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I am struggling for a while in proving the positive-semidefiniteness
( in the usual sense, i.e., $A$ is Hermitian and $x^*Ax\geq 0\,$ for all $\,x \in \mathbb{C}^2\,\big)$
of the following matrix $$A :=\,BC^\top + CB^\top+BB^\top,$$ under the hypotheses that both $B$ and $C$ have real entries and are positive-semidefinite in $\mathbb R^2 $: in the sense that $$x^\top B x\geq 0\quad\forall\, x\in\mathbb{R}^2$$ and likewise for $C$. Furthermore, $B,C$ are not necessarily symmetric.

Numerical tests suggest that this holds true whenever the corresponding non-diagonal elements of $B$ and $C$ share the same sign, that is, if we set $B=\begin{bmatrix} b_1 & b_2\\b_3 & b_4\end{bmatrix}$ and $C=\begin{bmatrix} c_1 & c_2\\c_3 & c_4\end{bmatrix}$ then $sign(b_2)=sign(c_2)$ and $sign(b_3)=sign(c_3)$.

My most promising attempt has been effectively computing matrix $A$ and checking the principal minors. We have that $$A=\begin{bmatrix} 2(l_1u_1+l_2u_2)+l_1^2+l_2^2 & l_1u_3+l_3u_1+l_4u_2+l_2u_4+l_1l_3+l_2l_4\\ l_1u_3+l_3u_1+l_4u_2+l_2u_4+l_1l_3+l_2l_4 & 2(l_3u_3+l_4u_4)+l_3^2+l_4^2 \end{bmatrix},$$ and the non-negativity of the diagonal elements of $A$ comes from the shared sign of the off-diagonal elements of $B$ and $C$. It remains to ensure that $\det(A)\geq 0$. My bet is on the following computations:

\begin{equation*} \begin{aligned} \det(A)&=4(b_1c_1+b_2c_2(b_3c_3+b_4c_4)+2b_1^2(b_3c_3+b_4c_4)+2b_2^2(b_3c_3+b_4c_4)+2b_3^2(b_1c_1+b_2c_2)+2b_4^2(b_1c_1+b_2c_2)\\ &\quad +b_1^2b_3^2+b_1^2b_4^2+b_2^2b_3^2+b_2^2b_4^2-(b_1c_3+b_3c_1+b_4c_2+b_2c_4+b_1b_3+b_2b_4)^2\\ &=4(b_1c_1+b_2c_2)(b_3c_3+b_4c_4)+2b_1^2({b_3c_3}+b_4c_4)+2b_2^2(b_3c_3+{b_4c_4})+2b_3^2({b_1c_1}+b_2c_2)+2b_4^2(b_1c_1+{b_2c_2})\\ &\quad +{b_1^2b_3^2}+b_1^2b_4^2+b_2^2b_3^2+{b_2^2b_4^2}-b_1^2c_3^2-b_3^2c_1^2-b_4^2c_2^2-b_2^2c_4^2-{b_1^2b_3^2}-{b_2^2b_4^2}\\ &\quad-2b_1c_3b_3c_1-2b_1c_3b_4c_2-2b_1c_3b_2c_4-{2b_1^2c_3b_3}-2b_1c_3b_2b_4-2b_3c_1b_4c_2-2b_3c_1b_2c_4-{2b_3^2b_1c_1}-2b_3c_1b_2b_4\\ &\quad-2b_4c_2b_2c_4-2b_4c_2b_1b_3-{2b_4^2c_2b_2}-2b_2c_4b_1b_3-{2b_2^2c_4b_2}-2b_1b_3b_2b_4\\ &=4b_1c_1b_3c_3+4b_1c_1b_4c_4+4b_2c_2b_3c_3+4b_2c_2b_4c_4+2b_1^2b_4c_4+2b_2^2b_3c_3+2b_3^2b_2c_2+2b_4^2b_1c_1+b_1^2b_4^2+b_2^2b_3^2\\ &\quad -b_1^2c_3^2-b_3^2c_1^2-b_4^2c_2^2-b_2^2c_4^2-2b_1c_3b_3c_1-2b_1c_3b_4c_2-2b_1c_3b_2c_4-2b_1c_3b_2b_4-2b_3c_1b_4c_2-2b_3c_1b_2c_4\\ &\quad -2b_3c_1b_2b_4-2b_4c_2b_2c_4-2b_4c_2b_1b_3-2b_2c_4b_1b_3-2b_1b_3b_2b_4\\ &=2b_1c_1b_3c_3+4b_1c_1b_4c_4+4b_2c_2b_3c_3+2b_2c_2b_4c_4+\underline{2b_1^2b_4c_4+2b_2^2b_3c_3+2b_3^2b_2c_2+2b_4^2b_1c_1}+\underline{(b_1b_4-b_2b_3)^2}\\ &\quad -b_1^2c_3^2-b_3^2c_1^2-b_4^2c_2^2-b_2^2c_4^2-2b_1c_3b_4c_2-2b_1c_3b_2c_4-2b_3c_1b_4c_2-2b_3c_1b_2c_4\\ &\quad \underline{-2b_1c_3b_2b_4-2b_3c_1b_2b_4-2b_4c_2b_1b_3-2b_2c_4b_1b_3}\\ &=2b_1c_1b_3c_3+4b_1c_1b_4c_4+4b_2c_2b_3c_3+2b_2c_2b_4c_4-b_1^2c_3^2-b_3^2c_1^2-b_4^2c_2^2-b_2^2c_4^2\\ &\quad -2b_1c_3b_4c_2-2b_1c_3b_2c_4-2b_3c_1b_4c_2-2b_3c_1b_2c_4\\ &\quad +(b_1b_4-b_2b_3)(b_1b_4-b_2b_3+2b_1c_4+2c_1b_4-2b_2c_3-2c_2b_3), \end{aligned} \end{equation*} where I have underlined the terms involved in the final equality in order to facilitate the reading.

