I am dealing with a matrix $H\in\mathbb{R}^{N \times N}$ that is the result of a vector $S\in\mathbb{R}^{1\times P}$ multiplied onto an order-3 tensor $T\in\mathbb{R}^{P\times N \times N}$. I am not precisely sure how to write this, but I have numerically confirmed that the $i^{th}$ row/column of $H$ (it's symmetric) is formed by the multiplication of $S$ and the $i^{th}$ "layer" of a order-3 tensor:

Using $\otimes$ as the tensor outer product (which might be the wrong type of tensor multiplication- I'm honestly not sure), $$H = S \otimes T.$$ To clarify, in Matlab notation we have $$H(i,:) = S ^* T(:,:,i),$$ where $T(:,:,i)\in\mathbb{R}^{P\times N}$.

Note that $H$ is symmetric, and that $T$ is the collection of second derivatives of a vector valued function $f:\mathbb{R}^N\rightarrow\mathbb{R}^P$ with respect to $x\in\mathbb{R}^N$. So it has some symmetric properties as well.

I would like to characterize the eigenvalues of $H$ (in particular, bound from below). This multiplication scheme is reminsicent of a Kronecker Product, which gives me some hope: maybe I can write it as an actual Kronecker Product?

The reason I relate to Kronecker is the nice eigenvalue property: the eigenvalues of $A\otimes B$ are $eig(A)_ieig(B)_j$ for each $i,j$, where I'm now using $\otimes$ for the Kronecker Product. Since $T$ has symmetry properties I might be able to deal with the eigenvalues of $T$ itself, but what happens to the eigenvalues when multiplied by this vector $S$?

  1. Is there some work on this type of matrix formula?
  2. Is there a better way to write $H$ that might be more enlightening (i.e. Kronecker products..)?
  3. What happens to the eigenvalues of $T$ after multiplication by $S$?
  • $\begingroup$ Actually, that is not tensor outer product, but inner product. In other words $$ H = \sum_i s_i T(i, :, :) $$ or if you treat $T$ as a collection of matrices $T_i(:, :) = T(i, :, :)$ that would be simply some linear combination of matrices $$ H = \sum_i s_i T_i $$ $\endgroup$ – uranix Aug 22 '16 at 16:32
  • $\begingroup$ @uranix, for safekeeping I have numerically verified your insight. This is actually very helpful for me... Including that it has made me realize how deep question 3 is. If you post as an answer I will accept it and edit out Question 3. $\endgroup$ – BenMooMath Aug 27 '16 at 18:27

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.