Let $N \in \mathbb{N}$, and $A \in \mathbb{C}^{N\times N}$ be a complex matrix of size $N\times N$. I want to prove that if $$A = \begin{pmatrix} A_{r_{1}\times s_{1}} && && && \\ && A_{r_{2} \times s_{2}} && && \\ && && \ddots && && \\ && && && A_{r_{m} \times s_{m}} \end{pmatrix}$$ where $A_{r_{i} \times s_{i}}$ is a complex matrix of size $r_{i} \times s_{i}$ with $r_{1} + \cdots + r_{m} = N = s_{1} + \cdots s_{m}$, then if $r_{i} \not= s_{i}$ for some $i$ implies that $\det(A) = 0$.
I have an intuitive idea of how to prove it; I believe that the rows become linearly dependent if any $r_{i}$ is different from $s_ {i}$, but I'm struggling to formalize it.
Suppose that $r_i \neq s_i$ for some $i$. We can assume without loss of generality (i.e. up to a suitable permutation similarity) that this holds for $i = 1$.
If $r_1 < s_1$, verify that the first $s_1$ columns of $A$ must be linearly dependent, which means that $A$ has determinant zero. Similarly, if $r_1 > s_1$, then the first $r_1$ rows of $A$ are linearly dependent, which means that $A$ has determinant zero.
So, $A$ has determinant zero, which was what we wanted.
If you want to make an argument without appealing to permutation similarity, note that if $r_i < s_i$, then the columns of $A$ that contain the block-entry $A_{r_i \times s_i}$ will be linearly dependent. However, this is a bit less convenient to specify formally.
Also, because $r_1 + \cdots + r_m = s_1 + \cdots + s_m$, there exists an $i$ for which $r_i < s_i$ if and only if there exists a $j \neq i$ for which $r_j > s_j$.
An alternative proof: note that $A^*A$ (with $A^*$ denoting a conjugate-transpose) is block diagonal with square blocks on the diagonal, with diagonal blocks $A_{r_i\times s_i}^*A_{r_i\times s_i}$ of shapes $s_i \times s_i$. As discussed above, if $r_i \neq s_i$ holds for at least one $i$, then there is a $j$ such that $r_j < s_j$, which implies that the rank of $A_{r_j \times s_j}^*A_{r_j \times s_j}$ is less than $s_i$, which implies that its determinant is zero. Because the blocks are square, the determinant of $A^*A$ is the product of the determinant of the blocks, which means that $\det(A^*A) = |\det(A)|^2$ must be zero.
The conclusion follows.