In spatial statistics, I am trying to deal with the following topic:

I have a real-valued, symmetric, full-rank matrix $\textbf{A}$, say $N \times N$. It's a connectivity matrix, i.e. $a_{ij}$ values are either 1 or 0. Eigenvectors of $\textbf{A}$ are orthogonal (and real). Now, for an empirical analysis, I need to generalize my $\textbf{A}$ matrix into a block-diagonal matrix such as $\mathcal{A}= (I_T \otimes \textbf{A})$ - also real-valued, symmetric and full rank. I assume it also has orthogonal eigenvectors (of a corresponding lenght $NT$.

My question is, are the eigenvectors of $\textbf{A}$ and $\mathcal{A}$ related? E.g. through some basic transformation or "rule"?

I searched this web for questions on eigenvectors in block diagonal matrices, but did not find an answer for this one... Thank you.

  • $\begingroup$ Could you clarify what you mean by $\mathcal A = (I_T \otimes \mathbf A)$? Is $\otimes$ the Kronecker product? Is $I_T$ an identity matrix? $\endgroup$ – Omnomnomnom Nov 19 '18 at 23:24
  • $\begingroup$ Sorry, yes, it's a Kroenecker product and identity matrix. $\endgroup$ – Tomas Nov 19 '18 at 23:25

In general: if $P$ and $Q$ are diagonalizable, then whenever $Px = \lambda x$ and $Qy = \mu y$, we will have $(P \otimes Q)(x \otimes y) = \mu \lambda (x \otimes y)$. Moreover, we can form an eigenbasis out of the Kronecker products $x \otimes y$.

For your particular example: $I_T \otimes A$ will have the same eigenvalues as $A$, but each will have its multiplicity multiplied by $T$. For any eigenvector $v$ of $A$, the vectors $e_1 \otimes v, \dots, e_n \otimes v$ will be eigenvectors of $I_T \otimes A$ (here $e_1,\dots,e_n$ is the canonical basis; so $e_1 = (1,0,\dots,0)^T$).

Another general point: the eigenvalues of any block-diagonal matrix is simply the union (with multiplicity) of the eigenvalues of each individual block.


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