I am interested to know if there is a spectral relationship between a matrix $A = \begin{bmatrix}a & b \\ b & a\end{bmatrix} \in \mathbb{R}^{2\times 2}$ and the block constant matrix
$$
A' = \left[
\begin{matrix}
a & \cdots & a & b & \cdots & b \\
\vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\
a & \cdots & a & b & \cdots & b \\
b & \cdots & b & a & \cdots & a \\
\vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\
b & \cdots & b & a & \cdots & a \\
\end{matrix}
\right] \in \mathbb{R}^{(p+q) \times (p+q)}
$$
where the top left block is $p\times p$ and the bottom right block is $q\times q$.
We can also explicitly calculate the eigenvalues of $A$ to be $\lambda_1 = a+b, \lambda_2 = a – b$ and then eigenvectors to be $v_1 = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix}, v_2 = \frac{1}{\sqrt{2}} \begin{bmatrix} -1 \\ 1 \end{bmatrix}$.
Clearly $A'$ is rank 2 so will have at most 2 non-zero eigenvectors. Is there any theorem which describes the eigenvalues and vectors in terms of $p$ and $q$?
Thank you Omnomnomnom for the eigenvalues. I've calculated the eigenvectors.
Let $v = (\underbrace{v_1,\dots v_1}_{p},\underbrace{v_2,\dots,v_2}_{q})$ be an eigenvalue then it must satisfy $A'v= \lambda v$ so
- $pav_1+qbv_2 = \lambda v_1$
- $pbv_1 + qav_2 = \lambda v_2$
Thus letting $v_2 = 1$ we have $v_1 = \frac{qb}{\lambda -pa}$. This is easily normalised by dividing by $\sqrt{p\left(\frac{qb}{\lambda – p}\right)^2 + q}$.
Best Answer
Let $1_n \in \Bbb R^n$ denote the column vector $(1,\dots,1)$. We can write $A'$ in the form $$ A' = \pmatrix{a1_p1_p^T & b 1_p1_q^T\\ b1_q1_p^T & a 1_q 1_q^T} $$ Now, select orthogonal matrices $P$ of size $p \times p$ and $Q$ of size $q \times q$ such that $P1_p = \sqrt p e_p$ and $Q 1_q = \sqrt q e_q$, where $e_n \in \Bbb R^n$ is the column vector $(1,0,\dots,0)$. Take $M$ to be the orthogonal matrix $\operatorname{diag}(P,Q)$. With block-matrix multiplication, we compute $$ MA'M^T = \pmatrix{P&0\\0&Q} \pmatrix{a1_p1_p^T & b 1_p1_q^T\\ b1_q1_p^T & a 1_q 1_q^T} \pmatrix{P^T&0\\0&Q^T}\\ = \pmatrix{a(P1_p)(P1_p)^T & b (P1_p)(Q1_q)^T\\ b(Q1_q)(P1_p)^T & a (Q1_q) (Q1_q)^T}\\ = \pmatrix{ap\,e_p e_p^T & b\sqrt{pq}\, e_pe_q^T\\ b \sqrt{pq}\, e_q e_p^T & bq\, e_qe_q^T}. $$ There exists a permutation matrix $R$ such that $R[MA'M^T]R^T = \operatorname{diag}(B,0)$, where $$ B = \pmatrix{ap & b\sqrt{pq}\\ b \sqrt{pq} & a q}. $$ So, the non-zero eigenvalues $A'$ are the eigenvalues of the matrix $B$ above. We can find the eigenvalues of $A'$ using the eigenvalues of $B$ as well.
These eigenvalues turn out to be $$ \lambda = \frac 12 \left(a (p + q) \pm \sqrt{[a(p-q)]^2 + 4 b^2 p q}\right) $$ which simplifies to $p(a\pm b)$ in the case that $p = q$.