Need help proving $n(T)=n(T^*)$ for finite dimensions.

Question

In my book this is showed:

Let H and K be complex Hilbert spaces and let $T\in B(H,K)$. There exists a unique operator $T^* \in B(K,H)$ such that

$(Tx,y)=(x,T^*y)$

for all $x\in H$ and all $y \in K$.

Now this must ofcourse also be the case when $H$ and $K$ are finite dimensional. The book later use in a proof that if both $H$ and $K$ are finite dimensional, then the dimension of the nullspace $\dim(\text{Ker T})=\dim(\text{Ker T}^*)$, that is $n(T)=n(T^*)$. Do you guys have any hints on how to show this?

I was thinking maybe I could create a bijection between the nullspaces, but I am not sure how.

PS: Strictly speaking the book only used it when we had one space H, and both $T,T^*$ was operators on that space, maybe we I have to use that aswell? Then since we have that if $H$ is $k$ dimensional then: $n(T)+r(T)=n(T^*)+r(T^*)=k$, where $r(T)$ denotes the dimension of the image.

Any hints or tips?

I like the following argument: $$ r(T) = r(T^T) \leq r(T^) $$ so that $r(T) = r(T^*)$. The conclusion follows. — Ben Grossmann, Aug 08 '15 at 22:07
@Omnomnomnom Thank you, so you used that we only have one Hilbert space, not two? But how do you get that $r(T)=r(T^*T)$? — user119615, Aug 08 '15 at 22:20
Yeah, but I wasn't able to generalize from regular matrices to regular vector spaces, but I see you made an answer, thanks. — user119615, Aug 08 '15 at 22:31

Ben Grossmann · Accepted Answer · 2015-08-08T22:54:51.737

2

One argument is as follows: we can simply state that $r(T) \leq r(T^*T) \leq r(T^*)$, so that $r(T) = r(T^*)$.

To prove that $r(T) \leq r(T^*T)$, it suffices to prove that $n(T^*T) \leq n(T)$. To prove this, note that $$ T^*Tx = 0 \implies\\ (x,T^*Tx) = 0 \implies\\ (Tx,Tx) = 0 \implies\\ Tx = 0 $$ so that $\ker(T^*T) \subseteq \ker(T)$.

edited Aug 08 '15 at 22:54

answered Aug 08 '15 at 22:27

Ben Grossmann

225,327

Thank you, but you had to implicitely use that we only have one space H, not two, H and K right? – user119615 Aug 08 '15 at 22:34
@user119615 nope, still works with two – Ben Grossmann Aug 08 '15 at 22:46

Need help proving $n(T)=n(T^*)$ for finite dimensions.

1 Answers1