Proving $\dim\ker(I-T)=\dim\ker(I-T^*)<\infty$ for finite-rank operator on a Hilbert space

In my functional analysis class, I have encountered the following problem

Let $$H$$ be a Hilbert space and $$T$$ a finite-rank (its range is finite-dimensional) and bounded linear operator on $$H$$. We are asked to show $$\dim\ker(I-T)=\dim\ker(I-T^*)<\infty.$$

I know that the fact that $$T$$ is finite-rank (its range is finite-dimensional) is supposed to help, but I cannot see how to use it. I do know that $$T$$ is compact because it is finite-rank and maybe we need to use spectral theory, but I have no other idea how to show this problem and I am stuck. Any help is appreciated and I thank all helpers.

• Can you show that $\ker(I-T)$ is a subset of the range of $T$? Oct 4 '20 at 1:00
• @Aweygan thank you, no I do not see it. Could you please explain why? Oct 4 '20 at 1:06
• If $x\in\ker(I-T)$, then $x=Tx\in TH$. This gives $\dim\ker(I-T)<\infty$. Oct 4 '20 at 1:12
• @Aweygan thank you and the same for $\dim\ker(I-T^*)$ ? Oct 4 '20 at 1:15
• I'm still thinking about that myself Oct 4 '20 at 3:16

Because $$T$$ is finite-rank, it is compact. Both $$I-T$$ and $$I-T^*$$ are Fredholm, of index zero. So the dimension of each kernel is the codimension of the range, or the dimension of the orthocomplement of the range, and both are finite-dimensional. But if $$x$$ is perpendicular to everything in the range of $$I-T$$, a formal computation shows that \begin{align*} 0&=\langle x,(I-T)y\rangle\\ &=\langle(I-T^*)x,y\rangle \end{align*} so $$x$$ is in the kernel of $$(I-T^*)$$. So the codimension of the range of $$I-T$$ is at most the dimension of the kernel of $$(I-T^*)$$, and, symmetrically, the codimension of the range of $$(I-T)$$ is at most the dimension of the kernel of $$(I-T)$$. But using the fact that the dimensions of the kernels and the codimensions of the ranges are the same, we get equality. \begin{align*} \text{codim} \text{ Ran}(I-T)&\le\dim\text{ Ker}(I-T^*)\\ \text{codim}\text{ Ran}(I-T^*)&\le\dim\text{ Ker}(I-T)\\ &\implies\\ \dim\text{ Ker}(I-T)&\le\dim\text{ Ker}(I-T^*)\\ \dim\text{ Ker}(I-T^*)&\le\dim\text{ Ker}(I-T) \end{align*} so they are equal.
• This is the basic thrust of Fredholm theory. Define an operator $S:X\to Y$, where $X$ and $Y$ are two Banach spaces, to be Fredholm if there is another operator $S':Y\to X$ such that $S'S+F_1=I$, $SS'+F_2=I$, where $F_1$ and $F_2$ are finite rank operators on $X$ and $Y$ respectively. Then $S$ and $S'$ both have finite dimension kernel and cokernel. Define the Fredholm index of $S$ to be $\dim\text{ Ker}S-\text{codim}\text{ Ran}S$, or the negative if you feel so inclined. Moreover, if $K$ is a compact operator, then $S+K$ is also Fredholm, and the index doesn't change. $I$ has index $0$. Oct 5 '20 at 2:41
• terrytao.wordpress.com/2011/04/10/… is a good introduction to this topic. The "third proof" is the one that gives what you want, particularly that the index doesn't change as you change the parameter. Also, the case you want is $\lambda=1$, which is easier. Oct 5 '20 at 4:43