Determinant of a special block matrix

Question

I have a special $NM \times NM$ matrix of the form

\begin{align*} S = \left[ \begin{array}{cccc} V + \lambda I & V & \cdots & V \\ V & V + \lambda I & \cdots & V \\ \cdots & \cdots & \cdots & \cdots \\ V & V & \cdots & V + \lambda I \end{array} \right] \end{align*} where $V$ is a symmetric $N\times N$ matrix and $I$ is an identity matrix of size $N$.

I want to know if there is any way to express $\text{det}(S)$ as a function of $\text{det}(V)$ and $\lambda$.

Answer 1

Consider the $M\times M$ block matrix

$$ \begin{pmatrix} v+\mu & v & v& \cdots & v \\ v & v+\mu & v& \cdots & v \\ \vdots&&&&\\ v && \cdots &v& v+\mu \end{pmatrix} = v 11^T + \mu I$$

By the matrix deteminant lemma, $$\det(\mu I + v11^T )=(1+v1^T(\mu I)^{-1}1)\det(\mu I) = \mu^M(1+Mv\mu^{-1}) = \mu^M +vM\mu^{M-1}$$

Now since in your $NM\times NM$ block matrix all blocks commute pairwise, the determinant of $S$ can be obtained by taking the determinant of the above expression after substituting $v$ with $V$ and $\mu$ with $\lambda I$. (cf. https://en.wikipedia.org/wiki/Determinant#Block_matrices).

$$ \det(S) = \det((\lambda I)^M + VM (\lambda I)^{M-1}) =\det( \lambda^MI + VM\lambda^{M-1}) = \lambda^{NM}\det(I + \tfrac{M}{\lambda} V) $$

The last expression, afaik, cannot be further simplified. However, asymptotically we have $\det(I+\epsilon V) = I +trace(V)\epsilon$

Answer 2

Let

$$R:=\begin{align*} \left[ \begin{array}{cccc} V & V & \cdots & V \\ V & V & \cdots & V \\ \cdots & \cdots & \cdots & \cdots \\ V & V & \cdots & V \end{array} \right] \end{align*}=\Omega \otimes V$$

where $\Omega$ is the $M \times M$ matrix with all its entries equal to $1$ and $\otimes$ denotes Kronecker product (https://en.wikipedia.org/wiki/Kronecker_product).

Thus, the eigenvalues of $R$ are all possible products between eigenvalues of $\Omega$ (which are $(M,0...0)$) and eigenvalues of $V$ (call them $(\lambda_1,...\lambda_N)$):

$$\text{spectrum}(R)=\left(\underbrace{M\lambda_1,...M \lambda_N,}_{M \times \text{spectrum(V)}} \underbrace{0,0...0}_{(M-1)\times N \ \text{times}}\right).$$

Therefore, as $S:=R+\lambda I_{MN}$ :

$$\text{spectrum}(S)=\left((M\lambda_1+\lambda),... (M \lambda_N+\lambda),\underbrace{\lambda,\lambda,...\lambda}_{(M-1)\times N \ \text{times}}\right).$$

The determinant of a matrix being equal to the product of its eigenvalues :

$$\det(S)=\left(\Pi_{k=1...N}(M\lambda_k+\lambda)\right) \lambda^{(M-1)\times N}$$

Otherwise said :

$$\det(S)=\det(M V+\lambda I_N) \lambda^{(M-1)\times N}$$

but no direct connection with $\det(V)$.