Let us consider a Hermitian block matrix \begin{align} A = \begin{bmatrix} A_{1,1} & \cdots & A_{1,n}\\ \vdots & \ddots & \vdots\\ A_{n,1} & \cdots & A_{n,n} \end{bmatrix} \end{align} with $A_{i,j} = a_{i,j} J_{d_i,d_j}$ where $a_{i,j}\in \mathbb{C}$ and $J_{d_i,d_j}$ is the all-ones matrix of size $d_i\times d_j$. Each block of the matrix is essentially a scalar in that $A$ shares the eigenvalues with \begin{align} A^{\sf small} = \begin{bmatrix} d_1 a_{1,1} & \cdots & \sqrt{d_1d_n}a_{1,n}\\ \vdots & \ddots & \vdots\\ \sqrt{d_nd_1} a_{n,1} & \cdots & d_n a_{n,n} \end{bmatrix}. \end{align} It can be easily derived from the observation that \begin{align} A = DA^{\sf small}D^\top \end{align} where $D = \operatorname{blkdiag}\Big(\frac{J_{d_1,1}}{\sqrt{d_1}},\ldots,\frac{J_{d_n,1}}{\sqrt{d_n}}\Big)$ is a block diagonal matrix.
Plugging it into the spectral decomposition $A^{\sf small} =\sum_n \lambda_n u_n u_n^*$, we obtain \begin{align} A = \sum_n \lambda_n (Du_n) (Du_n)^*. \end{align} Since $\{Du_n\}$ are orthonormal from $D^TD = I$, it is an eigendecomposition of $A$, which concludes the proof.
I need this small result as a lemma but it is rather trivial and I do not want to reinvent the wheel. Could you recommend a reference?