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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.

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    $\begingroup$ 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. $\endgroup$ – angryavian Aug 3 '17 at 21:54
  • $\begingroup$ To find the determinant of $A$, we could reduce it to its row echelon form $R$. Since R is a triangular matrix, its determinant is simply the product of the entries in the main diagonal. Then det A = k(det R) for some suitable for $k$. The row echelon form of M is the same as M with A replaced by R and B replaced by some other matrix. Then M is a triangular matrix and the product of the entries in the main diagonal remain the same since the entries in the diagonal of the identity matrix are all 1s. $\endgroup$ – Artus Aug 4 '17 at 1:12
  • $\begingroup$ Since adding a multiple of one row to another doesn't change the value of the determinant we can use the bottom half of the matrix to reduce $B$ to the zero matrix without altering the determinant. As you noted it's easy to prove from there. $\endgroup$ – CyclotomicField Aug 4 '17 at 2:02
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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.

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