Prove an equality between dimensions of kernels

Question

Let $V$ be a inner product space over field $\mathbb{R}$ with $\dim(V)<\infty$, and $T\in \text{Hom}(V,V)$. I'm trying to prove:$$\dim(\ker T)=\dim(\ker T^*)=\dim(\ker TT^*)$$ Also, as a conclusion from it I need to show that: $$\text{Im}(T)=\text{Im}(TT^*)$$
Note: $T^*$ is the adjoint operator (aka: Hermitian conjugate)

I would be very happy to receive some guidance for it, detailed as much as possible, since this subject is new to me and I am really struggling with it. Thanks!

Answer 1

Hints:

$$w\in \ker T^*\implies T^*w=0\implies TT^*w=T(0)=0\implies$$

$$\implies w\in\ker TT^*\implies \dim\ker T^*\le\dim\ker TT^*$$

OTOH, we also have (since $\;T^{**}=T\;$):

$$w\in\ker TT^*\implies TT^*w=0\implies\;\forall\,v\in V\;,\;0=\langle v\,,\,TT^*w\rangle=\langle T^*v\,,\,T^*w\rangle\implies$$

$$\implies \;\;\text{in particualr, we get for}\;\;v=w:\;\;\langle T^*w\,,\,T^*w\rangle=0\implies$$

$$\implies T^*w=0\implies w\in\ker T^*\implies \dim\ker TT^*\le\dim\ker T^*$$

Now just put $\,S=T^*\;$ and apply the above for $\;S\;$:

$$\dim\ker T=\dim \ker T^{**}=\dim\ker S^*=\dim \ker SS^*=\dim\ker T^*T^{**}=\dim\ker T^*T\;\ldots$$

Finally, use the dimension theorem:

$$\begin{align*}(1)&\;\;\dim V=\dim\ker T+\dim\text{Im}T=\dim\ker TT^*+\dim\text{Im}\,T\\ (2)&\;\;\dim V=\dim\ker TT^*+\dim\text{Im}\,TT^*\;\;\ldots\end{align*}$$

Note: you still need to complete some stuff above!

Answer 2

I will assume that $V$ is a vector space with an inner product. Then $\langle Tv,w\rangle = \langle v,T^* w \rangle$ for all $v,w\in V$. Now, $v\in\ker T$ if and only if $\langle Tv,w\rangle=0$ for all $w\in V$ if and only if $\langle Tv,Tv\rangle=0$. Try to use this to show that $\ker T = \ker T^* T$ (not only that the dimensions match).

For the conclusion try to prove that $\ker T = (\textrm{im } T^*)^\perp$.

Answer 3

If you prove that $\dim\ker T=\dim\ker TT^*$, you have that $$ \dim(\operatorname{im}T)=\dim V-\dim(\ker T)= \dim V-\dim(\ker TT^*)=\dim(\operatorname{im}TT^*). $$ But $\operatorname{im}T\supseteq\operatorname{im}TT^*$, so they're equal.

Let's tackle the other problem. If $v\in\ker TT^*$, then $$ 0=\langle v,0\rangle=\langle v,TT^*v\rangle= \langle T^*v,T^*v\rangle $$ so that $T^*v=0$. Hence $\ker TT^*\subseteq\ker T^*$. On the other hand, it's obvious that $\ker T^*\subseteq\ker TT^*$. Therefore $\ker TT^*=\ker T^*$.

Now recall that the ranks of $T$ and $T^*$ are equal to conclude that $\dim\ker T=\dim\ker ^*$ and you're done.