$$

Block matrix

Problem

Let

$A$ be an

$n\times n$ matrix,

$D$ an

$m\times m$ matrix, and

$B,C$ matrices of appropriate sizes. Show the following.

1. $det\left[\begin{array}{cc}A& O\\ O& D\end{array}\right]=det(A)det(D)$.
2. $det\left[\begin{array}{cc}A& B\\ O& D\end{array}\right]=det(A)det(D)$.
3. $det\left[\begin{array}{cc}A& B\\ C& D\end{array}\right]=det(A)det(D-C{A}^{-1}B)$ if
   $A$ is invertible.

Thought

Let us focus on a simple case first. What is the determinant of

$\left[\begin{array}{cc}A& O\\ O& I_m\end{array}\right]$? If

$A$ can be written as a product of elementary matrices

$A=F_1\cdots F_s$, then

$$\left[\begin{array}{cc}A& O\\ O& I_m\end{array}\right]=\left[\begin{array}{cc}{F}_1& O\\ O& I_m\end{array}\right]\cdots \left[\begin{array}{cc}{F}_s& O\\ O& I_m\end{array}\right].$$

If we could show that

$det\left[\begin{array}{cc}E& O\\ O& I_m\end{array}\right]=det(E)$, then we are almost done. Note that if we want to use the Laplacian expansion here, then we have to show it first, which might not be a good idea.

The second key here is to do the row operations carefully and record their elementary matrices. For examples, we can envision that

$$\left[\begin{array}{cc}A& B\\ O& D\end{array}\right]\stackrel{{\rho }_1:-\frac{B}{D}{\rho }_2}{\to }\left[\begin{array}{cc}A& O\\ O& D\end{array}\right],$$

but we have to be careful that the second row is multiplied by

$D^{-1}$ and then by

$-B$. Note that

$B{D}^{-1}$ and

$D^{-1}B$ could be different. Thus, we get the equality

$$\left[\begin{array}{cc}{I}_n& -B{D}^{-1}\\ O& I_m\end{array}\right]\left[\begin{array}{cc}A& B\\ O& D\end{array}\right]=\left[\begin{array}{cc}A& O\\ O& D\end{array}\right].$$

Similarly, we can envision that

$$\left[\begin{array}{cc}A& B\\ C& D\end{array}\right]\stackrel{{\rho }_2:-\frac{C}{A}{\rho }_1}{\to }\left[\begin{array}{cc}A& B\\ O& D-C{A}^{-1}B\end{array}\right].$$

Following the same philosophy, we have

$$\left[\begin{array}{cc}{I}_n& O\\ -C{A}^{-1}& I_m\end{array}\right]\left[\begin{array}{cc}A& B\\ C& D\end{array}\right]=\left[\begin{array}{cc}A& B\\ O& D-C{A}^{-1}B\end{array}\right].$$

These decompositions help us to derive the desired results.

Sample answer

For Problem 1, we consider two cases.

Case 1:

$A$ and

$D$ are invertible.

We first analyse the

$(n+m)\times (n+m)$ matrix

$$\hat{E}=\left[\begin{array}{cc}E& O\\ O& I_m\end{array}\right]$$

when

$E$ is an

$n\times n$ elementary matrix. If

$E$ is the elementary matrix of the row operation

$P$ on

$n\times n$ matrices, then

$\hat{E}$ is the elementary matrix of the same row operation

$P$ on

$(n+m)\times (n+m)$ matrices. Therefore,

$det(E)=det(\hat{E})$. For similar reason,

$$det\left[\begin{array}{cc}{I}_n& O\\ O& E\end{array}\right]=det(E)$$

as well.

Since both

$A$ and

$B$ are invertible, we may write

$$\begin{array}{rl}A& =F_1\cdots F_a\text{ and }\\ D& =G_1\cdots G_d\end{array}$$

as products of elementary matrices. Now it is easy to see that

$$\left[\begin{array}{cc}A& O\\ O& D\end{array}\right]=\left[\begin{array}{cc}{F}_1& O\\ O& I_m\end{array}\right]\cdots \left[\begin{array}{cc}{F}_a& O\\ O& I_m\end{array}\right]\left[\begin{array}{cc}{I}_n& O\\ O& G_1\end{array}\right]\cdots \left[\begin{array}{cc}{I}_n& O\\ O& G_d\end{array}\right]$$

and it is a product of elementary matrices. Thus,

$$\begin{array}{rl}det\left[\begin{array}{cc}A& O\\ O& D\end{array}\right]& =det\left[\begin{array}{cc}{F}_1& O\\ O& I_m\end{array}\right]\cdots det\left[\begin{array}{cc}{F}_a& O\\ O& I_m\end{array}\right]det\left[\begin{array}{cc}{I}_n& O\\ O& G_1\end{array}\right]\cdots det\left[\begin{array}{cc}{I}_n& O\\ O& G_d\end{array}\right]\\ & =det(F_1)\cdots det(F_a)det(G_1)\cdots det(G_d)=det(A)det(D).\end{array}$$

Case 2:

$A$ or

$D$ is singular.

If

$A$ or

$D$ is singular, then

$\left[\begin{array}{cc}A& O\\ O& D\end{array}\right]$ is singular. So the equality holds as well.

For Problem 2, we consider two cases.

Case 1:

$D$ is invertible.

If

$D$ is invertible, then

$$\left[\begin{array}{cc}{I}_n& -B{D}^{-1}\\ O& I_m\end{array}\right]\left[\begin{array}{cc}A& B\\ O& D\end{array}\right]=\left[\begin{array}{cc}A& O\\ O& D\end{array}\right].$$

The first matrix in the equality has determinant

$1$ since one may apply only row operations to it to obtain

$I_{n+m}$. Since the determinant is multiplicative, we get the desired equality.

Case 2:

$D$ is singular.

If

$A$ or

$D$ is singular, then

$\left[\begin{array}{cc}A& B\\ O& D\end{array}\right]$ is singular. So the equality holds as well.

For Problem 3, it follows immediately from the decomposition

$$\left[\begin{array}{cc}{I}_n& O\\ -C{A}^{-1}& I_m\end{array}\right]\left[\begin{array}{cc}A& B\\ C& D\end{array}\right]=\left[\begin{array}{cc}A& B\\ O& D-C{A}^{-1}B\end{array}\right].$$

and the fact that the determinant is multiplicative.

This note can be found at Course website > Learning resources.