I'm studying for my qualifying exams, and I came across the following question. I want to make sure I'm on the right track with my proof:
Suppose $(V, \left\langle , \right \rangle)$ is an inner product space over $F$ and $T\in \mathcal{L}(V)$ is a linear operator. Prove that $\ker(T^*T) = \ker(T)$.
First, the inclusion $\supseteq$ is easy to show. Since $T^*$ is linear, if $v \in \ker(T)$, then $(T^*T)(v) = T^*(T(v)) = T^*(0) = 0$. Now, to see $\subseteq$, let $w$ be an arbitrary vector in $V$. Then $0 = \left\langle 0,w \right\rangle = \left\langle (T^* T)(v),w \right\rangle = \left\langle T(v), T(w) \right\rangle = 0$, and since $w$ was arbitrary, that implies $T(v) = 0$ by the properties of an inner product. So then if $v \in \ker(T^*T)$, $v \in \ker(T)$.
That's what I was able to come up with, but I don't really know if it's correct because that really only shows it for $w' \in R(T)$ not $w \in V$. Any advice?
Thanks in advance!
You were close. take $$v\in ker T^* T.$$ Then, $$0=\langle 0,v \rangle=\langle T^* T(v),v \rangle=\langle T(v),T(v) \rangle.$$ And hence, $\langle T(v),T(v) \rangle=0$. By the properties of inner products that happens if and only if $T(v)=0$, in other words, $$v\in ker(T).$$