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 $S$?
- Is there some work on this type of matrix formula?
- Is there a better way to write $H$ that might be more enlightening (i.e. Kronecker products..)?
- What happens to the eigenvalues of $T$ after multiplication by $S$?