$V$ is an inner product space over $\mathbb {C}$. $T:V\to V, \alpha \in \mathbb {C}$ such that $|\alpha|\neq 1$, $T$ is a normal, $S=T-\alpha T^*$. 
$V$ is an inner product space over $\mathbb {C}$.


$T:V\to V, \alpha \in \mathbb {C}$ such that $|\alpha|\neq 1$.


$T$ is a normal linear transformation, denote $S=T-\alpha T^*$.


Prove $\ker T = \ker S$.

My solution :
Suppose $v\in \ker S$.
$0=\langle v,Sv\rangle = \langle v,(T-\alpha T^*)v\rangle=\langle v,Tv \rangle - \langle v, \alpha T^* v\rangle= \langle v,Tv \rangle - \overline{\alpha} \langle v,T^*v\rangle = \langle v,Tv \rangle- \overline{\alpha} \langle Tv,v\rangle = 0$
Since $|\alpha|\neq 1 \implies \langle Tv,v\rangle =0 \implies Tv=0 \implies v\in \ker T$.
Is it correct? Why is given that $T$ is a normal linear transformation ?
 A: 
$V$ an inner product space over $\Bbb{C}$.
$T:V\to V, \alpha \in \mathbb {C},$ $|\alpha|\neq 1$
$T$ is normal linear transformation, denote $ S=T−\alpha T^{∗}$

Claim : $ \ker T = \ker S$
Proof:

$\ker T \subset \ker S$

$v\in \ker(T) $
$\begin{align}\|Sv\|&=\|(T−\alpha T^{∗})(v)\|\\ &\le \|Tv\|+|\alpha|\|T^{*}v\|\space \space [ \color{blue}{1}]\\&=\|Tv\|+|\alpha|\|Tv\|\space \space [ \color{blue}{2}]\\&=0\space \space [ \color{blue}{3}]
\end{align}$
$\|Sv\|=0 \implies v\in \ker(S) $
$[\color{blue}{1}]:$ Triangle inequality of Norm
$[\color{blue}{2}]:\|Tv\|=\|T^{*}v\| \space \space , T$ is normal.
$[\color{blue}{3}]:\|Tv\|=0,v\in \ker(T) $

$\ker(S) \subset \ker(T) $

$v\in \ker(S) $
$\begin{align}\|Tv\|&=\|Tv-\alpha T^{*}v +\alpha T^{*}v\|\\&=\|(T-\alpha T^{*})v+\alpha T^{*}v\|\\&=\|Sv+\alpha T^{*}v \|\\&=\| \alpha T^{*}v \|\\&=|\alpha|\|Tv\|\end{align}$
$|\alpha|\neq 1\implies \|Tv\|=0$
$\implies v\in \ker(T) $
Hence,  $ \ker T = \ker S$.
A: Not sure how from $\langle Tv,v\rangle$ you concluded that $Tv=0$. You used just a single vector. What is true is that if $V$ is a complex inner product space and $\langle Tv,v\rangle$ for all $v\in V$ then $T=0$. And you can indeed use this fact in your exercise.
Since $T$ is normal it follows that $TS=ST$. Thus, $Ker(S)$ is a $T$-invariant subspace. So we can look at the restriction $T|_{Ker(S)}:Ker(s)\to Ker(S)$. Using what you did, we know that $\langle Tv,v\rangle=0$ for all $T\in Ker(S)$. This shows the restriction $T|_{Ker(S)}$ is the zero operator, i.e $Ker(S)\subseteq Ker(T)$.
For the other direction, note that since $T$ is normal, we have $Ker(T)=Ker(T^*)$. So clearly $Ker(T)\subseteq Ker(S)$.
