I want to investigate what kind of eigen values cat tensor product $A \otimes B$ of two linear transformations have, s.t $(A \otimes B)z = \lambda z$.
If $z$ is a simple tensor then $\lambda = \alpha \beta$, where $\alpha,\beta$ are eigen values of $A,B$ respectively - easy to prove.
How can I find eigen values if $z = \sum_{ij}\gamma_{ij}x_i \otimes y_j$ is not a simple tensor, where $\{x_i\}$ and $\{y_j\}$ are the bases?
My proof is not "good looking":
- Assume an eigen vector $z = x \otimes y$ is a simple tensor, then $\lambda z = (A \otimes B)(z)$ and:
$\lambda(x \otimes y) = \lambda z = (A \otimes B)(z) = (A \otimes B)(x \otimes y) = Ax \otimes By$. Since both $\lambda (x \otimes y)$ and $Ax \otimes By$ are simple tensors, $Ax = \alpha x$ and $By = \beta y$, for some scalars $\alpha, \beta$, where $\alpha \beta = \lambda$. Therefore $\alpha$ and $\beta$ are proper values of $A$ and $B$ respectively.
- Assume an eigen vector $z$ is not a simple tensor and a linear combination $z = \sum_{pq}\gamma_{pq} x_p \otimes y_q$.
Equation of a proper value:
$\lambda z = (A \otimes B)(z) = (A \otimes B)(\sum_{ij}\gamma_{ij} x_i \otimes y_j) = \sum_{ij}\gamma_{ij} Ax_i \otimes By_j$.
Let $Ax_i = \sum_{p}\alpha_{pi}x_p$ and $By_j = \sum_{q}\beta_{qj}y_q$.
Then:
$\lambda z = \sum_{ij}\gamma_{ij} (\sum_{p}\alpha_{pi}x_p) \otimes (\sum_{q}\beta_{qj}y_q) = \sum_{pq} (\sum_{ij}\gamma_{ij}\alpha_{pi}\beta_{qj})x_p \otimes y_q$
For all $p$ and $q$ we have $\sum_{ij}\gamma_{ij}\alpha_{pi}\beta_{qj} = \lambda \gamma_{pq}$.
Small observation: $f_{pq}(z) = \sum_{ij}\gamma_{ij}\cdot (\alpha_{pi}\beta_{qj}/\lambda) = \gamma_{pq}$ looks like a dual basis.
What's next?
If $z$ can be any vector, then I solve this equation for all $z \in \{ x_p \otimes y_q : p = 1.. n, q = 1..m \}$ - basis of a tensor product, and get the following result: $A = \alpha \cdot 1$ and $B = \beta \cdot 1$ and $\alpha \beta = \lambda$.