Let $F$ is some field. Suppose that $M \in M_{n \times n}(F)$ can be written as $$M = \begin{bmatrix} A & B \\ 0 & I \\ \end{bmatrix},$$ where $A$ is a square matrix. Prove that $\det(M) = \det(A)$.
Is there a way to fix the following proof? First note that it is easy to prove that $\det( \begin{bmatrix} P & 0 \\ 0 & I \\ \end{bmatrix}) = \det(P)$. Let $P$ and $Q$ be those invertible matrices such that $PAQ = D$, where $D$ is a diagonal matrix having amount of $1$'s along its diagonal equal to rank of $A$. Then
$$ \det(M) = \det(\begin{bmatrix} A & B \\ 0 & I \\ \end{bmatrix})$$
$$= \det(\begin{bmatrix} P^{-1}DQ^{-1} & B \\ 0 & I \\ \end{bmatrix})$$
$$=\det(\begin{bmatrix} P^{-1} & 0 \\ 0 & I \\ \end{bmatrix} \begin{bmatrix} D & PB \\ 0 & I \\ \end{bmatrix} \begin{bmatrix} Q^{-1} & 0 \\ 0 & I \\ \end{bmatrix}$$
$$=\det(P^{-1}) \det(\begin{bmatrix} D & PB \\ 0 & I \\ \end{bmatrix}) \det(Q^{-1})$$
Since the middle determinant is clearly the determinant of an upper triangular matrix, it will be product of $D$'s diagonal elements and $I$'s elements; i.e., $\det(P^{-1}) \det(D) \det(Q^{-1}) = \det(P^{-1}DQ^{-1}) = \det(A)$, which proves the theorem.
Okay. I realize I am being rather fast and loose with the dimension of the matrices. I am trying to avoid as much induction as possible; so far as I can tell, the only thing that needs to be proven through induction is $\det( \begin{bmatrix} P & 0 \\ 0 & I \\ \end{bmatrix}) = \det(P)$ My only problem is that $PB$ may not be defined, but I am hoping that this doesn't prove t be an insuperable problem for my proof.
An easier approach is to do the Laplace expansion on the bottom row of the original matrix. This yields the result without appealing to the SVD decomposition. Regarding your question about whether $PB$ exists or not, I think the dimension-matching is fine because $A$ is a square matrix.