Realness of eigenvalues of $\left(E[xx']\otimes I + I\otimes E[xx']\right)^{-1} E[xx'\otimes xx']$

Question

Suppose $x$ is a random variable in $\mathbb{R}^d$ and $I$ is a $d\times d$ identity. What's the easiest way of showing that eigenvalues of the following matrix are real?

$$\left(E[xx']\otimes I + I\otimes E[xx']\right)^{-1} E[xx'\otimes xx']$$

This might be equivalent to showing that the $\Lambda,\Phi$, solution to generalized problem is real:

$$E[xx'\otimes xx'] \Phi = (E[xx']\otimes I + I\otimes E[xx'])\Phi \Lambda$$

_^Notebook

It's a product of two symmetric matrices, so it's not symmetric — Yaroslav Bulatov, Dec 18 '23 at 21:26
Thanks, see also the last paragraph of https://math.stackexchange.com/q/113859 — user196574, Dec 18 '23 at 21:32
@user196574 Yes, there seems to be some confusion over whether "positive definite" requires symmetric. If you treat matrix as quadratic form, then it doesn't matter whether it's symmetric. If you treat it as linear operation, it matters. — Yaroslav Bulatov, Dec 18 '23 at 21:41

score 2 · Accepted Answer · answered Dec 18 '23 at 21:28

2

It suffices to note that all of your expectation matrices are symmetric and positive definite. Denote $P = \left(E[xx']\otimes I + I\otimes E[xx']\right), Q = E[xx' \otimes xx']$. The matrices $P,Q$ are symmetric positive definite. It follows that $P^{-1}Q$ is similar to the matrix $$ P^{1/2}(P^{-1}Q)P^{-1/2} = P^{-1/2}QP^{-1/2}. $$ This is a matrix of the form $M^TQM$ for a real matrix $M$. It follows that this matrix is positive semidefinite, and thus has real eigenvalues.

answered Dec 18 '23 at 21:28

Ben Grossmann

225,327

How do you know anticommutator {E[xx'],I} is positive definite? – Yaroslav Bulatov Dec 18 '23 at 21:29
1

A Kronecker product of positive definite matrices is positive definite, and a sum of positive definite matrices is positive definite. – Ben Grossmann Dec 18 '23 at 21:30
Also, I'm not used to your usage of "anticommutator". You could say that is is the Kronecker sum of $E[xx']$ with itself. The anticommutator more commonly refers to ${A,B} = AB + BA$. – Ben Grossmann Dec 18 '23 at 21:32
ok, Kronecker sum seems like a better term – Yaroslav Bulatov Dec 18 '23 at 21:37
By the way, it is notable that the product of symmetric matrices can potentially have complex (non-real) eigenvalues. For instance, $$ \pmatrix{-1&0\0&1}\pmatrix{0&1\1&0} = \pmatrix{0&-1\1&0}. $$ This is ruled out if one of the matrices is positive semidefinite. – Ben Grossmann Dec 18 '23 at 21:38
btw, any thoughts whether definition of "positive definite" should require symmetric or not? – Yaroslav Bulatov Dec 18 '23 at 21:42
@YaroslavBulatov Personally I think it should require symmetry, but that's not a universal convention so I include the descriptor "symmetric" for clarity. – Ben Grossmann Dec 18 '23 at 21:45
BTW, a related curious phenomenon is that multiplying random matrices together eventually gives a matrix with only real eigenvalues -- post – Yaroslav Bulatov Dec 18 '23 at 21:46

Realness of eigenvalues of $\left(E[xx']\otimes I + I\otimes E[xx']\right)^{-1} E[xx'\otimes xx']$

1 Answers1