T^{*}(x)=0$ (I don't follow how they get this)

Hence $N(T)=N(T^{*})$.

For the second one, I am not sure.

Answer 1

If $Tx=0$, then $||Tx||=0$, so $||T^*x||=0$ which implies $T^*x=0$. So, $$ Tx=0\iff T^*x=0 $$ which means exactly $\ker(T)=\ker(T^*)$.

For the second, use the first, $$ R(T)=N(T^*)^\perp\stackrel{\text{first part}}{=}N(T)^\perp=R(T^*) $$