1
$\begingroup$

Let $P$ be a markov transition matrix, that is, all entries of $P$ satisfies $0\le P_{ij}\le 1$, and row sums equal to $1$.

Do we have the property that: $v^T(I-P)^2v\ge 0$, for any vector $v$ and I a identity matrix? And if the answer is yes, do we also have $v^T(I-P)M(I-P)v\ge 0$ where M is an arbitrary positive semidefinite matrix?

It seems true with some computer simulation, and I didn't find any counter example.

My attempt is to find the spectrum of $(I-P)^2+(I-P^T)^2$, but didn't get anything. Thanks.

$\endgroup$
  • $\begingroup$ If $P$ were symmetric then you would have $v^T(I-P)^2v=\| (I-P)v \|^2 \geq 0$, and similarly $v^T (I-P) M (I-P) v = \| M^{1/2} (I-P) v \|^2 \geq 0$ where $M^{1/2}$ denotes a matrix square root (non-unique but existent for positive semidefinite matrices). Can you try "adding and subtracting" the nonnegative quantity $v^T(I-P^T)(I-P)v$ to try to get an estimate? $\endgroup$ – Ian Mar 17 '17 at 4:20
  • $\begingroup$ Decompose the matrix into symmetric part $S$ and antisymmetric part $Q$, it is equivalent to show that $(I-S)^2+Q^2$ is positive semidefinite, where $P = S+Q$. Then I have no idea how to do the estimate. Thank you. $\endgroup$ – Simo Mar 17 '17 at 8:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.