2
$\begingroup$

I have got a question and I would appreciate if one could help with it.

Assume $S$ is a diagonal matrix (with diagonal entries $s_1, s_2, \cdots$) and $M$ is a positive symmetric matrix with eigen value decomposition as follows:

$$\mathrm{eig}(M) = ULU^T$$

where $U^T$ means the transpose of $U$. I am trying to find out about the eigenvalues of $SM$. In other words, is there any relation between the eigenvalues of the matrix $S$, the matrix $M$ and their product $SM?$

any help/hint is appreciated

$\endgroup$
  • $\begingroup$ $\mathrm{eig}(U^T M U) = L$ and $\mathrm{eig}(U^T S U) $ is another positive symmetric matrix. The $\mathrm{eig}(U^T S U U^T M U) = SM$. I think I do not understand what you mean, or I am missing something. Do you mean that since $S = \mathrm{eig}(U^T S U)$ and $\mathrm{eig}(M) = \mathrm{eig}(U^T M U)$, we can write $\mathrm{eig}(SM) = \mathrm{eig}(U^T S U U^T M U)$? If this is the case, I doubt to agree. I just checked it with matlab and this does not seem to be correct $\endgroup$ – M.X Jul 2 '16 at 14:05
  • 2
    $\begingroup$ If the entries of $S$ are permuted, say to $S_1$, which preserves the eigenvalues of $S$, then in general, the eigenvalues of $S_1 M$ are not equal to the eigenvalues of $S M$. So clearly, order counts for $S$, not just its eigenvalues. $\endgroup$ – Mark L. Stone Jul 2 '16 at 15:33
1
$\begingroup$

There's no immediate relationship, no. The largest (in absolute value) eigenvalue of the product is no larger than the product of the largest eigenvalues, and the smallest eigenvalue of the product is no smaller then the product of the smallest.

Take any unit vector $u$ in the plane, and let $u'$ be 90 degrees counterclockwise from $u$. And let $Q$ be the matrix whose rows are $u$ and $u'$. Then for any diagonal matrix $D$, $Q^t D Q$ represents stretching along $u$ by $d_1$ and along $u'$ by $d_2$ (where these are the diagonal entries of $D$). Hold that thought.

For $u_1 = e_1$ and $(d_1, d_2) = (1, 4)$, for instance, we get a matrix $M$ that's just diagonal, with $1$ and $4$ on the diagonal. If $u$ is a unit vector in the $45$-degree direction, we get something else, but again with eigenvalues $1$ and $4$. For each possible angle $\theta$, let $$ M_\theta = Q^t \begin{bmatrix} 1 & 0 \\ 0 & 4 \end{bmatrix}Q $$ where $Q$ is the the matrix made from $u$ and $u'$, and $u$ is the vector $$ \begin{bmatrix} \cos u \\ \sin u \end{bmatrix}. $$

So $M_\theta$ has eigenvalues $1$ and $4$. Let $$ S = \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix} $$ Then for $\theta = 0$, we have $SM_\theta$ is a diagonal matrix with eigenvalues 2 and 12, the product of smallest and product of largest eigenvalues. But for $\theta = \pi/2$, the product has eigenvalues $8 = 2 \cdot 4$ and $3 = 1 \cdot 3$, the "middle" two products of the eigenvalues of the two original matrices. For intermediate values of $\theta$, you get other possible eigenvalues. This shows that the eigenvalues of the product can range all the way from the smallest possible (the product two smallest evals of the factors) to the largest possible.

Of course, the PRODUCT of the eigenvalues of $SM$ is the products of those of $S$ with those of $M$, by determinants. So that's another samll constraint on them.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.