0
$\begingroup$

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 $n<m$. 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.

$\endgroup$

1 Answer 1

0
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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? $\endgroup$
    – Elarion
    Jul 9, 2020 at 21:54
  • $\begingroup$ @Elarion Yes, it is indeed true that if $A$ has rank $n$, then $A^TBA$ is positive definite. $\endgroup$ Jul 10, 2020 at 5:57
  • $\begingroup$ 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 $x \in \mathbb{R}^n$? $\endgroup$
    – Elarion
    Jul 10, 2020 at 6:55
  • $\begingroup$ @Elarion It is incorrect to say that the matrix $M = ABA^T$ is such that $x^TMx > 0$ for any non-zero $x \in \Bbb R^m$. $\endgroup$ Jul 10, 2020 at 6:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .