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# Comprehensive Guide on Block Matrices

schedule Aug 10, 2023
Last updated
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Linear Algebra
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Definition.

# Block matrices

A block matrix or a partitioned matrix is a matrix composed of sub-matrices. Block matrices can be constructed by slicing a matrix horizontally and vertically.

Example.

## Block matrix with four sub-matrices

Consider the block matrix $\boldsymbol{R}$ below:

$$\boldsymbol{R}= \begin{pmatrix} \color{red}3&\color{red}2&\color{red}5&\color{blue}1&\color{blue}2\\ \color{red}1&\color{red}6&\color{red}3&\color{blue}2&\color{blue}4\\ \color{red}1&\color{red}3&\color{red}0&\color{blue}0&\color{blue}3\\ \color{green}1&\color{green}9&\color{green}1&\color{purple}2&\color{purple}1\\ \color{green}0&\color{green}7&\color{green}2&\color{purple}9&\color{purple}0\\ \end{pmatrix}= \begin{pmatrix} \color{red}\boldsymbol{A}&\color{blue}\boldsymbol{B}\\ \color{green}\boldsymbol{C}&\color{purple}\boldsymbol{D} \end{pmatrix}$$

Here, we have sliced $\boldsymbol{R}$ horizontally and vertically thereby decomposing $\boldsymbol{R}$ into $4$ sub-matrices: $\boldsymbol{A}$, $\boldsymbol{B}$, $\boldsymbol{C}$ and $\boldsymbol{D}$.

Note that the matrix $\boldsymbol{R}$ defined below is not a block matrix because the shapes of the sub-matrices do not match up:

$$\boldsymbol{R}= \begin{pmatrix} \color{red}3&\color{red}2&\color{red}5&\color{blue}1&\color{blue}2\\ \color{red}1&\color{red}6&\color{red}3&\color{blue}2&\color{blue}4\\ \color{red}1&\color{red}3&\color{red}0&\color{blue}0&\color{blue}3\\ \color{green}1&\color{green}9&\color{purple}1&\color{purple}2&\color{purple}1\\ \color{green}0&\color{green}7&\color{purple}2&\color{purple}9&\color{purple}0\\ \end{pmatrix}$$

For a matrix to be a block matrix, the partitioning must be a straight vertical or horizontal line.

Example.

## Representing a matrix as a collection of column vectors

In linear algebra, we often represent matrices as a collection of column vectors. For instance, consider the following block matrix:

$$\boldsymbol{A}=\begin{pmatrix} 2&4&2&9\\ 3&1&8&3\\ 1&7&2&5 \end{pmatrix}= \begin{pmatrix} \vert&\vert&\vert&\vert\\ \boldsymbol{a}_1&\boldsymbol{a}_2&\boldsymbol{a}_3&\boldsymbol{a}_4\\ \vert&\vert&\vert&\vert \end{pmatrix}$$

Here, $\boldsymbol{A}$ is a block matrix because we have sliced the matrix vertically $4$ times.

Theorem.

# Transpose of a block matrix

Consider the following block matrix:

$$\boldsymbol{R}= \begin{pmatrix} \boldsymbol{A}&\boldsymbol{B}\\ \boldsymbol{C}&\boldsymbol{D} \end{pmatrix}$$

The transpose of $\boldsymbol{R}$ is:

$$\boldsymbol{R}^T= \begin{pmatrix} \boldsymbol{A}^T&\boldsymbol{C}^T\\ \boldsymbol{B}^T&\boldsymbol{D}^T \end{pmatrix}$$

Proof. Consider the following block matrix:

$$\boldsymbol{R}=\begin{pmatrix} \color{red}a_{11}&\color{red}a_{12}&\color{red}\cdots&\color{red}a_{1m}&\color{blue}b_{11}&\color{blue}b_{12}&\color{blue}\cdots&\color{blue}b_{1n}\\ \color{red}a_{21}&\color{red}a_{22}&\color{red}\cdots&\color{red}a_{2m}&\color{blue}b_{21}&\color{blue}b_{22}&\color{blue}\cdots&\color{blue}\color{blue}b_{2n}\\ \color{red}\color{red}\vdots&\color{red}\vdots&\color{red}\smash\ddots&\color{red}\vdots& \color{blue}\vdots&\color{blue}\vdots&\color{blue}\smash\ddots&\color{blue}\vdots\\ \color{red}a_{s1}&\color{red}a_{s2}&\color{red}\cdots&\color{red}a_{sm}& \color{blue}b_{s1}&\color{blue}b_{s2}&\color{blue}\cdots&\color{blue}b_{sn}\\ \color{green}c_{11}&\color{green}c_{12}&\color{green}\cdots&\color{green}c_{1m}& \color{purple}d_{11}&\color{purple}d_{12}&\color{purple}\cdots&\color{purple}d_{1n}\\ \color{green}c_{21}&\color{green}c_{22}&\color{green}\cdots&\color{green}c_{2m}& \color{purple}d_{21}&\color{purple}d_{22}&\color{purple}\cdots&\color{purple}d_{2n}\\ \color{green}\vdots&\color{green}\vdots&\color{green}\smash\ddots&\color{green}\vdots &\color{purple}\vdots&\color{purple}\vdots&\color{purple}\smash\ddots&\color{purple}\vdots\\ \color{green}c_{t1}&\color{green}c_{t2}&\color{green}\cdots&\color{green}c_{tm}& \color{purple}d_{t1}&\color{purple}d_{t2}&\color{purple}\cdots&\color{purple}d_{tn}\\ \end{pmatrix}$$

We now take the transpose:

\begin{align*} \boldsymbol{R}^T&=\begin{pmatrix} \color{red}a_{11}&\color{red}a_{21}&\color{red}\cdots&\color{red}a_{s1}& \color{green}c_{11}&\color{green}c_{21}&\color{green}\cdots&\color{green}c_{t1}\\ \color{red}a_{12}&\color{red}a_{22}&\color{red}\cdots&\color{red}a_{s2}& \color{green}c_{12}&\color{green}c_{22}&\color{green}\cdots&\color{blue}\color{green}c_{t2}\\ \color{red}\color{red}\vdots&\color{red}\vdots&\color{red}\smash\ddots&\color{red}\vdots& \color{green}\vdots&\color{green}\vdots&\color{green}\smash\ddots&\color{green}\vdots\\ \color{red}a_{1m}&\color{red}a_{2m}&\color{red}\cdots&\color{red}a_{sm}& \color{green}c_{1m}&\color{green}c_{2m}&\color{green}\cdots&\color{green}c_{tm}\\ \color{blue}b_{11}&\color{blue}b_{21}&\color{blue}\cdots&\color{blue}b_{s1}& \color{purple}d_{11}&\color{purple}d_{21}&\color{purple}\cdots&\color{purple}d_{t1}\\ \color{blue}b_{12}&\color{blue}b_{22}&\color{blue}\cdots&\color{blue}b_{s2}& \color{purple}d_{12}&\color{purple}d_{22}&\color{purple}\cdots&\color{purple}d_{t2}\\ \color{blue}\vdots&\color{blue}\vdots&\color{blue}\smash\ddots&\color{blue}\vdots &\color{purple}\vdots&\color{purple}\vdots&\color{purple}\smash\ddots&\color{purple}\vdots\\ \color{blue}b_{1n}&\color{blue}b_{2n}&\color{blue}\cdots&\color{blue}b_{sn}& \color{purple}d_{1n}&\color{purple}d_{2n}&\color{purple}\cdots&\color{purple}d_{tn}\\ \end{pmatrix}\\ &=\begin{pmatrix} \color{red}\boldsymbol{A}^T&\color{green}\boldsymbol{C}^T\\ \color{blue}\boldsymbol{B}^T&\color{purple}\boldsymbol{D}^T \end{pmatrix} \end{align*}

This completes the proof.

Theorem.

