0

Given a matrix $A$ can we write the eigendecomposition of $A \otimes A$ if we know the eigendecomposition of $A$.

I am trying to use this answer to solve $$((A \otimes A) + I)^{-1}$$ in an efficient manner.

Dylan Dijk
  • 323
  • 2
    If $\lambda_1,\dots,\lambda_n$ are eigenvalues of $A$, and $e_1,\dots,e_n$ is an eigenbasis, then any $\lambda_{ij} = \lambda_i \lambda_j$ is an eigenvalue of $A \otimes A$ with an eigenvector $e_i \otimes e_j$. The set of such vectors for all $1 \leq i, j \leq n$ form an eigenbasis of $A \otimes A$. – Oleksandr Kulkov Dec 03 '23 at 18:16
  • When you say $e_1 \otimes e_j$, what does this notation mean? As this should be a vector? – Dylan Dijk Dec 03 '23 at 18:22
  • 1
    I think you should define what you mean by tensor product $\otimes$ because you comment does not really make sense if $A \otimes A$ is defined the canonical way... – julio_es_sui_glace Dec 03 '23 at 18:25
  • Yep sorry stupid comment from me $e_1 \otimes e_j$ makes sense, this is a vector. – Dylan Dijk Dec 03 '23 at 18:28
  • @OleksandrKulkov I was wondering where you got this from, do you have a reference. Thanks. – Dylan Dijk Dec 03 '23 at 19:29
  • 1
    E.g. here. I mean, it's just $(A \otimes A) (e_i \otimes e_j) = (A e_i) \otimes (A e_j) = (\lambda_i e_i) \otimes (\lambda_j e_j) = \lambda_i \lambda_j (e_i \otimes e_j)$. – Oleksandr Kulkov Dec 03 '23 at 19:38
  • I wonder if we know $E^{-1}$ in the eigendecomposition $A = EDE^{-1}$ then is there a good way of using this to compute $P^{-1}$, where $A \otimes A = P \Lambda P^{-1}$ using what you have said? – Dylan Dijk Dec 08 '23 at 13:52

0 Answers0