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I am trying to find properties or constraints on a $(p \times n) $matrix $U$ such that upon left multiplying a real symmetric square positive semi-definite matrix $V$ with $U$ the resulting matrix $W$ of dimensions $(p \times n) $ is still real positive semi-definite matrix. In other words i am trying to establish that there is $U$ such that $$W = U V$$ where $W$ and $V$ are real positive semi-definite matrices.

If $W=V$ then obviously $U=I$ fits the equation.

I have a suspicion that this should be possible as by definition of real symmetric positive semi-definite matrix, $V$ should have non-negative eigenvalues, which implies that it should have a right inverse which I suspect would also be positive semi-definite matrix, so if $W$ (some positive semi-definite matrix) is multiplied with the right inverse we should be able to compute $U$.

Any help would be much appreciated.

edit I suspect if my reasoning about computing $U$ above is correct then it should also be positive semi-definite matrix. So perhaps a way to construct $U$ may be by applying Gram-Schmidt process and finding new eigenvectors for W.

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  • $\begingroup$ There is something I am missing here. If $p\neq n$, the product $UV$ is not square, so it certainly cannot be symmetric positive definite. $\endgroup$
    – Michael Grant
    Apr 13, 2013 at 22:47
  • $\begingroup$ @MichaelGrant I believe a non-square matrix can be symmetric semi-positive definite. $\endgroup$
    – Comic Book Guy
    Apr 13, 2013 at 22:49
  • $\begingroup$ @MichaelGrant I am looking to try find an appropriate example for my claim. I think covariance matrices fit this category. $\endgroup$
    – Comic Book Guy
    Apr 13, 2013 at 22:52
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    $\begingroup$ No, a non-square matrix cannot be symmetric positive definite. Either property separately implies the matrix is square. $\endgroup$
    – Michael Grant
    Apr 13, 2013 at 22:56
  • $\begingroup$ @MichaelGrant I could be wrong but i would like to clarify here. Firstly I am interested in semi-Positive definite matrices not strictly positive definite. Secondly it is well known condition that covariance matrices must be semi-positive definite matrices. As they are allowed to be non square i can only conclude from this that it is possible to have non-square semi-positive definite matrices. If i am in correct in my understanding then please elaborate why this is not possible. $\endgroup$
    – Comic Book Guy
    Apr 13, 2013 at 23:01

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I am unsure how we should properly resolve this question.

As we have discussed at length in the comments, only square matrices can be positive semidefinite. Therefore, if $p\neq n$, there is no way that the product matrix $W=UV$ can be positive semidefinite, because it will also be non-square.

For grins, let's assume that $p=n$. Under what conditions is $W=UV$ positive semidefinite? This requires that $x^HUVx\geq 0$ for all complex vectors $x$. Alternatively, this is true if and only if $UV+VU^H=Q$ where $Q=W+W^H$ has nonnegative eigenvalues (and is therefore PSD itself). If $U$, $V$ are real, then you can relax the Hermitian transposes to real tranposes, and consider only reall vectors $x$.

Now for one special case, I know the answer. If $V$ is positive definite---i.e., not just PSD but nonsingular---and we require $W$ to be positive definite as well, then the answer follows from Lyapunov's theorem applied to linear systems:

  • The eigenvalues of $U$ must have positive real part.

What about the more relaxed cases? That is, what if $V$ is only positive semidefinite? What if $W$ is only required to be positive semidefinite? I am afraid I do not know. I'm sure people who study Lyapunov's theorem for linear systems in some depth know...

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