For a self-adjoint $T$, if $T^k$ is compact, then so is $T$.

Question

I'm studying functional analysis. Let $T:\mathcal{H}\rightarrow\mathcal{H}$ be a bounded self-adjoint linear operator on a Hilbert space $\mathcal{H}$. The problem is showing if $T^k$ is a compact operator for some $n\in\mathbb{N}$, then $T$ is also compact.

Since $T^k$ is compact and self-adjoint, we may choose a countable orthonomal system $\{x_n\}$ and real eigenvalues $\{\lambda_n\}$ such that $T^k=\displaystyle\sum_n \lambda_n x_n\otimes\overline{x_n}.$ Then I guess $T$ should be the form of $S:=\sum_n (\lambda_n)^{1/k} x_n\otimes\overline{x_n}.$ Hm, if $n$ is even, we should consider the sign more carefully. $S^k=T^k$ is trivial. But.. how can I prove the uniqueness?

In general, $T$ will not have a unique $k$th root. However, if $T^k$ is compact, then all self-adjoint $k$th roots of $T$ are as well. — Ben Grossmann, Jun 24 '15 at 05:44

score 1 · Accepted Answer · answered Jun 24 '15 at 06:24

$T^k$ being compact there exists a sequence of distinct real numbers $\lambda_n \to 0$ and a sequence of pairwise orthogonal projectors of finite rank $P_n$ so that $$T^k = \sum \lambda_n P_n$$

Since $T$ commutes with $T^k$ it will invariate the eigenspaces $H_n \colon = \text{Im}\ P_n$ of $T^k$. The restriction of $T$ to the finite dimensional subspace $H_n$ is self-adjoint so diagonalizable. Since $T^k= T^k$ we have $$H_n = H'_n \oplus H''_n$$ corresponding to the roots (at most $2$) of the equation $\mu_n^k = \lambda_n$.

We get a decomposition $$T= \sum \mu_n Q_n$$ $\mu_n \to 0$, hence a compact operator.

For a self-adjoint $T$, if $T^k$ is compact, then so is $T$.

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