Let $\mathcal{H}$ be a Hilbert space. Assume $x, y \in B(\mathcal{H})$ are self-adjoint. We say $x \geq y$ if $x-y$ is positive semi-definite. We define positive semi-definite as $\langle x \xi, \xi \rangle \geq 0$ for all $\xi \in \mathcal{H}$.

I am looking for specific examples to illustrate the following statements:

(1) $x,y \geq 0 \nRightarrow xy \geq 0.$

I looked for an example of two matrices such that both are positive semi-definite but their product is not, but I couldn't find an example where the matrices are symmetric (since I want $x,y$ self-adjoint). I only saw examples online where the matrices were not symmetric.

(2) $x \geq y \geq 0 \nRightarrow x^2 \geq y^2.$

By the definition above, we are looking for $x$ and $y$ self-adjoint such that $(x-y)$ and $y$ are both positive semi-definite, but $(x^2-y^2)$ is not. Again I looked for matrices that would satisfy this but had no luck.

Are matrices over $\mathbb{C}$ a bad example of $B(\mathcal{H})$ to take for these situations?


1 Answer 1


2x2 matrices are enough to see these things.

For the first one: consider $$ A = \begin{pmatrix} 2 & 1 \\ 1 & 1\end{pmatrix} \text{ and } B = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}. $$ Both are positive, but $$AB = \begin{pmatrix} 2 & 0 \\ 1 & 0 \end{pmatrix}, $$ which is not self-adjoint, hence not positive.

For the second one, lets use the same matrices! Clearly $A \geq B \geq 0$ (as $A - B$ is a positive multiple of a projection), but $$ A^2 - B^2 = \begin{pmatrix} 4 & 3 \\ 3 & 2 \end{pmatrix}, $$ which has eigenvalue $3 - \sqrt{10} < 0$, so it cannot be positive.

  • $\begingroup$ Thank you so much!!! Quick question, the AB you gave in the first part is not symmetric but it does have only non-negative eigenvalues right? So isn't it positive semi-definite still, or MUST we have symmetric? $\endgroup$ Commented Aug 31, 2020 at 14:09
  • 1
    $\begingroup$ Positive semi-definite here requires being self-adjoint (symmetric when the entries are real). An operator $T$ is positive (i.e., $\langle T\xi,\xi \rangle \geq 0 \forall \xi$) if and only if $T$ is self-adjoint and the spectrum of $T$ is is contained in $[0,\infty)$. In $M_2$, the spectrum is just the set of eigenvalues, so having non-negative eigenvalues is not enough to ensure positivity of $AB$. $\endgroup$
    – PStheman
    Commented Aug 31, 2020 at 16:56

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