Proving Kernel Equality of a Linear Transformation and Adjoint

Question

Let $T$ be a linear transformation in an inner product space $V$.

Determine if the following it true or false:

$$Ker (T)= Ker (T^*T)$$

Where $*$ donates the adjoint operator.

Would it help proving $KerT=KerT^*$? I think I can do that.

Thanks!

Answer 1

Clearly $\ker(T) \subseteq \ker(T^*T)$. For the other inclusion,

$$\begin{align*} x \in \ker(T^* T) &\Leftrightarrow (T^*T x,v)=0\mbox{ for all }v\\ &\Leftrightarrow (Tx,Tv)=0 \mbox{ for all }v\\ &\Rightarrow (Tx,Tx)=0\\ &\Leftrightarrow x\in \ker(T)\end{align*}$$

Therefore $\ker(T^*T)\subset \ker(T)$.

An alternate proof is to show (similarly) that $\ker(T^*)=R(T)^{\perp}$. Therefore $T^*T x=0$ if and only $Tx \in \ker (T^*)$ if and only if $x \in \ker(T)$.

Answer 2

Cetrtainly $\ker T \subset \ker T^*T$. The other way, consider $a \in \ker (T^* T)$. Note that $T^*Ta=0 \implies \langle T^*T(a),a\rangle =0 \implies \langle T(a),T(a)\rangle =0 \implies ||Ta||=0 \implies T(a)=0$, so $Ker(T)=Ker(T^*T)$.