Eigenvalues of $3 \times 3$ block matrix

Question

What are the eigenvalues of the following block matrix?

$$\begin{bmatrix} A & I_n & I_n \\ I_n & I_n & O_n \\ I_n & O_n & I_n \end{bmatrix}$$

Here, $A$ is any square matrix of order $n$ whose eigenvalues are denoted by $\lambda_1, \lambda_2, \dots, \lambda_n$, $I_n$ is an identity matrix of order $n$ and $O_n$ is a zero matrix of order $n$.

Answer 1

Call your matrix $M$. The eigenvalues of $M$ are the roots of $\det(tI_{3n}-M)$. You may evaluate $\det(tI_{3n}-B)$ in many ways:

1. By considering the Schur complement of $D=\pmatrix{(t-1)I_n&0\\ 0&(t-1)I_n}$ in $tI_{3n}-M$, we have \begin{align} \det(tI_{3n}-M) &= \det(D)\det\left(tI_n-A-\pmatrix{-I&-I}D^{-1}\pmatrix{-I\\ -I}\right)\\ &= \det((t-1)I_n)^2\det\left(tI_n-A-2(t-1)^{-1}I_n\right)\\ &= \det((t-1)I_n)\det\left((t-1)(tI_n-A)-2I_n\right)\\ &= (t-1)^n\det\left((t-1)(tI_n-A)-2I_n\right). \end{align}
2. By using simultaneous row/column reductions. Obviously, $tI_{3n}-M$ is congruent to $$ \pmatrix{ tI-A+\frac2{t-1}I_n&0&0\\ 0&(t-1)I_n&0\\ 0&0&(t-1)I_n} $$ via applications of row/column additions. As determinant is preserved under row/column additions, we get $\det(tI_{3n}-M) = (t-1)^n\det\left((t-1)(tI_n-A)-2I_n\right)$.
3. By noting that all sub-blocks in $M$ commute. View $tI_{3n}-M$ as a $3\times3$ matrix over the commutative ring $R$ of all polynomials in $A$. We can then calculate the determinant of this $3\times3$ matrix as if each sub-block is a scalar (actually, it is a scalar for $M_3(R)$ because it is an element of $R$): $$ {\det}_R\pmatrix{tI_n-A&-I_n&-I_n\\ -I_n&(t-1)I_n&0\\ -I_n&0&(t-1)I_n} =(t-1)^2(tI_n-A)-(t-1)I_n. $$ The result in the above, being an element of $R$, is an $n\times n$ matrix over the ground field $F$. Now, by taking the determinant of this $n\times n$ matrix, we obtain the determinant of the original $3n\times3n$ matrix (cf. theorem 1 of John Silvester, Determinants of Block Matrices) \begin{align} \det(tI_{3n}-M) &=\det\left({\det}_R(tI_{3n}-M)\right)\\ &= \det\left((t-1)^2(tI_n-A)-2(t-1)I_n\right)\\ &= (t-1)^n\det\left((t-1)(tI_n-A)-2I_n\right)\\ &= (t-1)^n\det\left(t^2I_n - t(A+I_n)+(A-2I_n)\right). \end{align}

So, we have $\det(tI_{3n}-M)=(t-1)^n\det\left(t^2I_n - t(A+I_n)+(A-2I_n)\right)$. By triangularising $A$, we see that the eigenvalues of $M$ are given by $n$ copies of $1$ as well as the roots of the quadratic polynomial $t^2-(\lambda+1)t+(\lambda-2)$ for every eigenvalue $\lambda$ of $A$.

You may verify the above results as follows. Let $\{e_1,e_2,\ldots,e_n\}$ be the standard basis of $F^n$ (where $F$ is the underlying field) and let $e=\sum_je_j$ (i.e. the all-one vector in $F^n$). If $\lambda$ is an eigenvalue of $A$ and $(t,v)$ is an eigenpair of the matrix $$ N(\lambda)=\pmatrix{\lambda&1&1\\ 1&1&0\\ 1&0&1}, $$ then $(t,v\otimes e)$ is an eigenpair of $M$. Hence the eigenvalues of $M$ are the roots of the characteristic polynomial $p_{N(\lambda)}(t)=(t-1)(t^2-(\lambda+1)t+(\lambda-2))$ for each $\lambda$. However, a more careful analysis is needed to reason about the multiplicities of the eigenvalues in this case.