# Product of block matrices (1)

Suppose we have $m\times{n}$ matrix $\boldsymbol{A}$ and $n\times{m}$ matrix $\boldsymbol{B}$ where $\boldsymbol{B}$ is a block matrix composed of sub-matrices $\boldsymbol{B}_1$ and $\boldsymbol{B}_2$ like so:

$$\boldsymbol{B}= \begin{pmatrix} \boldsymbol{B}_1\\\boldsymbol{B}_2 \end{pmatrix}$$

The matrix product $\boldsymbol{BA}$ can be expressed as:

$$\boldsymbol{BA}= \begin{pmatrix} \boldsymbol{B}_1\boldsymbol{A}\\ \boldsymbol{B}_2\boldsymbol{A} \end{pmatrix}$$

Proof. Let $m\times{n}$ matrix $\boldsymbol{A}$ be:

$$\boldsymbol{A}= \begin{pmatrix} \color{blue}\vert&\color{blue}\vert&\color{blue}\cdots&\color{blue}\vert\\ \color{blue}\boldsymbol{a}_1&\color{blue}\boldsymbol{a}_2& \color{blue}\cdots&\color{blue}\boldsymbol{a}_n\\ \color{blue}\vert&\color{blue}\vert&\color{blue}\cdots&\color{blue}\vert\\ \end{pmatrix}$$

Let $n\times{m}$ matrix $\boldsymbol{B}$ be:

$$\boldsymbol{B}= \begin{pmatrix} \color{purple}-&\color{purple}\boldsymbol{b}_1&\color{purple}-\\ \color{purple}-&\color{purple}\boldsymbol{b}_2&\color{purple}-\\ \color{purple}\vdots&\color{purple}\vdots&\color{purple}\vdots\\ \color{purple}-&\color{purple}\boldsymbol{b}_k&\color{purple}-\\ \color{green}-&\color{green}\boldsymbol{b}_{k+1}&\color{green}-\\ \color{green}-&\color{green}\boldsymbol{b}_{k+2}&\color{green}-\\ \color{green}\vdots&\color{green}\vdots&\color{green}\vdots\\ \color{green}-&\color{green}\boldsymbol{b}_n&\color{green}-\\ \end{pmatrix}$$

Where the sub-matrices are:

$${\color{purple}\boldsymbol{B}_1}= \begin{pmatrix} \color{purple}-&\color{purple}\boldsymbol{b}_1&\color{purple}-\\ \color{purple}-&\color{purple}\boldsymbol{b}_2&\color{purple}-\\ \color{purple}\vdots&\color{purple}\vdots&\color{purple}\vdots\\ \color{purple}-&\color{purple}\boldsymbol{b}_k&\color{purple}- \end{pmatrix},\;\;\;\;\;\; {\color{green}\boldsymbol{B}_2}= \begin{pmatrix} \color{green}-&\color{green}\boldsymbol{b}_{k+1}&\color{green}-\\ \color{green}-&\color{green}\boldsymbol{b}_{k+2}&\color{green}-\\ \color{green}\vdots&\color{green}\vdots&\color{green}\vdots\\ \color{green}-&\color{green}\boldsymbol{b}_n&\color{green}- \end{pmatrix}$$

The matrix product $\boldsymbol{BA}$ is therefore:

$$\boldsymbol{BA}= \begin{pmatrix} {\color{purple}\boldsymbol{b}_1}\cdot\color{blue}\boldsymbol{a}_1&{\color{purple}\boldsymbol{b}_1}\cdot\color{blue}\boldsymbol{a}_2&\cdots&{\color{purple}\boldsymbol{b}_1}\cdot\color{blue}\boldsymbol{a}_n\\ {\color{purple}\boldsymbol{b}_2}\cdot\color{blue}\boldsymbol{a}_1&{\color{purple}\boldsymbol{b}_2}\cdot\color{blue}\boldsymbol{a}_2&\cdots&{\color{purple}\boldsymbol{b}_2}\cdot\color{blue}\boldsymbol{a}_n\\ \vdots&\vdots&\smash\ddots&\vdots\\ {\color{purple}\boldsymbol{b}_k}\cdot\color{blue}\boldsymbol{a}_1&{\color{purple}\boldsymbol{b}_k}\cdot\color{blue}\boldsymbol{a}_2&\cdots&{\color{purple}\boldsymbol{b}_k}\cdot\color{blue}\boldsymbol{a}_n\\ {\color{green}\boldsymbol{b}_{k+1}}\cdot\color{blue}\boldsymbol{a}_1&{\color{green}\boldsymbol{b}_{k+1}}\cdot\color{blue}\boldsymbol{a}_2&\cdots&{\color{green}\boldsymbol{b}_{k+1}}\cdot\color{blue}\boldsymbol{a}_n\\ {\color{green}\boldsymbol{b}_{k+2}}\cdot\color{blue}\boldsymbol{a}_1&{\color{green}\boldsymbol{b}_{k+2}}\cdot\color{blue}\boldsymbol{a}_2&\cdots&{\color{green}\boldsymbol{b}_{k+2}}\cdot\color{blue}\boldsymbol{a}_n\\ \vdots&\vdots&\smash\ddots&\vdots\\ {\color{green}\boldsymbol{b}_{n}}\cdot\color{blue}\boldsymbol{a}_1&{\color{green}\boldsymbol{b}_{n}}\cdot\color{blue}\boldsymbol{a}_2&\cdots&{\color{green}\boldsymbol{b}_{n}}\cdot\color{blue}\boldsymbol{a}_n \end{pmatrix}= \begin{pmatrix} \color{purple}\boldsymbol{B}_1\color{blue}\boldsymbol{A}\\ \color{green}\boldsymbol{B}_2\color{blue}\boldsymbol{A} \end{pmatrix}$$

This completes the proof.

Theorem.

# Product of block matrices (2)

Consider an $m\times{n}$ block matrix $\boldsymbol{A}$ and an $n\times{m}$ block matrix $\boldsymbol{B}$ below:

$$\boldsymbol{A}= \begin{pmatrix} \boldsymbol{A}_1&\boldsymbol{A}_2 \end{pmatrix},\;\;\;\;\;\;\; \boldsymbol{B}= \begin{pmatrix} \boldsymbol{B}_1\\\boldsymbol{B}_2 \end{pmatrix}$$

Where:

• $\boldsymbol{A}_1$ is an $m\times{k}$ sub-matrix.

• $\boldsymbol{A}_2$ is an $m\times(n-k)$ sub-matrix.

• $\boldsymbol{B}_1$ is an $k\times{m}$ sub-matrix

• $\boldsymbol{B}_2$ is an $(n-k)\times{m}$ sub-matrix.

The product $\boldsymbol{BA}$ can be expressed as:

$$\boldsymbol{BA} =\begin{pmatrix} \boldsymbol{B}_1\boldsymbol{A}_1& \boldsymbol{B}_1\boldsymbol{A}_2\\ \boldsymbol{B}_2\boldsymbol{A}_1& \boldsymbol{B}_2\boldsymbol{A}_2 \end{pmatrix}$$

Proof. Let $\boldsymbol{A}$ be the following $m\times{n}$ block matrix:

$$\boldsymbol{A}= \begin{pmatrix} \color{blue}\vert&\color{blue}\vert&\color{blue}\cdots&\color{blue}\vert&\color{red}\vert&\color{red}\vert&\color{red}\cdots&\color{red}\vert\\ \color{blue}\boldsymbol{a}_1&\color{blue}\boldsymbol{a}_2& \color{blue}\cdots&\color{blue}\boldsymbol{a}_k&\color{red}\boldsymbol{a}_{k+1}&\color{red}\boldsymbol{a}_{k+2} &\color{red}\cdots&\color{red}\boldsymbol{a}_{n}\\ \color{blue}\vert&\color{blue}\vert&\color{blue}\cdots&\color{blue}\vert& \color{red}\vert&\color{red}\vert&\color{red}\cdots&\color{red}\vert\\ \end{pmatrix}$$

Where:

$${\color{blue}\boldsymbol{A}_1}= \begin{pmatrix} \color{blue}\vert&\color{blue}\vert&\color{blue}\cdots&\color{blue}\vert\\ \color{blue}\boldsymbol{a}_1&\color{blue}\boldsymbol{a}_2& \color{blue}\cdots&\color{blue}\boldsymbol{a}_k\\ \color{blue}\vert&\color{blue}\vert&\color{blue}\cdots&\color{blue}\vert \\ \end{pmatrix},\;\;\;\;\;\; {\color{red}\boldsymbol{A}_2}= \begin{pmatrix} \color{red}\vert&\color{red}\vert&\color{red}\cdots&\color{red}\vert\\ \color{red}\boldsymbol{a}_{k+1}&\color{red}\boldsymbol{a}_{k+2}&\color{red}\cdots&\color{red}\boldsymbol{a}_{n}\\ \color{red}\vert&\color{red}\vert&\color{red}\cdots&\color{red}\vert\\ \end{pmatrix}$$

