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If $B^{m\times m}$ is positive semidefinite, is it true that $A^TBA$ is positive semidefinite with $A^{m\times n}$? I think it is, because all of my counterexamples have failed, but I don't know how to prove it. I know that $x^TBx\ge 0$ if B is positive semidefinite. Is this the same? Do I need to prove it or is it well-known?

Vons
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  • The level of detail you'd need to give in a proof of this result definitely depends upon the definition of "positive semidefinite," and there are a number of definitions floating around (all equivalent, but maybe you'd want to only use one and not the others, depending on your audience). The "if" property you recite involving $B$ is an "if and only if" and often the definition. As the answer below explains carefully, it solves the problem. – leslie townes Feb 11 '21 at 01:02
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    As the comment above states, whether the property used is a definition or a theorem about psd matrices depends on your definition. But it is well known, and very often used as definition. – Clement C. Feb 11 '21 at 01:03
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    Gotcha, thanks! I was somehow thinking that there was a matrix version of that, e.g. $A$ serves the function of $x$ in $A^TBA$, but I guess your answer makes much more sense. – Vons Feb 11 '21 at 01:05
  • Another characterization of positive semidefiniteness of a square matrix is the existence of the factorization of the form $X^T X$. You can see how this would lead to another proof of the result. – leslie townes Feb 11 '21 at 01:43

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First, note that $M := A^\top B A$ is an $n$-by-$n$ square matrix, and further it is symmetric: $$ M^\top = (A^\top B A)^\top = A^\top B^\top A = A^\top B A = M $$ since $B$ is symmetric. Now, indeed, a (real) symmetric $n$-by-$n$ matrix $M$ is psd if, and only if, $x^\top M x \geq 0$ for every $x\in\mathbb{R}^n$. In our case, we can check that, for every $x\in\mathbb{R}^n$, $$ x^\top M x = x^\top A^\top B A x = (Ax)^\top B (Ax) \geq 0 $$ since $Ax\in\mathbb{R}^m$ and $B$ itself is psd, so $y^\top B y \geq 0$ for every $y\in \mathbb{R}^m$.

Clement C.
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