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 $\lambda$.
Best Answer
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 $\det(I+\epsilon V) = I +trace(V)\epsilon$