# Eigenvectors in a block diagonal matrix

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.

• 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? – Omnomnomnom Nov 19 '18 at 23:24
• Sorry, yes, it's a Kroenecker product and identity matrix. – 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 $$e_1 = (1,0,\dots,0)^T$$).