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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 $\sqrt{p\left(\frac{qb}{\lambda - p}\right)^2 + q}$.

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  • $\begingroup$ The properties of the Kronecker product give you the answer in the case that $p = q$. $\endgroup$ – Omnomnomnom Nov 11 '19 at 16:48
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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 $p = q$.

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  • $\begingroup$ Thank you! I've calculated the eigenvalues too - I'll add to the original question. $\endgroup$ – Alex Modell Nov 12 '19 at 14:01
  • $\begingroup$ Is it true that there exists an orthogonal matrix $P$ such that $P1_p = \sqrt{p}e_p$? $\endgroup$ – Alex Modell Nov 13 '19 at 12:13
  • $\begingroup$ @AlexModell yes. For instance, we can use a suitable Householder matrix. There will be an orthogonal $P$ satisfying $Pu = v$ whenever $\|u\| = \|v\|$. $\endgroup$ – Omnomnomnom Nov 13 '19 at 12:48

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