# A restriction $A_2$ of a compact self-adjoint compact linear operator $A$ is also compact and self-adjoint?

Let $X$ be an inner product space and let $A$ be a compact and self-adjoint linear operator. Let $p_1$ be an eigenvector of $A$. Let $A_2$ be the restriction of $A$ to $X_2$ where $X_2$ is given by $$X_2 = \{x \in X: (x, p_1) = 0\}$$

Then clearly $A_2$ is again a compact and self-adjoint linear operator.

Question

So $X_2$ is a space that is orthogonal to the eigenvector $p_1$ of $A$. But I don't see why its clear that it is again compact and self-adjoint? Can it be shown formally?

First, note that $X_2$ is an $A$-invariant subspace, so that $A_2:X_2\to X_2$.
We can show formally that the adjoint of $A_2$ should be the restriction of $A^*$ to $X_2$, which is again $A^*$.
It then suffices to note that the image of the closed unit ball under $A_2$ is a closed subset of the image of the closed unit ball under $A$ and is therefore compact.