Let $V$ an inner product space and $T:V \to V$ a linear transformation. Prove $\operatorname{Ker} (T^*T)=\operatorname{Ker} T$

Question

So I started to write proof using two-sided containment.

So for $v$ $\in$ $\operatorname{Ker} T$ we have $(T^*T)(v) = T^*(T(v)) = T^*(0) = 0$

Therefore we have $\operatorname{Ker} T \subseteq \operatorname{Ker} T^*T$

Now if $\operatorname{Ker} T^*T = \{0\}$ then we have proof. Then let's assume we have $\operatorname{Ker} T^*T \neq \{0\}$ then we have $v \neq 0$ for some $v \in \operatorname{Ker}T^*T$, hence $(T^*T)(v) = 0$.

Now we can deduce using inner product that $0=\langle 0,v \rangle = \langle T^*T(v),v \rangle = \langle v,TT^*(v) \rangle$

Since $v \neq 0$ it means that $TT^*(v)=0$ and therefore we have $v \in \operatorname{Ker} TT^*$ and therefore we have $\operatorname{Ker} T^*T \subseteq \operatorname{Ker} TT^*$.

This part is where I'm currently stuck since I'm not sure how to transition to $\operatorname{Ker} T^*T \subseteq \operatorname{Ker} T$.

Answer 1

You have two errors by my accounting:

• $\langle T^*T(v),v \rangle$ isn't $\langle v,TT^*(v) \rangle$
• and if we had $\langle v, TT^*(v)\rangle = 0$ for some $v\neq 0$, we cannot conclude that $TT^*(v) =0$.

To fix this, we can let $v \in \operatorname{Ker}(T^*T)$ be arbitrary (we don't need to know whether $v$ is nonzero or not) and then consider $$0 = \langle 0, v\rangle = \langle T^*T(v), v\rangle = \langle T(v), T(v)\rangle$$

so that $T(v) = 0$ and therefore $v\in \operatorname{Ker}(T)$.

Answer 2

As you have said, clearly $\ker(T)\subseteq\ker(T^*T)$. No, remember that the definition of $T^*$ implies that $$ \langle T^* u,v\rangle = \langle u,Tv\rangle $$ But then, let $v \in \ker(T^*T)$. Then $T^*Tv=0$. But then $$ \langle T^* Tv,v\rangle = \langle Tv,Tv\rangle = 0 \iff Tv=0 $$ Thus $\ker(T^*T) \subseteq \ker(T)$