# Show that the eigenvalues of a compact and self-adjoint bounded linear operator are summable

Let

• $H$ be a $\mathbb R$-Hilbert space
• $A\in\mathfrak L(H)$ be compact and self-adjoint
• $I:=\mathbb N\cap[0,\operatorname{rank}A]$

By the Hilbert-Schmidt theorem, there is a $(\lambda_i)_{i\in I}\subseteq\mathbb R\setminus\left\{0\right\}$ with $$\left|\lambda_i\right|\ge\left|\lambda_{i+1}\right|\;\;\;\text{for all }i\in I\tag1$$ and $$Ae_i=\lambda_ie_i\;\;\;\text{for all }i\in I\tag2$$ for some orthonormal basis $(e_i)_{i\in I}$ of $\overline{AH}$. Moreover, If $\operatorname{rank}A=\infty$, then $$\lim_{n\to\infty}\lambda_n=\inf_{i\in I}\left|\lambda_i\right|=0\tag3\;.$$

Now, I've read that $$\left\|A\right\|_{\mathfrak L(H)}=\sup_{i\in I}\left|\lambda_i\right|\tag4$$ and $$\sum_{i\in I}\left|\lambda_i\right|<\infty\;.\tag5$$

1. What am I missing? In light of $(1)$, the supremum on the right-hand side of $(4)$ should simply be $\left|\lambda_1\right|$.
2. If $\operatorname{rank}A=\infty$, I don't see why $(5)$ is correct, i.e. why $\left(\sum_{n=1}^N\left|\lambda_n\right|\right)_{N\in\mathbb N}$ is bounded. So, how do we see that?

$1)$ Where you read your statement, it probably referred to the set of eigenvalues, without being ordered.
$2)$ The second statement is not true. If you take a diagonal operator with diagonal entries real numbers (thus self-adjoint) a necessary and sufficient condition to be compact is that the entries converge to $0$. The statement probably referred to compact, self-adjoint, trace-class operators.