Let $\lambda\neq 0 \in\mathbb{C}$.My professor mentioned in a lecture that $\dim \ker (T-\lambda) = \text{codim(ran} (T-\lambda)$ holds for $T$ compact operator on a Banach space. We only discussed the proof for when $T$ is self adjoint on a hilbert space. I believe this is called "Riesz-Schauder" theorem but I haven't had any luck finding it online. I already know that both sides of the equation are finite. Does anyone know a proof that could go here or a reference?
Proof $\dim \ker (T-\lambda) = \text{codim(ran} (T-\lambda)$ for $T$ compact
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The key is to show that $S = (\lambda I - T)\lvert_{V_\lambda^\perp}$ has the property
$$\bigl(\forall x \in V_\lambda^\perp\bigr)\bigl(\lVert Sx\rVert \geqslant \delta \lVert x\rVert\bigr)\tag{1}$$
for some $\delta > 0$. Everything follows quite easily from that.
To show the existence of such a $\delta > 0$, suppose the contrary. Then there is a sequence $(x_n)$ in $V_\lambda^\perp$ with $\lVert x_n\rVert = 1$ for all $n$ and $Sx_n \to 0$. Use the compactness of $T$ and then $\lambda \neq 0$ to obtain a contradiction.
I assume the OP has meanwhile filled in the details, so let me add them here too:
If we had a sequence $x_n \in V_\lambda^\perp$ with $\lVert x_n\rVert = 1$, and $z_n = Sx_n \to 0$, by the compactness of $T$ there is a subsequence such that $Tx_{n_k}$ converges. Suppose without loss of generality that the subsequence is the entire sequence, so $y_n = Tx_n \to y$. Then
$$\lambda x_n = Sx_n + Tx_n = z_n + y_n \to 0 + y = y.$$
Since $\lambda\neq 0$, we have $\lVert y\rVert = \lvert\lambda\rvert > 0$, it follows that $x_n \to \lambda^{-1}y \in V_\lambda^\perp\setminus\{0\}$ and
$$S(\lambda^{-1}y) = \lim_{n\to\infty} Sx_n = \lim_{n\to\infty} z_n = 0,$$
contradicting $V_\lambda^\perp \cap V_\lambda = \{0\}$.
So the existence of a $\delta > 0$ with $\lVert Sx\rVert \geqslant \delta \lVert x\rVert$ for all $x\in V_\lambda^\perp$ is established.
Since $\mathcal{H} = V_\lambda \oplus V_\lambda^\perp$, we have the (linear) bijection $S\colon V_\lambda^\perp \to \mathcal{R}(\lambda I - T)$. Now if $(y_n)$ is a Cauchy sequence in $\mathcal{R}(\lambda I-T)$, then $(S^{-1}y_n)$ is a Cauchy sequence in $V_\lambda^\perp$ by $(1)$, and we have
$$\lim_{n\to\infty} y_n = \lim_{n\to\infty} S(S^{-1}y_n) = S(\lim_{n\to\infty} S^{-1}y_n) \in \mathcal{R}(S) = \mathcal{R}(\lambda I - T).$$
If we had not assumed $\lambda\neq 0$, we couldn't have deduced the convergence of $x_n$ from the convergence of $Sx_n\to 0$ and $Tx_n \to y$, and in fact the range of a compact operator is in general not closed.
It should be noted that the result does not depend on $\mathcal{H}$ being a Hilbert space, that only makes the proof a little bit more easy. When working with an arbitrary Banach space, the proof is exactly the same if one can establish a closed complement of $V_\lambda$. Although generally closed subspaces need not be complemented in Banach spaces, the compactness of $T$ implies the finite-dimensionality of $V_\lambda$, and finite-dimensional subspaces (as well as subspaces of finite codimension) are always complemented in Banach spaces. A slight variation of the proof considers the quotient $\mathcal{H}/V_\lambda$ instead of a complementing subspace.
Consider the linear operators
$$i: \ker A \to X: x \mapsto x$$ and
$$\pi: X \to X/V: x \mapsto [x].$$
$i$ is the inclusion map and $\pi$ is the projection of $X$ on $V$ and therefore these are continous maps. Furthermore $y \in \ker \pi$ iff $y \in V$ and equivalent $y \notin \ker A$ as by assumption $X = \ker A \oplus V$.
The composition $\Phi := \pi \circ i$ is the isomorphism you are looking for: Let $x \in \ker \Phi$. Then $\pi (ix) = [0]$ and therefore $ix \in V$. Since $X = \ker A \oplus V$ this means $x = 0$ and hence $\Phi$ is injective.
It is also surjective. Let $[x] \in X/V$. If $[x] = [0]$ then $\Phi(0) = [x]$. If $[x] \neq [0]$ then $x \notin V$ and therefore $x \in \ker A$ and $\Phi(x) = [x]$.
Best Answer
The proof is in rudin functional analysis in the middle of the compact operators section