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# 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. $\det\begin{bmatrix} A & O \\ O & D \end{bmatrix} = \det(A)\det(D)$.
2. $\det\begin{bmatrix} A & B \\ O & D \end{bmatrix} = \det(A)\det(D)$.
3. $\det\begin{bmatrix} A & B \\ C & D \end{bmatrix} = \det(A)\det(D - CA^{-1}B)$ if $A$ is invertible.

## Thought

Let us focus on a simple case first. What is the determinant of $\begin{bmatrix} A & O \\ O & I_m \end{bmatrix}$? If $A$ can be written as a product of elementary matrices $A = F_1\cdots F_s$, then

$$
\begin{bmatrix} A & O \\ O & I_m \end{bmatrix} =
\begin{bmatrix} F_1 & O \\ O & I_m \end{bmatrix} \cdots
\begin{bmatrix} F_s & O \\ O & I_m \end{bmatrix}.
$$

If we could show that $\det\begin{bmatrix} E & O \\ O & I_m \end{bmatrix} = \det(E)$, 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

$$
\begin{bmatrix} A & B \\ O & D \end{bmatrix}
\xrightarrow{\rho_1:-\frac{B}{D}\rho_2}
\begin{bmatrix} A & O \\ O & D \end{bmatrix},
$$

but we have to be careful that the second row is multiplied by $D^{-1}$ and then by $-B$. Note that $BD^{-1}$ and $D^{-1}B$ could be different. Thus, we get the equality

$$
\begin{bmatrix} I_n & -BD^{-1} \\ O & I_m \end{bmatrix}
\begin{bmatrix} A & B \\ O & D \end{bmatrix} =
\begin{bmatrix} A & O \\ O & D \end{bmatrix}.
$$

Similarly, we can envision that

$$
\begin{bmatrix} A & B \\ C & D \end{bmatrix}
\xrightarrow{\rho_2:-\frac{C}{A}\rho_1}
\begin{bmatrix} A & B \\ O & D - CA^{-1}B \end{bmatrix}.
$$

Following the same philosophy, we have

$$
\begin{bmatrix} I_n & O \\ -CA^{-1} & I_m \end{bmatrix}
\begin{bmatrix} A & B \\ C & D \end{bmatrix} =
\begin{bmatrix} A & B \\ O & D - CA^{-1}B \end{bmatrix}.
$$

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

$$
\hat{E} = \begin{bmatrix} E & O \\ O & I_m \end{bmatrix}
$$

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,

$$
\det\begin{bmatrix} I_n & O \\ O & E \end{bmatrix} = \det(E)
$$

as well.

Since both $A$ and $B$ are invertible, we may write

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

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

$$
\begin{bmatrix} A & O \\ O & D \end{bmatrix} =
\begin{bmatrix} F_1 & O \\ O & I_m \end{bmatrix}
\cdots
\begin{bmatrix} F_a & O \\ O & I_m \end{bmatrix}
\begin{bmatrix} I_n & O \\ O & G_1 \end{bmatrix}
\cdots
\begin{bmatrix} I_n & O \\ O & G_d \end{bmatrix}
$$

and it is a product of elementary matrices. Thus,

$$
\begin{aligned}
\det\begin{bmatrix} A & O \\ O & D \end{bmatrix} &=
\det\begin{bmatrix} F_1 & O \\ O & I_m \end{bmatrix}
\cdots
\det\begin{bmatrix} F_a & O \\ O & I_m \end{bmatrix}
\det\begin{bmatrix} I_n & O \\ O & G_1 \end{bmatrix}
\cdots
\det\begin{bmatrix} I_n & O \\ O & G_d \end{bmatrix} \\
&= \det(F_1)\cdots\det(F_a)\det(G_1)\cdots\det(G_d) = \det(A)\det(D).
\end{aligned}
$$

**Case 2**: $A$ or $D$ is singular.

If $A$ or $D$ is singular, then $\begin{bmatrix} A & O \\ O & D \end{bmatrix}$ 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

$$
\begin{bmatrix} I_n & -BD^{-1} \\ O & I_m \end{bmatrix}
\begin{bmatrix} A & B \\ O & D \end{bmatrix} =
\begin{bmatrix} A & O \\ O & D \end{bmatrix}.
$$

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 $\begin{bmatrix} A & B \\ O & D \end{bmatrix}$ is singular. So the equality holds as well.

For Problem 3, it follows immediately from the decomposition

$$
\begin{bmatrix} I_n & O \\ -CA^{-1} & I_m \end{bmatrix}
\begin{bmatrix} A & B \\ C & D \end{bmatrix} =
\begin{bmatrix} A & B \\ O & D - CA^{-1}B \end{bmatrix}.
$$

and the fact that the determinant is multiplicative.

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