3
$\begingroup$

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!

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.