# Eigenvalues and eigenvectors of block constant matrix

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

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

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 $$p = q$$.
• Is it true that there exists an orthogonal matrix $P$ such that $P1_p = \sqrt{p}e_p$? – Alex Modell Nov 13 '19 at 12:13
• @AlexModell yes. For instance, we can use a suitable Householder matrix. There will be an orthogonal $P$ satisfying $Pu = v$ whenever $\|u\| = \|v\|$. – Omnomnomnom Nov 13 '19 at 12:48