Similarly, let $\boldsymbol{B}$ be the following $n\times{m}$ matrix:

$$\boldsymbol{B}= \begin{pmatrix} \color{purple}-&\color{purple}\boldsymbol{b}_1&\color{purple}-\\ \color{purple}-&\color{purple}\boldsymbol{b}_2&\color{purple}-\\ \color{purple}\vdots&\color{purple}\vdots&\color{purple}\vdots\\ \color{purple}-&\color{purple}\boldsymbol{b}_k&\color{purple}-\\ \color{green}-&\color{green}\boldsymbol{b}_{k+1}&\color{green}-\\ \color{green}-&\color{green}\boldsymbol{b}_{k+2}&\color{green}-\\ \color{green}\vdots&\color{green}\vdots&\color{green}\vdots\\ \color{green}-&\color{green}\boldsymbol{b}_n&\color{green}-\\ \end{pmatrix}$$

Where:

$${\color{purple}\boldsymbol{B}_1}= \begin{pmatrix} \color{purple}-&\color{purple}\boldsymbol{b}_1&\color{purple}-\\ \color{purple}-&\color{purple}\boldsymbol{b}_2&\color{purple}-\\ \color{purple}\vdots&\color{purple}\vdots&\color{purple}\vdots\\ \color{purple}-&\color{purple}\boldsymbol{b}_k&\color{purple}- \end{pmatrix},\;\;\;\;\;\; {\color{green}\boldsymbol{B}_2}= \begin{pmatrix} \color{green}-&\color{green}\boldsymbol{b}_{k+1}&\color{green}-\\ \color{green}-&\color{green}\boldsymbol{b}_{k+2}&\color{green}-\\ \color{green}\vdots&\color{green}\vdots&\color{green}\vdots\\ \color{green}-&\color{green}\boldsymbol{b}_n&\color{green}- \end{pmatrix}$$

The product $\boldsymbol{BA}$ is:

\begin{align*} \boldsymbol{BA}&= \begin{pmatrix} \color{purple}-&\color{purple}\boldsymbol{b}_1&\color{purple}-\\\color{purple}-&\color{purple}\boldsymbol{b}_2&\color{purple}-\\ \color{purple}\vdots&\color{purple}\vdots&\color{purple}\vdots\\\color{purple}-&\color{purple}\boldsymbol{b}_k&\color{purple}-\\ \color{green}-&\color{green}\boldsymbol{b}_{k+1}&\color{green}-\\\color{green}-&\color{green}\boldsymbol{b}_{k+2}&\color{green}-\\ \color{green}\vdots&\color{green}\vdots&\color{green}\vdots\\\color{green}-&\color{green}\boldsymbol{b}_n&\color{green}-\\ \end{pmatrix}\begin{pmatrix} \color{blue}\vert&\color{blue}\vert&\color{blue}\cdots&\color{blue}\vert&\color{red}\vert&\color{red}\vert&\color{red}\cdots&\color{red}\vert\\ \color{blue}\boldsymbol{a}_1&\color{blue}\boldsymbol{a}_2&\color{blue}\cdots&\color{blue}\boldsymbol{a}_k&\color{red}\boldsymbol{a}_{k+1}&\color{red}\boldsymbol{a}_{k+2} &\color{red}\cdots&\color{red}\boldsymbol{a}_{n}\\\color{blue}\vert&\color{blue}\vert&\color{blue}\cdots&\color{blue}\vert& \color{red}\vert&\color{red}\vert&\color{red}\cdots&\color{red}\vert\\ \end{pmatrix} \\&=\begin{pmatrix} {\color{purple}\boldsymbol{b}_1}\cdot\color{blue}\boldsymbol{a}_1&{\color{purple}\boldsymbol{b}_1}\cdot\color{blue}\boldsymbol{a}_2& \cdots&{\color{purple}\boldsymbol{b}_1}\cdot\color{blue}\boldsymbol{a}_k&{\color{purple}\boldsymbol{b}_1}\cdot\color{red}\boldsymbol{a}_{k+1}& \cdots&{\color{purple}\boldsymbol{b}_1}\cdot\color{red}\boldsymbol{a}_{n}\\ {\color{purple}\boldsymbol{b}_2}\cdot\color{blue}\boldsymbol{a}_1&{\color{purple}\boldsymbol{b}_2}\cdot\color{blue}\boldsymbol{a}_2& \cdots&{\color{purple}\boldsymbol{b}_2}\cdot\color{blue}\boldsymbol{a}_k&{\color{purple}\boldsymbol{b}_2}\cdot\color{red}\boldsymbol{a}_{k+1}& \cdots&{\color{purple}\boldsymbol{b}_2}\cdot\color{red}\boldsymbol{a}_{n}\\\vdots&\vdots&\smash\ddots&\vdots&\vdots&\smash\ddots&\vdots\\ {\color{purple}\boldsymbol{b}_k}\cdot\color{blue}\boldsymbol{a}_1&{\color{purple}\boldsymbol{b}_k}\cdot\color{blue}\boldsymbol{a}_2& \cdots&{\color{purple}\boldsymbol{b}_k}\cdot\color{blue}\boldsymbol{a}_k&{\color{purple}\boldsymbol{b}_k}\cdot\color{red}\boldsymbol{a}_{k+1}& \cdots&{\color{purple}\boldsymbol{b}_k}\cdot\color{red}\boldsymbol{a}_{n}\\ {\color{green}\boldsymbol{b}_{k+1}}\cdot\color{blue}\boldsymbol{a}_1&{\color{green}\boldsymbol{b}_{k+1}}\cdot\color{blue}\boldsymbol{a}_2& \cdots&{\color{green}\boldsymbol{b}_{k+1}}\cdot\color{blue}\boldsymbol{a}_k&{\color{green}\boldsymbol{b}_{k+1}}\cdot\color{red}\boldsymbol{a}_{k+1}&\cdots&{\color{green}\boldsymbol{b}_{k+1}}\cdot\color{red}\boldsymbol{a}_{n}\\ {\color{green}\boldsymbol{b}_{k+2}}\cdot\color{blue}\boldsymbol{a}_1&{\color{green}\boldsymbol{b}_{k+2}}\cdot\color{blue}\boldsymbol{a}_2& \cdots&{\color{green}\boldsymbol{b}_{k+2}}\cdot\color{blue}\boldsymbol{a}_k&{\color{green}\boldsymbol{b}_{k+2}}\cdot\color{red}\boldsymbol{a}_{k+1}&\cdots&{\color{green}\boldsymbol{b}_{k+2}}\cdot\color{red}\boldsymbol{a}_{n}\\ \vdots&\vdots&\smash\ddots&\vdots&\vdots&\smash\ddots&\vdots\\ {\color{green}\boldsymbol{b}_n}\cdot\color{blue}\boldsymbol{a}_1&{\color{green}\boldsymbol{b}_n}\cdot\color{blue}\boldsymbol{a}_2& \cdots&{\color{green}\boldsymbol{b}_n}\cdot\color{blue}\boldsymbol{a}_k& {\color{green}\boldsymbol{b}_n}\cdot\color{red}\boldsymbol{a}_{k+1}& \cdots&{\color{green}\boldsymbol{b}_n}\cdot\color{red}\boldsymbol{a}_{n} \end{pmatrix}\\ &=\begin{pmatrix} \color{purple}\boldsymbol{B}_1\color{blue}\boldsymbol{A}_1& \color{purple}\boldsymbol{B}_1\color{red}\boldsymbol{A}_2\\ \color{green}\boldsymbol{B}_2\color{blue}\boldsymbol{A}_1& \color{green}\boldsymbol{B}_2\color{red}\boldsymbol{A}_2 \end{pmatrix} \end{align*}

This completes the proof.

