Let $H_{1}$ and $H_{2}$ be two Hilbert spaces. Now let $A \in B(H_{1})$ and $B_{ij} \in B(H_{2})$ for all i and j be positive operators. i.e. $\forall \phi \in H_{1}$ and $\forall \psi \in H_{2}$

i.e. $\langle \phi ,A \phi\rangle \geq 0 $ and $\langle \psi, B_{ij}\psi\rangle\geq 0$ for all i and j. I am wondering if the following operator is also a positive operator. Let us write $A$ using some arbitrary basis that spans $H_{1} \{\nu_{n}\}_{n}$. $$ A \rightarrow \sum_{nm}a_{nm}\nu_{n}\otimes \nu_{m}.$$

I am wondering if the following operator acting over $H_{1}\otimes H_{2}$ is also a positive operator.

$$OP:=\sum_{nm}a_{nm}\big(\nu_{n}\otimes \nu_{m}\big)\otimes B_{nm}.$$

I have tried starting with an arbitrary element of $H_{1}\otimes H_{2}$ call it $\omega$ and using the fact that it may be represented as a schmidt decomposition. Let us assume infinite dimensional Hilbert spaces.

$$\omega = \sum_{i} \beta_{i}f_{i}\otimes g_{i}$$ where $\{f_{i}\}_{i}$ and $\{g_{i}\}_{i}$ are respectively basis whose span is respectively dense in $H_{1}$ and $H_{2}$.

Computing $\langle \omega , OP \omega\rangle$ I am not able to attain anythig useful. I get $$ \langle \omega , OP \omega\rangle = \sum_{i,n,m} a_{nm}|\beta_{i}|^{2}\langle f_{i},\nu_{n}\rangle \langle \nu_{m},f_{i}\rangle \langle g_{i}, B_{nm}, g_{i}\rangle $$

Thank you very much in advance for any help.


1 Answer 1


In case $H_1=\mathbb C^n$, $A$ is the matrix with $a_{i,j}=1$, for all $i$ and $j$, and $H_2=\mathbb C$, your question is asking whether a matrix with positive entries is a positive operator. The answer to this is of course no!

  • $\begingroup$ A is a positive operator. Also, all of the operators $B_{ij}$ are positive. I never said anything about their entries. The A you defined is not a positive operator while the one I have defined is so your example is urelated. Thank you for your attention nevertheless. $\endgroup$
    – Hldngpk
    Jun 19, 2022 at 0:35
  • $\begingroup$ Wait! I beleive that I have made a mistake. The all ones matrix is positive semidefinite or am I mistaken? $\endgroup$
    – Hldngpk
    Jun 19, 2022 at 0:37
  • 1
    $\begingroup$ Yes, the all ones matrix is a positive multiple of a self adjoint projection, hence positive. $\endgroup$
    – Ruy
    Jun 19, 2022 at 0:52

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