And now I am stuck. I don't know if there is any obscure equality to gather the remaining terms, and can not even ensure that $$b_1b_4-b_2b_3+2b_1c_4+2c_1b_4-2b_2c_3-2c_2b_3\geq 0,$$ although intuition tells me it should be true.

If you have any ideas for the follow up from here, I would be happy to hear them.

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  • $\begingroup$ Welcome to math.SE ! If you do not mind, then I would edit your post lateron to rectify the spelling. $\endgroup$
    – Hanno
    Jan 26, 2021 at 13:42
  • $\begingroup$ @Hanno Be my guest! And thank you for your answer, I will comment on it there $\endgroup$
    – obol
    Jan 26, 2021 at 14:59

2 Answers 2

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In case $B$ and $C$ are positive semidefinite in mathematicians' definition, they are symmetric and $A=X^2-Y^2$, where $X:=B+C\succeq Y:=C\succeq0$. However, it is known that $X\succeq Y\succeq0$ in general does not imply that $X^2\succeq Y^2$ (see here for instance). Therefore $A$ is not always positive semidefinite.

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  • $\begingroup$ +1 for your conceptual explanation! $\endgroup$
    – Hanno
    Jan 26, 2021 at 21:35
  • $\begingroup$ Thank you, I know see I was trying to go for something impossible $\endgroup$
    – obol
    Feb 3, 2021 at 10:33
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A counter-example is proposed, in a rush after the lunch break, hence check carefully please.

Consider the real matrices $$B\:=\:\begin{pmatrix}1& 0\\ 0& \beta\end{pmatrix} \text{ with }\:0<\beta\qquad C\:=\:\begin{pmatrix}1& \gamma\\ \gamma& 1\end{pmatrix} \text{ with }\:0<\gamma <1\,,$$ they are symmetric, thus Hermitian a fortiori. Within the indicated parameter ranges both matrices are positive-definite, either seen as members of $M_2(\mathbb R)$, either assuming that $B,C\in M_2(\mathbb C)$.

We compute that $$A\:=\:BC^\top + CB^\top+BB^\top \:=\: \begin{pmatrix}3& (1+\beta)\gamma\\ (1+\beta)\gamma& \beta (2+\beta)\end{pmatrix}\\[5ex] \implies \det A\:=\: 3\beta (2+\beta) - (1+\beta)^2\gamma^2$$ and conclude, if $\beta$ is sufficiently small, that $A$ has one positive and one negative eigenvalue. In other words, $A$ is (definitely!) an indefinite matrix.

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  • $\begingroup$ Thank you for your counter example. In fact the entries of these matrices are real valued functions, with $B=\begin{bmatrix}f_1(x) & f_2(x) \\ f_3(x) & f_4(x)\end{bmatrix}$ and with $C=\begin{bmatrix}f_1(y) & f_2(y) \\ f_3(y) & f_4(y)\end{bmatrix}$. So the upper left entries of $B$ and $C$ being equal while the lower right do not is not an example I encountered. But thank you again for your input, it provided a new valuable insight into my problem, namely on further hypothesis to take into account in this specific proof. $\endgroup$
    – obol
    Jan 26, 2021 at 15:32
  • $\begingroup$ @obol Notice that the entry $b_{11}$ may be shifted away from $1=c_{11}$ and remains being a counter-example. It is somehow stable, and not an isolated one. $\endgroup$
    – Hanno
    Jan 26, 2021 at 15:35
  • $\begingroup$ True that. In fact, the original matrices $B$ and $C$ are such that they somewhat depend on each other, as $B=D_1+D_2$ and $C=D_2+D_3$, each of the $D$s being semi-positive definite in $\mathbb{R}^2$ in the sense of the original question. I now see that this fact must be taken into consideration. $\endgroup$
    – obol
    Jan 26, 2021 at 16:23

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