Theorem.

# Product of block matrices (3)

Suppose we have two block matrices where each matrix is composed of four sub-matrices. The product of the two block matrices is:

$$\begin{pmatrix} \boldsymbol{A}&\boldsymbol{B}\\ \boldsymbol{C}&\boldsymbol{D} \end{pmatrix} \begin{pmatrix} \boldsymbol{E}&\boldsymbol{F}\\ \boldsymbol{G}&\boldsymbol{H} \end{pmatrix}= \begin{pmatrix} \boldsymbol{AE}+\boldsymbol{BG}& \boldsymbol{AF}+\boldsymbol{BH}\\ \boldsymbol{CE}+\boldsymbol{DG}& \boldsymbol{CF}+\boldsymbol{DH}\\ \end{pmatrix}$$

Note that the shape of the matrices must match for the matrix product to be valid. For instance, the number of columns of $\boldsymbol{A}$ must be equal to the number of rows of $\boldsymbol{E}$. This will be clear in the proof below.

Proof. Suppose we have the following two block matrices:

$$\begin{pmatrix} \color{purple}\boldsymbol{A}&\color{green}\boldsymbol{B}\\ \color{orange}\boldsymbol{C}&\color{yellow}\boldsymbol{D} \end{pmatrix}= \begin{pmatrix} \color{purple}-&\color{purple}\boldsymbol{a}_1&\color{purple}-&\color{green}-&\color{green}\boldsymbol{b}_1&\color{green}-\\ \color{purple}-&\color{purple}\boldsymbol{a}_2&\color{purple}-&\color{green}-&\color{green}\boldsymbol{b}_1&\color{green}-\\ \color{purple}\vdots&\color{purple}\vdots&\color{purple}\vdots&\color{green}\vdots&\color{green}\vdots&\color{green}\vdots\\ \color{purple}-&\color{purple}\boldsymbol{a}_m&\color{purple}-&\color{green}-&\color{green}\boldsymbol{b}_m&\color{green}-\\ \color{orange}-&\color{orange}\boldsymbol{c}_1&\color{orange}-&\color{yellow}-&\color{yellow}\boldsymbol{d}_1&\color{yellow}-\\ \color{orange}-&\color{orange}\boldsymbol{c}_2&\color{orange}-&\color{yellow}-&\color{yellow}\boldsymbol{d}_2&\color{yellow}-\\ \color{orange}\vdots&\color{orange}\vdots&\color{orange}\vdots&\color{yellow}\vdots&\color{yellow}\vdots&\color{yellow}\vdots\\ \color{orange}-&\color{orange}\boldsymbol{c}_p&\color{orange}-&\color{yellow}-&\color{yellow}\boldsymbol{d}_p&\color{yellow}-\\ \end{pmatrix},\;\;\;\;\; \begin{pmatrix} \color{blue}\boldsymbol{E}&\color{brown}\boldsymbol{F}\\ \color{red}\boldsymbol{G}&\boldsymbol{H} \end{pmatrix}= \begin{pmatrix} \color{blue}\vert&\color{blue}\vert&\color{blue}\cdots&\color{blue}\vert&\color{brown}\vert&\color{brown}\vert&\color{brown}\cdots&\color{brown}\vert\\ \color{blue}\boldsymbol{e}_1&\color{blue}\boldsymbol{e}_2& \color{blue}\cdots&\color{blue}\boldsymbol{e}_k&\color{brown}\boldsymbol{f}_{1}&\color{brown}\boldsymbol{f}_{2} &\color{brown}\cdots&\color{brown}\boldsymbol{f}_{r}\\ \color{blue}\vert&\color{blue}\vert&\color{blue}\cdots&\color{blue}\vert& \color{brown}\vert&\color{brown}\vert&\color{brown}\cdots&\color{brown}\vert\\ \color{red}\vert&\color{red}\vert&\color{red}\cdots&\color{red}\vert&\vert&\vert&\cdots&\vert\\ \color{red}\boldsymbol{g}_1&\color{red}\boldsymbol{g}_2& \color{red}\cdots&\color{red}\boldsymbol{g}_k&\boldsymbol{h}_{1}&\boldsymbol{h}_{2} &\cdots&\boldsymbol{h}_{r}\\ \color{red}\vert&\color{red}\vert&\color{red}\cdots&\color{red}\vert& \vert&\vert&\cdots&\vert\\ \end{pmatrix}$$

Now, let's multiply the two matrices:

$$$$\label{eq:vkEm638LwWTxTnRMSdn} \begin{pmatrix} \color{purple}\boldsymbol{A}&\color{green}\boldsymbol{B}\\ \color{orange}\boldsymbol{C}&\color{yellow}\boldsymbol{D} \end{pmatrix} \begin{pmatrix} \color{blue}\boldsymbol{E}&\color{brown}\boldsymbol{F}\\ \color{red}\boldsymbol{G}&\boldsymbol{H} \end{pmatrix}= \begin{pmatrix} \boldsymbol{P}_1&\boldsymbol{P}_2 \end{pmatrix}$$$$

Where $\boldsymbol{P}_1$ is:

