# Orthogonality of eigenvectors of a positive operator in the polar decomposition

Let $T : H \rightarrow H$ be a compact operator ($H$: complex Hilbert space). Then $|T| = (T^* T)^{1/2}$ is a compact self-adjoint operator.

If $\{x_j\}$ is an orthonormal basis of $(\ker |T|)^{\perp}$, is it true that $\langle Tx_i, x_j \rangle = 0$ for $i\neq j$?

I tried to check this by using the polar decomposition of $T$, that is, $T = U|T|$ where $U$ is a partial isometry.

One can obtain that $\langle Tx_i x_j \rangle = \lambda_i \langle Ux_i, x_j\rangle = \lambda_i \langle Ux_i, U^*Ux_j \rangle = \lambda_i \langle U^2 x_i, Ux_j\rangle .$ From this, is it possible to deduce that $\langle Tx_i, x_j\rangle = 0$ for $i\neq j$?

Any help will be appreciated!

Start with the example $Tx = \sum_{j=1}^{\infty}\lambda_j\langle x,e_j\rangle e_j$ for some orthonormal basis $\{ e_j \}$ of $H$, and where $\lambda_j$ is a sequence of positive real numbers that tends to $0$. This is a selfadjoint compact operator $T$, and $\mathcal{N}(|T|)=\{0\}$. Choose another orthonormal basis $\{ e_j'\}$ of $\mathcal{N}(|T|)^{\perp}=H$. Suppose $e_1'=\frac{1}{\sqrt{2}}(e_1+e_2)$ and $e_2'=\frac{1}{\sqrt{2}}(e_1-e_2)$. Then, $$\langle Te_1',e_2'\rangle = \frac{1}{2}(\lambda_1 -\lambda_2) \ne 0.$$ Your statement does hold for the orthonormal basis eigenvectors of this $T$, but not for every orthonormal basis, unless all of the eigenvalues of $T$ are the same, which would imply that all eigenvalues would be $0$ because $T$ is compact.

Posting another answer as the counter-example in my first one was incorrect.

The statement in question in general is not true. For this consider the Schmidt decomposition of the compact operator $T$

$$T=\sum_{n\in\mathfrak N} s_n(T)\langle f_n,\cdot\rangle g_n$$

with $\mathfrak N\subseteq\mathbb N$, orthonormal systems $(f_n)_{n\in\mathfrak N}$, $(g_n)_{n\in\mathfrak N}$ in $\mathcal H$, and the positive singular values $s_n(T)\overset{n\to\infty}\longrightarrow0$. This implies $|T|=\sum_{n}s_n(T)\langle f_n,\cdot\rangle f_n$ so extending the ONS $(f_n)_{n\in\mathfrak N}$ to an orthonormal basis $({\hat f}_j)_{j\in\mathfrak N}$ of $\mathcal H$ yields that $|T|$ is diagonal with respect to said ONB. Thus

$$\operatorname{ker}|T|=\overline{\operatorname{span}\lbrace \tilde f_j \,|\, j\in\mathbb N, \tilde f_j\notin(f_n)_{n\in\mathfrak N}\rbrace}$$

and

$$(\operatorname{ker}|T|)^\perp=\overline{\operatorname{span}\,(f_n)_{n\in\mathfrak N}}.$$

Let $m,n\in\mathfrak N$, $m\neq n$ (which of course assumes that $|\mathfrak N|>1$; your statement is true for rank-1-operators as in that case $(\operatorname{ker}|T|)^\perp$ is one-dimensional so there are no two distinct basis elements to choose from) then

$$\langle f_m,Tf_n\rangle = s_n(T)\langle f_n,g_m\rangle\tag{1}$$

which in general obviously is not zero unless $T\geq 0$ from the start so $T=|T|$ and $g_n=f_n$ for all $n\in\mathfrak N$. As a conclusion, your statement holds (within the boundries of the "choice of the basis"-problem which @DisintegratingByParts pointed out in his answer) if either $T$ is positive semi-definite or if $T$ has rank one. Otherwise, generally speaking, it is not true as it's characterized by equation (1).

For a concrete counter-example consider some orthonormal basis $(e_n)_{n\in\mathbb N}$ of any separable Hilbert space. Then

$$T:=\langle e_1,\cdot\rangle e_2+\langle e_2,\cdot\rangle e_3$$

is compact because finite-rank so $|T|=\langle e_1,\cdot\rangle e_1+\langle e_2,\cdot\rangle e_2$ and $(\operatorname{ker}|T|)^\perp=\operatorname{span}(e_1,e_2)$. But

$$\langle e_2,Te_1\rangle=\Vert e_1\Vert^2\Vert e_2\Vert^2=1\neq 0.$$