Why is it true that for every eigen value a: rank $(T^*- a^*I) =$ rank $(T-aI)$?

Question

I've seen a lemma: rank$(T^*- a^*I) =$ rank$(T-aI)$ where $T^*$ is $T$ adjoint and $a^*$ is the adjoint eigen value.

I know that $T$ and $T^*$ has the same rank, but can someone help me understand how they have the same rank if I subtract $a$ or $a^*$ multiply $I$?

Answer 1

To be more explicit: For the matrix $S=T-aI$ you have $S^*=T^*-a^*I$ and by your own insight, the rank of $S^*$ is the same as the rank of $S$.

Answer 2

If you already know that an operator and its adjoint have the same rank, then observe that

$$\left(T-aI\right)^*=T^*-\left(aI\right)^*=T^*-\overline aI\;\ldots$$