\begin{align*} \boldsymbol{P}_1&= \begin{pmatrix} {\color{purple}\boldsymbol{a}_1}\cdot{\color{blue}\boldsymbol{e}_1}+{\color{green}\boldsymbol{b}_1}\cdot{\color{red}\boldsymbol{g}_1}& {\color{purple}\boldsymbol{a}_1}\cdot{\color{blue}\boldsymbol{e}_2}+{\color{green}\boldsymbol{b}_1}\cdot{\color{red}\boldsymbol{g}_2}& \cdots&{\color{purple}\boldsymbol{a}_1}\cdot{\color{blue}\boldsymbol{e}_k}+{\color{green}\boldsymbol{b}_1}\cdot{\color{red}\boldsymbol{g}_k} \\{\color{purple}\boldsymbol{a}_2}\cdot{\color{blue}\boldsymbol{e}_1}+{\color{green}\boldsymbol{b}_2}\cdot{\color{red}\boldsymbol{g}_1}& {\color{purple}\boldsymbol{a}_2}\cdot{\color{blue}\boldsymbol{e}_2}+{\color{green}\boldsymbol{b}_2}\cdot{\color{red}\boldsymbol{g}_2}& \cdots&{\color{purple}\boldsymbol{a}_2}\cdot{\color{blue}\boldsymbol{e}_k}+{\color{green}\boldsymbol{b}_2}\cdot{\color{red}\boldsymbol{g}_k}\\\vdots&\vdots&\smash\ddots&\vdots\\{\color{purple}\boldsymbol{a}_m}\cdot{\color{blue}\boldsymbol{e}_1}+{\color{green}\boldsymbol{b}_m}\cdot{\color{red}\boldsymbol{g}_1}&{\color{purple}\boldsymbol{a}_m}\cdot{\color{blue}\boldsymbol{e}_2}+{\color{green}\boldsymbol{b}_m}\cdot{\color{red}\boldsymbol{g}_2}&\cdots&{\color{purple}\boldsymbol{a}_m}\cdot{\color{blue}\boldsymbol{e}_k}+{\color{green}\boldsymbol{b}_m}\cdot{\color{red}\boldsymbol{g}_k}\\{\color{orange}\boldsymbol{c}_1}\cdot{\color{blue}\boldsymbol{e}_1}+{\color{yellow}\boldsymbol{d}_1}\cdot{\color{red}\boldsymbol{g}_1}&{\color{orange}\boldsymbol{c}_1}\cdot{\color{blue}\boldsymbol{e}_2}+{\color{yellow}\boldsymbol{d}_1}\cdot{\color{red}\boldsymbol{g}_2}& \cdots&{\color{orange}\boldsymbol{c}_1}\cdot{\color{blue}\boldsymbol{e}_k}+{\color{yellow}\boldsymbol{d}_1}\cdot{\color{red}\boldsymbol{g}_k} \\{\color{orange}\boldsymbol{c}_2}\cdot{\color{blue}\boldsymbol{e}_1}+{\color{yellow}\boldsymbol{d}_2}\cdot{\color{red}\boldsymbol{g}_1}& {\color{orange}\boldsymbol{c}_2}\cdot{\color{blue}\boldsymbol{e}_2}+{\color{yellow}\boldsymbol{d}_2}\cdot{\color{red}\boldsymbol{g}_2}& \cdots&{\color{orange}\boldsymbol{c}_2}\cdot{\color{blue}\boldsymbol{e}_k}+{\color{yellow}\boldsymbol{d}_2}\cdot{\color{red}\boldsymbol{g}_k}\\ \vdots&\vdots&\smash\ddots&\vdots\\{\color{orange}\boldsymbol{c}_p}\cdot{\color{blue}\boldsymbol{e}_1}+{\color{yellow}\boldsymbol{d}_p}\cdot{\color{red}\boldsymbol{g}_1}& {\color{orange}\boldsymbol{c}_p}\cdot{\color{blue}\boldsymbol{e}_2}+{\color{yellow}\boldsymbol{d}_p}\cdot{\color{red}\boldsymbol{g}_2}& \cdots&{\color{orange}\boldsymbol{c}_p}\cdot{\color{blue}\boldsymbol{e}_k}+{\color{yellow}\boldsymbol{d}_p}\cdot{\color{red}\boldsymbol{g}_k}\\ \end{pmatrix}\\&=\begin{pmatrix}{\color{purple}\boldsymbol{a}_1}\cdot{\color{blue}\boldsymbol{e}_1}&{\color{purple}\boldsymbol{a}_1}\cdot{\color{blue}\boldsymbol{e}_2}&\cdots&{\color{purple}\boldsymbol{a}_1}\cdot{\color{blue}\boldsymbol{e}_k}\\{\color{purple}\boldsymbol{a}_2}\cdot{\color{blue}\boldsymbol{e}_1}& {\color{purple}\boldsymbol{a}_2}\cdot{\color{blue}\boldsymbol{e}_2}&\cdots&{\color{purple}\boldsymbol{a}_2}\cdot{\color{blue}\boldsymbol{e}_k}\\\vdots&\vdots&\smash\ddots&\vdots\\{\color{purple}\boldsymbol{a}_m}\cdot{\color{blue}\boldsymbol{e}_1}& {\color{purple}\boldsymbol{a}_m}\cdot{\color{blue}\boldsymbol{e}_2}&\cdots&{\color{purple}\boldsymbol{a}_m}\cdot{\color{blue}\boldsymbol{e}_k}\\{\color{orange}\boldsymbol{c}_1}\cdot{\color{blue}\boldsymbol{e}_1}& {\color{orange}\boldsymbol{c}_1}\cdot{\color{blue}\boldsymbol{e}_2}&\cdots&{\color{orange}\boldsymbol{c}_1}\cdot{\color{blue}\boldsymbol{e}_k} \\{\color{orange}\boldsymbol{c}_2}\cdot{\color{blue}\boldsymbol{e}_1}&{\color{orange}\boldsymbol{c}_2}\cdot{\color{blue}\boldsymbol{e}_2}& \cdots&{\color{orange}\boldsymbol{c}_2}\cdot{\color{blue}\boldsymbol{e}_k}\\\vdots&\vdots&\smash\ddots&\vdots\\{\color{orange}\boldsymbol{c}_p}\cdot{\color{blue}\boldsymbol{e}_1}& {\color{orange}\boldsymbol{c}_p}\cdot{\color{blue}\boldsymbol{e}_2}&\cdots&{\color{orange}\boldsymbol{c}_p}\cdot{\color{blue}\boldsymbol{e}_k}\\ \end{pmatrix}+\begin{pmatrix} {\color{green}\boldsymbol{b}_1}\cdot{\color{red}\boldsymbol{g}_1}& {\color{green}\boldsymbol{b}_1}\cdot{\color{red}\boldsymbol{g}_2}& \cdots&{\color{green}\boldsymbol{b}_1}\cdot{\color{red}\boldsymbol{g}_k} \\{\color{green}\boldsymbol{b}_2}\cdot{\color{red}\boldsymbol{g}_1}& {\color{green}\boldsymbol{b}_2}\cdot{\color{red}\boldsymbol{g}_2}& \cdots&{\color{green}\boldsymbol{b}_2}\cdot{\color{red}\boldsymbol{g}_k}\\ \vdots&\vdots&\smash\ddots&\vdots\\{\color{green}\boldsymbol{b}_m}\cdot{\color{red}\boldsymbol{g}_1}& {\color{green}\boldsymbol{b}_m}\cdot{\color{red}\boldsymbol{g}_2}& \cdots&{\color{green}\boldsymbol{b}_m}\cdot{\color{red}\boldsymbol{g}_k}\\ {\color{yellow}\boldsymbol{d}_1}\cdot{\color{red}\boldsymbol{g}_1}& {\color{yellow}\boldsymbol{d}_1}\cdot{\color{red}\boldsymbol{g}_2}& \cdots&{\color{yellow}\boldsymbol{d}_1}\cdot{\color{red}\boldsymbol{g}_k} \\{\color{yellow}\boldsymbol{d}_2}\cdot{\color{red}\boldsymbol{g}_1}& {\color{yellow}\boldsymbol{d}_2}\cdot{\color{red}\boldsymbol{g}_2}& \cdots&{\color{yellow}\boldsymbol{d}_2}\cdot{\color{red}\boldsymbol{g}_k}\\ \vdots&\vdots&\smash\ddots&\vdots\\{\color{yellow}\boldsymbol{d}_p}\cdot{\color{red}\boldsymbol{g}_1}& {\color{yellow}\boldsymbol{d}_p}\cdot{\color{red}\boldsymbol{g}_2}& \cdots&{\color{yellow}\boldsymbol{d}_p}\cdot{\color{red}\boldsymbol{g}_k}\\ \end{pmatrix}\\ &=\begin{pmatrix} {\color{purple}\boldsymbol{A}}{\color{blue}\boldsymbol{E}}\\ {\color{orange}\boldsymbol{C}}{\color{blue}\boldsymbol{E}}\\ \end{pmatrix}+ \begin{pmatrix} {\color{green}\boldsymbol{B}}{\color{red}\boldsymbol{G}}\\ {\color{yellow}\boldsymbol{D}}{\color{red}\boldsymbol{G}}\\ \end{pmatrix}\\ &=\begin{pmatrix} {\color{purple}\boldsymbol{A}}{\color{blue}\boldsymbol{E}}+{\color{green}\boldsymbol{B}}{\color{red}\boldsymbol{G}}\\ {\color{orange}\boldsymbol{C}}{\color{blue}\boldsymbol{E}}+{\color{yellow}\boldsymbol{D}}{\color{red}\boldsymbol{G}}\\ \end{pmatrix} \end{align*}

While $\boldsymbol{P}_2$ is:

