4.3: Exercise 21
Prove that if \(M \in M_{n \times n}(\mathbf{F})\) can be written in the form $$ \begin{align*} M = \begin{pmatrix} A & B \\ O & C \end{pmatrix} \end{align*} $$ where \(A\) and \(C\) are square square matrices. Prove that \(\det(M) = \det(A)\det(C)\)


Proof:

Suppose \(A\) is of size \(k \times k\) and \(C\) is of size \(m \times m\), then we know that \(k+m=n\) and \(O\) is of size \(m \times k\). We have two cases. If \(A\) is not invertible, then we know that \(\det(A) = 0\). Moreover, the columns of \(A\) are linearly dependent and the rank of \(A\) is less than \(k\). But \(M\) contains \(k+m\) columns and since the first \(k\) columns are linearly dependent, then the rank of \(M\) is less than \(k+m\). Therefore \(\det(M) = 0 = \det(A)\det(C)\) and we are done.

Suppose that \(A\) is invertible. Then define

$$ \begin{align*} P &= \begin{pmatrix} A & O' \\ O & I_m \end{pmatrix}, Q = \begin{pmatrix} I_k & A^{-1}B \\ O & C \end{pmatrix} \end{align*} $$

Where \(O'\) is of size \(k \times m\). Note here that \(PQ = M\). We claim that \(\det(P) = \det(A)\) and \(\det(Q)=C\). Once we show this claim is true, then we can conclude that \(\det(M) = \det(P)\det(Q) = \det(A)\det(Q)\).

To show that \(\det(P)=\det(O)\), We’ll prove this by induction on \(m\).
Base Case \(m=1\): \(I_m\) is of size \(1 \times 1\). \(O\) is of size \(1 \times k\). The last row of \(P\) therefore contains a row of zeros followed by 1. Computing the determinant of \(P\) by cofactor expansion along this last row yields exactly \(\det(A)\) as required.

Inductive Case: Suppose this is true for \(m-1\). We will show that this is true for \(m\). In this case \(O\) is of size \(m \times k\) and \(I_m\) is of size \(m \times m\). The last row of this matrix is also a sequence of zeros followed by a 1 by the definition of \(O\) and \(I_m\). So compute the determinant by cofactor expansion along the last row to see that \(\det(P) = 1 \det(P')\). where \(P'\) is the matrix \(P\) but with the last column and the last row. We can now apply the inductive hypothesis to conclude that \(\det(P') = \det(A)\) and therefore \(\det(P) = 1\det(A) = \det(A)\) as we wanted to show.

We can use the same inductive proof to prove that \(\det(Q) = C\). Once we do so, we are done.

\(\ \blacksquare\)

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