# Transformation of PD matrix with rank deficient other matrix

This should be easy for you.

I have an intuitive feeling for why the following should be correct, but I would like something more rigorous than my feeling :)

Consider the $$m\times m$$-matrix $$B$$, which is symmetric and positive definite (full rank). Now this matrix is transformed using another matrix, say $$A$$, in the following manner: $$A B A^T$$. The matrix $$A$$ is $$n\times m$$ with $$n. Furthermore the constraint $$rank(A) < n$$ is imposed.

My intuition tells me that $$A B A^T$$ must be symmetric and positive semi-definite, but what is the mathematical proof for this? (why exactly does the transformation preserve symmetry and why is it that possibly negative eigenvalues in $$A$$ still result in the transformation to be PSD? Or is my intuition wrong)?

Edit: please exclude the case of A=0.

For symmetry: note that in general, we have $$(AB)^T = B^TA^T$$, hence $$(ABC)^T = C^TB^TA^T$$. With that, we see that $$(ABA^T)^T = A^{TT}B^TA^T = ABA^T.$$ For positive semidefiniteness: an $$n \times n$$ symmetric matrix $$M$$ is positive semidefinite if (and only if) $$x^TMx \geq 0$$ whenever $$x \in \Bbb R^n$$. We note that $$x^T(ABA^T)x = (x^TA)B(A^Tx) = (A^Tx)^T B (A^Tx).$$ Because $$B$$ is positive definite (and hence positive semidefinite), we must have $$y^TBy \geq 0$$ for $$y = A^Tx$$. Thus, $$x^T(ABA^T)x \geq 0$$, so that $$ABA^T$$ is indeed positive semidefinite.
• This is great! I am lacking one point though: You enforced $x^T M x \geq 0$ right away because being PD implies being PSD. But doesn't the equal sign arise because of the rank deficiency of $A$? -> if $A$ had rank $n$, then would't the transformed matrix be PD? Jul 9, 2020 at 21:54
• @Elarion Yes, it is indeed true that if $A$ has rank $n$, then $A^TBA$ is positive definite. Jul 10, 2020 at 5:57
• So the more thorough reasoning is that $x^T M x > 0$ for any nonzero $x \in \mathbb{R}^m$ and that due to A being rank deficient, $y$ may be zero for a nonzero $x$. Hence it holds that $x^T M x \geq 0$ for any $x \in \mathbb{R}^n$? Jul 10, 2020 at 6:55
• @Elarion It is incorrect to say that the matrix $M = ABA^T$ is such that $x^TMx > 0$ for any non-zero $x \in \Bbb R^m$. Jul 10, 2020 at 6:59