\begin{align*} \boldsymbol{P}_2&= \begin{pmatrix} {\color{purple}\boldsymbol{a}_1}\cdot{\color{brown}\boldsymbol{f}_1}+{\color{green}\boldsymbol{b}_1}\cdot{\boldsymbol{h}_1}& {\color{purple}\boldsymbol{a}_1}\cdot{\color{brown}\boldsymbol{f}_2}+{\color{green}\boldsymbol{b}_1}\cdot{\boldsymbol{h}_2}& \cdots&{\color{purple}\boldsymbol{a}_1}\cdot{\color{brown}\boldsymbol{f}_r}+{\color{green}\boldsymbol{b}_1}\cdot{\boldsymbol{h}_r} \\{\color{purple}\boldsymbol{a}_2}\cdot{\color{brown}\boldsymbol{f}_1}+{\color{green}\boldsymbol{b}_2}\cdot{\boldsymbol{h}_1}& {\color{purple}\boldsymbol{a}_2}\cdot{\color{brown}\boldsymbol{f}_2}+{\color{green}\boldsymbol{b}_2}\cdot{\boldsymbol{h}_2}& \cdots&{\color{purple}\boldsymbol{a}_2}\cdot{\color{brown}\boldsymbol{f}_r}+{\color{green}\boldsymbol{b}_2}\cdot{\boldsymbol{h}_r}\\ \vdots&\vdots&\smash\ddots&\vdots\\{\color{purple}\boldsymbol{a}_m}\cdot{\color{brown}\boldsymbol{f}_1}+{\color{green}\boldsymbol{b}_m}\cdot{\boldsymbol{h}_1}&{\color{purple}\boldsymbol{a}_m}\cdot{\color{brown}\boldsymbol{f}_2}+{\color{green}\boldsymbol{b}_m}\cdot{\boldsymbol{h}_2}& \cdots&{\color{purple}\boldsymbol{a}_m}\cdot{\color{brown}\boldsymbol{f}_r}+{\color{green}\boldsymbol{b}_m}\cdot{\boldsymbol{h}_r}\\ {\color{orange}\boldsymbol{c}_1}\cdot{\color{brown}\boldsymbol{f}_1}+{\color{yellow}\boldsymbol{d}_1}\cdot{\boldsymbol{h}_1}& {\color{orange}\boldsymbol{c}_1}\cdot{\color{brown}\boldsymbol{f}_2}+{\color{yellow}\boldsymbol{d}_1}\cdot{\boldsymbol{h}_2}& \cdots&{\color{orange}\boldsymbol{c}_1}\cdot{\color{brown}\boldsymbol{f}_r}+{\color{yellow}\boldsymbol{d}_1}\cdot{\boldsymbol{h}_r}\\ {\color{orange}\boldsymbol{c}_2}\cdot{\color{brown}\boldsymbol{f}_1}+{\color{yellow}\boldsymbol{d}_2}\cdot{\boldsymbol{h}_1}& {\color{orange}\boldsymbol{c}_2}\cdot{\color{brown}\boldsymbol{f}_2}+{\color{yellow}\boldsymbol{d}_2}\cdot{\boldsymbol{h}_2}& \cdots&{\color{orange}\boldsymbol{c}_2}\cdot{\color{brown}\boldsymbol{f}_r}+{\color{yellow}\boldsymbol{d}_2}\cdot{\boldsymbol{h}_r}\\ \vdots&\vdots&\smash\ddots&\vdots\\{\color{orange}\boldsymbol{c}_p}\cdot{\color{brown}\boldsymbol{f}_1}+{\color{yellow}\boldsymbol{d}_p}\cdot{\boldsymbol{h}_1}& {\color{orange}\boldsymbol{c}_p}\cdot{\color{brown}\boldsymbol{f}_2}+{\color{yellow}\boldsymbol{d}_p}\cdot{\boldsymbol{h}_2}&\cdots&{\color{orange}\boldsymbol{c}_p}\cdot{\color{brown}\boldsymbol{f}_r}+{\color{yellow}\boldsymbol{d}_p}\cdot{\boldsymbol{h}_r} \end{pmatrix}\\&=\begin{pmatrix}{\color{purple}\boldsymbol{a}_1}\cdot{\color{brown}\boldsymbol{f}_1}&{\color{purple}\boldsymbol{a}_1}\cdot{\color{brown}\boldsymbol{f}_2}&\cdots&{\color{purple}\boldsymbol{a}_1}\cdot{\color{brown}\boldsymbol{f}_r}\\{\color{purple}\boldsymbol{a}_2}\cdot{\color{brown}\boldsymbol{f}_1}& {\color{purple}\boldsymbol{a}_2}\cdot{\color{brown}\boldsymbol{f}_2}&\cdots&{\color{purple}\boldsymbol{a}_2}\cdot{\color{brown}\boldsymbol{f}_r}\\\vdots&\vdots&\smash\ddots&\vdots\\{\color{purple}\boldsymbol{a}_m}\cdot{\color{brown}\boldsymbol{f}_1}& {\color{purple}\boldsymbol{a}_m}\cdot{\color{brown}\boldsymbol{f}_2}&\cdots&{\color{purple}\boldsymbol{a}_m}\cdot{\color{brown}\boldsymbol{f}_r}\\{\color{orange}\boldsymbol{c}_1}\cdot{\color{brown}\boldsymbol{f}_1}&{\color{orange}\boldsymbol{c}_1}\cdot{\color{brown}\boldsymbol{f}_2}& \cdots&{\color{orange}\boldsymbol{c}_1}\cdot{\color{brown}\boldsymbol{f}_r}\\{\color{orange}\boldsymbol{c}_2}\cdot{\color{brown}\boldsymbol{f}_1}& {\color{orange}\boldsymbol{c}_2}\cdot{\color{brown}\boldsymbol{f}_2}&\cdots&{\color{orange}\boldsymbol{c}_2}\cdot{\color{brown}\boldsymbol{f}_r}\\ \vdots&\vdots&\smash\ddots&\vdots\\{\color{orange}\boldsymbol{c}_p}\cdot{\color{brown}\boldsymbol{f}_1}&{\color{orange}\boldsymbol{c}_p}\cdot{\color{brown}\boldsymbol{f}_2}& \cdots&{\color{orange}\boldsymbol{c}_p}\cdot{\color{brown}\boldsymbol{f}_r}\end{pmatrix}+\begin{pmatrix} {\color{green}\boldsymbol{b}_1}\cdot{\boldsymbol{h}_1}& {\color{green}\boldsymbol{b}_1}\cdot{\boldsymbol{h}_2}& \cdots &{\color{green}\boldsymbol{b}_1}\cdot{\boldsymbol{h}_r} \\{\color{green}\boldsymbol{b}_2}\cdot{\boldsymbol{h}_1}& {\color{green}\boldsymbol{b}_2}\cdot{\boldsymbol{h}_2}& \cdots &{\color{green}\boldsymbol{b}_2}\cdot{\boldsymbol{h}_r}\\ \vdots&\vdots&\smash\ddots&\vdots\\ {\color{green}\boldsymbol{b}_m}\cdot{\boldsymbol{h}_1}& {\color{green}\boldsymbol{b}_m}\cdot{\boldsymbol{h}_2}& \cdots &{\color{green}\boldsymbol{b}_m}\cdot{\boldsymbol{h}_r}\\ {\color{yellow}\boldsymbol{d}_1}\cdot{\boldsymbol{h}_1}& {\color{yellow}\boldsymbol{d}_1}\cdot{\boldsymbol{h}_2}& \cdots &{\color{yellow}\boldsymbol{d}_1}\cdot{\boldsymbol{h}_r} \\ {\color{yellow}\boldsymbol{d}_2}\cdot{\boldsymbol{h}_1}& {\color{yellow}\boldsymbol{d}_2}\cdot{\boldsymbol{h}_2}& \cdots &{\color{yellow}\boldsymbol{d}_2}\cdot{\boldsymbol{h}_r}\\ \vdots&\vdots&\smash\ddots&\vdots\\ {\color{yellow}\boldsymbol{d}_p}\cdot{\boldsymbol{h}_1}& {\color{yellow}\boldsymbol{d}_p}\cdot{\boldsymbol{h}_2}& \cdots&{\color{yellow}\boldsymbol{d}_p}\cdot{\boldsymbol{h}_r} \end{pmatrix}\\ &=\begin{pmatrix} {\color{purple}\boldsymbol{A}}{\color{brown}\boldsymbol{F}}\\{\color{orange}\boldsymbol{C}}{\color{brown}\boldsymbol{F}} \end{pmatrix}+ \begin{pmatrix} {\color{green}\boldsymbol{B}}\boldsymbol{H}\\{\color{yellow}\boldsymbol{D}}\boldsymbol{H} \end{pmatrix}\\ &= \begin{pmatrix} {\color{purple}\boldsymbol{A}}{\color{brown}\boldsymbol{F}}+{\color{green}\boldsymbol{B}}\boldsymbol{H}\\ {\color{orange}\boldsymbol{C}}{\color{brown}\boldsymbol{F}}+{\color{yellow}\boldsymbol{D}}\boldsymbol{H} \end{pmatrix} \end{align*}

