Fast way to compute eigenvalue of block identity matrices

Question

Let $M \in \mathbb{R^{2n \times 2n}}$ have the form \begin{equation} M = \begin{pmatrix} \mathbb{I}_{n} & - a \mathbb{I}_{n} \\ b \mathbb{I}_{n} & \mathbb{I}_{n} \end{pmatrix} , \end{equation} for $a,b >0$ be such that $0 < ab < 1$.

My aim is to find the eigenvalues of $M^{T}M$.

My approach is first to compute $M^{T}M$, which is \begin{equation} M^{T} M = \begin{pmatrix} \left( 1 + a^{2} \right) \mathbb{I}_{n} & \left( b - a \right) \mathbb{I}_{n} \\ \left( b - a \right) \mathbb{I}_{n} & \left( 1 + b^{2} \right) \mathbb{I}_{n} \end{pmatrix} , \end{equation} then using the Schur Complement to compute $\det \left( M^{T}M - \mathbb{I}_{2n} \right)$ to deduce the eigenvalues.

However this seems to be problematic, especially when we have a very specific form of $M^{T}M$ which includes only the identity matrices. I think that there should be a faster and much easier way to compute the eigenvalue but I don't know how.

Answer 1

First, let's address the trivial case of $a=b$. In that case, the matrix is diagonal and its eigenvalues are $(1+a^2)$ repeated $2n$ times.

Now, let's assume $a \neq b$. Let $w=(u,v)^T$ be an eigenvector with eigenvalue $\lambda$, where both $u$ and $v$ are of dimension $n$. In other words, $$ M^TM w =\lambda w$$ Translating into 2 equations of dimension $n$ each, means $$ \begin{cases} (1+a^2)u+(b-a)v = \lambda u\\ (1+b^2)v+(b-a)u = \lambda v \end{cases}\tag{1}$$ On the surface, this has 3 unknowns ($u$, $v$ and $\lambda$).

Clearly, $u$ and $v$ are colinear. You can also check that $u\neq 0$ and $v\neq 0$ (otherwise we would have both of them $0$ and $w=0$). So we can write $v=\mu u$, and system $(1)$ now has 2 unknowns (not 3), namely $\mu$ and $\lambda$. Indeed, the system can be rewritten $$ \begin{cases} (1+a^2)+(b-a)\mu = \lambda \\ (1+b^2)\mu+(b-a) = \lambda \mu \end{cases}\tag{$1^\prime$}$$ Solving this system yields two pairs of solutions $(\lambda, \mu)$ $$\begin{cases} \lambda=1+\frac{a^2+b^2\pm(b-a)\sqrt{(b+a)^2+4}}{2}\\ \mu=\frac{a+b\pm\sqrt{(b+a)^2+4}}{2} \end{cases}$$ Which gives you the eigenvalues as $$\lambda=1+\frac{a^2+b^2\pm(b-a)\sqrt{(b+a)^2+4}}{2}$$ And each eigenvalue has multiplicity $n$ because you can take an arbitrary vector $u\in\mathbb R^n$, compute $\mu$ with the formula above and deduce $v$.

Answer 2

You can write $M^T M$ as a Kronecker product $A \otimes I_n$ where $A = \pmatrix{1+a^2 & b-a\cr b-a & 1+b^2\cr}$, therefore the eigenvalues of $M^T M$ are the eigenvalues of $A$ times the eigenvalues of $I_n$, i.e. the eigenvalues of $A$ repeated $n$ times.