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Let $(a_{m,n})$ be a double sequence of complex numbers such that $\sum_{m\geq1}\sum_{n\geq1}|a_{m,n}|^2<\infty$, and define the bounded linear map $T:\ell^2\to\ell^2$ by $$(Tx)(m)=x(m+1)+\sum_{n\geq1}a_{m,n}x(n).$$ I'm asked to prove that $\dim\ker T=\dim\ker T^*+1$ for any choice of $(a_{m,n})$. I calculated that $T^*$ is given by $$(T^*x)(n)=\begin{cases}\sum_{m\geq1}\overline{a_{m,1}}x(m)&\text{if }n=1;\\x(n-1)+\sum_{m\geq1}\overline{a_{m,n}}x(m)&\text{if }n\geq1.\end{cases}$$ So $x\in \ker T$ if and only if for all $m\in\mathbb N$, $x(m+1)=-\sum_{n\geq1}a_{m,n}x(n)$, while $x\in\ker T^*$ if and only if $\sum_{m\geq1}\overline{a_{m,1}}y(m)=0$ and $y(n)=-\sum_{m\geq1}\overline{a_{m,n+1}}y(m)$. I do not see how $\dim\ker T=\dim\ker T^*+1$ follows from this.

Another approach is to use $\ker T^*=(\mathrm{ran}\ T)^\perp$, but this approach does not seem helpful either.

Note that I'm asked to prove that the Fredholm index of $T$ equals $1$. Note that is $S$ is the right unilateral shift, then $ST-1$ is compact, so $T$ is left semi-Fredholm. However, I don't see how this can be helpful either.

Any help, on any of these three approaches, is much appreciated.

Václav Mordvinov
  • 3,268

1 Answers1

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My third approach worked. Note that $T$ is the sum of the backward shift and the compact operator $A$ defined by $(Ax)(m)=\sum_{n\geq1}a_{m,n}x(n)$. By Fredholm index theory, the index of $T$ equals the index of the backward shift, which is easily seen to be $1$.

Václav Mordvinov
  • 3,268