Note that the reason why we split up the resulting matrix into sub-matrices $\boldsymbol{P}_1$ and $\boldsymbol{P}_2$ is that it did not fit on an A4 page. Just think of the resulting matrix as $\boldsymbol{P}_1$ and $\boldsymbol{P}_2$ stacked horizontally. Substituting $\boldsymbol{P}_1$ and $\boldsymbol{P}_2$ into \eqref{eq:vkEm638LwWTxTnRMSdn} gives us:

\begin{align*} \begin{pmatrix} \color{purple}\boldsymbol{A}&\color{green}\boldsymbol{B}\\ \color{orange}\boldsymbol{C}&\color{yellow}\boldsymbol{D} \end{pmatrix} \begin{pmatrix} \color{blue}\boldsymbol{E}&\color{brown}\boldsymbol{F}\\ \color{red}\boldsymbol{G}&\boldsymbol{H} \end{pmatrix}&= \begin{pmatrix} \boldsymbol{P}_1&\boldsymbol{P}_2 \end{pmatrix}\\ &=\begin{pmatrix} {\color{purple}\boldsymbol{A}}{\color{blue}\boldsymbol{E}}+{\color{green}\boldsymbol{B}}{\color{red}\boldsymbol{G}} &{\color{purple}\boldsymbol{A}}{\color{brown}\boldsymbol{F}}+{\color{green}\boldsymbol{B}}\boldsymbol{H}\\ {\color{orange}\boldsymbol{C}}{\color{blue}\boldsymbol{E}}+{\color{yellow}\boldsymbol{D}}{\color{red}\boldsymbol{G}}& {\color{orange}\boldsymbol{C}}{\color{brown}\boldsymbol{F}}+{\color{yellow}\boldsymbol{D}}\boldsymbol{H} \end{pmatrix} \end{align*}

This completes the proof.

Theorem.

# Determinant of block matrices holding zero matrices and identity matrix

Consider the following block matrices:

$$\boldsymbol{B}= \begin{pmatrix} \boldsymbol{A}&\boldsymbol{0}\\ \boldsymbol{0}&\boldsymbol{I} \end{pmatrix},\;\;\;\;\;\;\; \boldsymbol{C}= \begin{pmatrix} \boldsymbol{I}&\boldsymbol{0}\\ \boldsymbol{0}&\boldsymbol{A} \end{pmatrix}$$

Where:

• $\boldsymbol{0}$ represents a zero matrix.

• $\boldsymbol{I}$ represents an identity matrix.

The determinant of $\boldsymbol{B}$ and $\boldsymbol{C}$ is:

\begin{align*} \det(\boldsymbol{B})&=\boldsymbol{A}\\ \det(\boldsymbol{C})&=\boldsymbol{A} \end{align*}

Proof. Suppose $\boldsymbol{A}$ is an $n\times{n}$ matrix. We will prove by induction that $\det(\boldsymbol{B})=\boldsymbol{A}$ - the proof for $\det(\boldsymbol{C})=\boldsymbol{A}$ is analogous. Consider the base case when $\boldsymbol{I}_1$ is an $1\times1$ identity matrix:

$$$$\label{eq:pqpB7ZttMZ8MIyBSkEd} \boldsymbol{B}_1= \begin{pmatrix} \boldsymbol{A}&\boldsymbol{0}\\ \boldsymbol{0}&\boldsymbol{I}_1 \end{pmatrix}= \begin{pmatrix} \boldsymbol{A}&\boldsymbol{0}\\ \boldsymbol{0}&1 \end{pmatrix}$$$$

Here, the subscript in $\boldsymbol{B}_1$ represents the size of the identity matrix.

We obtain the determinant of $\boldsymbol{B}_1$ by performing cofactor expansionlink along the last row. The last row is composed of all zeros except the last entry, which is $1$. This means that the determinant of $\boldsymbol{B}_1$ is:

$$\det(\boldsymbol{B}_1)= \pm\det(\boldsymbol{A})$$

Here, the sign of the determinant depends on the position of the $1$ in \eqref{eq:pqpB7ZttMZ8MIyBSkEd}. Recall the checkerboard pattern of signslink for determinants:

$$\begin{pmatrix} +&-&+&-&\cdots\\-&+&-&+&\cdots \\+&-&+&-&\cdots \\-&+&-&+&\cdots \\\vdots&\vdots&\vdots&\vdots&\smash\ddots\\ \end{pmatrix}$$

Observe how the sign corresponding to the position of $1$ is always $+$ because it is a diagonal entry. For instance, if $\boldsymbol{A}$ is an $2\times2$ matrix, then the position of $1$ is:

$$\begin{pmatrix} *&*&+\\ *&*&-\\ +&-&\color{blue}+ \end{pmatrix}$$

If $\boldsymbol{A}$ is an $3\times3$ matrix, then the position of $1$ is:

$$\begin{pmatrix} *&*&*&-\\ *&*&*&+\\ *&*&*&-\\ -&+&-&\color{blue}+ \end{pmatrix}$$

Therefore, the determinant of $\boldsymbol{B}_1$ is:

$$\det(\boldsymbol{B}_1)= \det(\boldsymbol{A})$$

We have just managed to prove the theorem for the base case when the identity matrix is of size $1\times1$. We now assume that the theorem holds when the identity matrix is of size $(k-1)\times(k-1)$, that is:

$$$$\label{eq:uEGEHog5T0Xt4gSIVsN} \boldsymbol{B}_{k-1}= \begin{pmatrix} \boldsymbol{A}&\boldsymbol{0}\\ \boldsymbol{0}&\boldsymbol{I}_{k-1} \end{pmatrix} \;\;\;\;\;\;\;{\color{blue}\implies}\;\;\;\;\;\;\; \det(\boldsymbol{B}_{k-1})=\det(\boldsymbol{A})$$$$

Our goal is to show that the theorem still holds when the identity matrix is of size $k\times{k}$. The matrix $\boldsymbol{B}_k$ can be written as:

$$\boldsymbol{B}_k= \begin{pmatrix} \boldsymbol{A}&\boldsymbol{0}\\ \boldsymbol{0}&\boldsymbol{I}_k \end{pmatrix}= \begin{pmatrix} \boldsymbol{A}&\boldsymbol{0}&\boldsymbol{0}\\ \boldsymbol{0}&\boldsymbol{I}_{k-1}&\boldsymbol{0}\\ \boldsymbol{0}&\boldsymbol{0}&1\\ \end{pmatrix}$$

Since the $1$ lies on the diagonal of $\boldsymbol{B}_k$, the corresponding sign on the checkerboard is positive. Once again, we compute the determinant of $\boldsymbol{B}_k$ by cofactor expansion along the last row:

\begin{align*} \det(\boldsymbol{B}_k)&= \det\left[\begin{pmatrix} \boldsymbol{A}&\boldsymbol{0}\\ \boldsymbol{0}&\boldsymbol{I}_{k-1} \end{pmatrix}\right]\\ &=\det(\boldsymbol{B}_{k-1})\\ \end{align*}

We now use the inductive hypothesis \eqref{eq:uEGEHog5T0Xt4gSIVsN} to get:

\begin{align*} \det(\boldsymbol{B}_k) &=\det(\boldsymbol{B}_{k-1})\\ &=\det(\boldsymbol{A})\\ \end{align*}

By the principle of mathematical induction, we conclude that the theorem holds for an identity matrix of any size. This completes the proof.

Theorem.

# Determinant of block matrices with a zero matrix (1)

Consider the following block matrix:

$$\boldsymbol{R}= \begin{pmatrix} \boldsymbol{A}&\boldsymbol{B}\\ \boldsymbol{0}&\boldsymbol{C} \end{pmatrix}$$

Where $\boldsymbol{0}$ represents a zero matrix.

The determinant of $\boldsymbol{R}$ is:

$$\det(\boldsymbol{R})= \det(\boldsymbol{A})\cdot\det(\boldsymbol{C})$$

Proof. Matrix $\boldsymbol{R}$ can be expressed:

\label{eq:ktfwG6UlaqfhoY7uWso} \begin{aligned}[b] \begin{pmatrix}\boldsymbol{A}&\boldsymbol{B}\\ \boldsymbol{0}&\boldsymbol{C}\end{pmatrix}&= \begin{pmatrix} \boldsymbol{I}_k\boldsymbol{A}+\boldsymbol{B}\boldsymbol{0}_{rk}& \boldsymbol{I}_k\boldsymbol{0}_{kr}+\boldsymbol{B}\boldsymbol{I}_{r}\\ \boldsymbol{0}_{rk}\boldsymbol{A}+\boldsymbol{C}\boldsymbol{0}_{rk} &\boldsymbol{0}_{rk}\boldsymbol{0}_{kr}+\boldsymbol{C}\boldsymbol{I}_r \end{pmatrix} \end{aligned}

Where the subscript of the matrices represents their shape. For instance, $\boldsymbol{0}_{rk}$ is a zero matrix with $r$ rows and $k$ columns.

Now, using theoremlink, we can express \eqref{eq:ktfwG6UlaqfhoY7uWso} as a product of two block matrices:

$$\begin{pmatrix}\boldsymbol{A}&\boldsymbol{B}\\ \boldsymbol{0}&\boldsymbol{C}\end{pmatrix} =\begin{pmatrix} \boldsymbol{I}_k&\boldsymbol{B}\\ \boldsymbol{0}_{rk}&\boldsymbol{C} \end{pmatrix} \begin{pmatrix} \boldsymbol{A}&\boldsymbol{0}_{kr}\\ \boldsymbol{0}_{rk}&\boldsymbol{I}_r \end{pmatrix}$$

Similar to what we did in \eqref{eq:ktfwG6UlaqfhoY7uWso}, we express the left matrix on the right-hand side as:

\begin{align*} \begin{pmatrix}\boldsymbol{A}&\boldsymbol{B}\\ \boldsymbol{0}&\boldsymbol{C}\end{pmatrix} &= \begin{pmatrix} \boldsymbol{I}_k\boldsymbol{I}_k+\boldsymbol{0}_{kr}\boldsymbol{0}_{rk} &\boldsymbol{I}_k\boldsymbol{B}+\boldsymbol{0}_{kr}\boldsymbol{I}_{r}\\ \boldsymbol{0}_{rk}\boldsymbol{I}_k+\boldsymbol{C}\boldsymbol{0}_{rk} &\boldsymbol{0}_{rk}\boldsymbol{B}+\boldsymbol{C}\boldsymbol{I}_{r} \end{pmatrix} \begin{pmatrix} \boldsymbol{A}&\boldsymbol{0}_{kr}\\ \boldsymbol{0}_{rk}&\boldsymbol{I}_r \end{pmatrix} \end{align*}

\begin{align*} \begin{pmatrix}\boldsymbol{A}&\boldsymbol{B}\\ \boldsymbol{0}&\boldsymbol{C}\end{pmatrix} &= \begin{pmatrix} \boldsymbol{I}_k&\boldsymbol{0}_{kr}\\ \boldsymbol{0}_{rk}&\boldsymbol{C} \end{pmatrix} \begin{pmatrix} \boldsymbol{I}_k&\boldsymbol{B}\\ \boldsymbol{0}_{rk}&\boldsymbol{I}_r \end{pmatrix} \begin{pmatrix} \boldsymbol{A}&\boldsymbol{0}_{kr}\\ \boldsymbol{0}_{rk}&\boldsymbol{I}_r \end{pmatrix} \end{align*}

Now, let's take the determinant of both sides:

\begin{align*} \det\left[\begin{pmatrix}\boldsymbol{A}&\boldsymbol{B}\\ \boldsymbol{0}&\boldsymbol{C}\end{pmatrix}\right] &= \det\left[\begin{pmatrix} \boldsymbol{I}_k&\boldsymbol{0}_{kr}\\ \boldsymbol{0}_{rk}&\boldsymbol{C} \end{pmatrix} \begin{pmatrix} \boldsymbol{I}_k&\boldsymbol{B}\\ \boldsymbol{0}_{rk}&\boldsymbol{I}_r \end{pmatrix} \begin{pmatrix} \boldsymbol{A}&\boldsymbol{0}_{kr}\\ \boldsymbol{0}_{rk}&\boldsymbol{I}_r \end{pmatrix}\right]\\ \end{align*}

By the multiplicative propertylink of determinant:

\begin{align*} \det\left[\begin{pmatrix}\boldsymbol{A}&\boldsymbol{B}\\ \boldsymbol{0}&\boldsymbol{C}\end{pmatrix}\right] &= \det\left[\begin{pmatrix} \boldsymbol{I}_k&\boldsymbol{0}_{kr}\\ \boldsymbol{0}_{rk}&\boldsymbol{C} \end{pmatrix}\right] \cdot\det\left[\begin{pmatrix} \boldsymbol{I}_k&\boldsymbol{B}\\ \boldsymbol{0}_{rk}&\boldsymbol{I}_r \end{pmatrix}\right] \cdot\det\left[\begin{pmatrix} \boldsymbol{A}&\boldsymbol{0}_{kr}\\ \boldsymbol{0}_{rk}&\boldsymbol{I}_r \end{pmatrix}\right]\\ \end{align*}

\begin{align*} \det\left[\begin{pmatrix}\boldsymbol{A}&\boldsymbol{B}\\ \boldsymbol{0}&\boldsymbol{C}\end{pmatrix}\right] &= \det(\boldsymbol{C}) \cdot\det\left[\begin{pmatrix} \boldsymbol{I}_k&\boldsymbol{B}\\ \boldsymbol{0}_{rk}&\boldsymbol{I}_r \end{pmatrix}\right] \cdot\det(\boldsymbol{A})\\ \end{align*}

Notice how the middle determinant is an upper triangular matrix. By theoremlink, this means that the determinant is equal to the product of its diagonals, which in this case are all ones. Therefore, the middle determinant is equal to one, thereby giving us the desired result:

\begin{align*} \det\left[\begin{pmatrix}\boldsymbol{A}&\boldsymbol{B}\\ \boldsymbol{0}&\boldsymbol{C}\end{pmatrix}\right] &=\det(\boldsymbol{C}) \cdot\det(\boldsymbol{A})\\ \det(\boldsymbol{R})&=\det(\boldsymbol{C}) \cdot\det(\boldsymbol{A})\\ \end{align*}

This completes the proof.

Theorem.

# Determinant of block matrices with a zero matrix (2)

Consider the following block matrix:

$$\boldsymbol{R}= \begin{pmatrix} \boldsymbol{A}&\boldsymbol{0}\\ \boldsymbol{B}&\boldsymbol{C} \end{pmatrix}$$

Where $\boldsymbol{0}$ is a zero matrix.

The determinant of $\boldsymbol{R}$ is:

$$\det(\boldsymbol{R})= \det(\boldsymbol{A})\cdot \det(\boldsymbol{C})$$

Proof. By theoremlink, the transpose of $\boldsymbol{R}$ is:

$$\boldsymbol{R}^T= \begin{pmatrix} \boldsymbol{A}^T&\boldsymbol{B}^T\\ \boldsymbol{0}^T&\boldsymbol{C}^T \end{pmatrix}= \begin{pmatrix} \boldsymbol{A}^T&\boldsymbol{B}^T\\ \boldsymbol{0}&\boldsymbol{C}^T \end{pmatrix}$$

Taking the determinant of both sides gives:

$$$$\label{eq:s5iBxmyJjR1NjxgRVbD} \det(\boldsymbol{R}^T)= \det\left[\begin{pmatrix} \boldsymbol{A}^T&\boldsymbol{B}^T\\ \boldsymbol{0}&\boldsymbol{C}^T \end{pmatrix}\right]$$$$

By theoremlink, the right-hand side is:

$$\det\left[\begin{pmatrix} \boldsymbol{A}^T&\boldsymbol{B}^T\\ \boldsymbol{0}&\boldsymbol{C}^T \end{pmatrix}\right]= \det(\boldsymbol{A}^T)\cdot \det(\boldsymbol{C}^T)$$

Now, by theoremlink, we know that the determinant of a matrix transpose is equal to the determinant of the matrix:

\begin{align*} \det\left[\begin{pmatrix} \boldsymbol{A}^T&\boldsymbol{B}^T\\ \boldsymbol{0}&\boldsymbol{C}^T \end{pmatrix}\right]&= \det(\boldsymbol{A})\cdot \det(\boldsymbol{C}) \end{align*}

Similarly, we have that $\det(\boldsymbol{R}^T)=\det(\boldsymbol{R})$. Therefore, \eqref{eq:s5iBxmyJjR1NjxgRVbD} reduces to:

$$\det(\boldsymbol{R})= \det(\boldsymbol{A})\cdot \det(\boldsymbol{C})$$

This completes the proof.

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