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# Distributive law

## Problem

Let

$$
A = \begin{bmatrix}
- & \ba & - \\
- & \br_2 & - \\
~ & \vdots & ~ \\
- & \br_n & - \\
\end{bmatrix},\
B = \begin{bmatrix}
- & \bb & - \\
- & \br_2 & - \\
~ & \vdots & ~ \\
- & \br_n & - \\
\end{bmatrix}, \text{ and }
M = \begin{bmatrix}
- & \ba + \bb & - \\
- & \br_2 & - \\
~ & \vdots & ~ \\
- & \br_n & - \\
\end{bmatrix}.
$$

Show that $\det(M) = \det(A) + \det(B)$.

## Thought

Essentially, this problem is asking us to show the _distributive law_ by definition. This is usually a theorem in the textbook, so it is good enough to understand the proof first and then prove it again by yourself.

Fortunately, there are some special cases that is not too difficult.

The first special case is when $\{\br_2, \ldots, \br_n\}$ is linearly dependent. In this case, we have $\det(A) = \det(B) = \det(M) = 0$, which easily make the equality hold.

The other special case is when both $\ba$ and $\bb$ are multiple of a vector $\bx$. We may assume $\ba = p\bx$ and $\bb = q\bx$. By definition, $\det(A) = p\Delta$, $\det(B) = q\Delta$, and $\det(M) = (p+q)\Delta$, where

$$
\Delta = \det \begin{bmatrix}
- & \bx & - \\
- & \br_2 & - \\
~ & \vdots & ~ \\
- & \br_n & - \\
\end{bmatrix}.
$$
Thus, the equality holds again.

It turns out these two special cases with some extra steps are enough for us to prove the whole statement.

## Sample answer

First, when $\{\br_2, \ldots, \br_n\}$ is linearly dependent, $\det(A) = \det(B) = \det(M)$, so obvsiouly $\det(M) = 0 = \det(A) + \det(B)$.

Suppose $\{\br_2, \ldots, \br_n\}$ is linearly independent. Then we may extend it into a basis $\beta = \{\bx, \br_2, \ldots, \br_n\}$ of $\mathbb{R}^n$[^extend]. Thus, every vector in $\mathbb{R}^n$ can be written as a linear combination of $\beta$[^representation]. We may assume
$$
\begin{aligned}
\ba &= p_1\bx + p_2\br_2 + \cdots + p_n\br_n, \text{ and} \\
\bb &= q_1\bx + q_2\br_2 + \cdots + q_n\br_n,
\end{aligned}
$$

for some real numbers $p_1, \ldots, p_n$ and $q_1, \ldots, q_n$. Consequently, we have

$$
\ba + \bb = (p_1 + q_1)\bx + (p_1 + q_2)\br_2 + \cdots + (p_n + q_n)\br_n.
$$

Applying appropriate row operations to $A$, $B$, and $M$, we have

$$
\det(A) = \det\begin{bmatrix}
- & p_1\bx + p_2\br_2 + \cdots + p_n\br_n & - \\
- & \br_2 & - \\
~ & \vdots & ~ \\
- & \br_n & - \\
\end{bmatrix} =
\det\begin{bmatrix}
- & p_1\bx & - \\
- & \br_2 & - \\
~ & \vdots & ~ \\
- & \br_n & - \\
\end{bmatrix},
$$

$$
\det(B) = \det\begin{bmatrix}
- & q_1\bx + q_2\br_2 + \cdots + q_n\br_n & - \\
- & \br_2 & - \\
~ & \vdots & ~ \\
- & \br_n & - \\
\end{bmatrix} =
\det\begin{bmatrix}
- & q_1\bx & - \\
- & \br_2 & - \\
~ & \vdots & ~ \\
- & \br_n & - \\
\end{bmatrix},
$$

and

$$
\det(M) = \det\begin{bmatrix}
- & (p_1 + q_1)\bx + (p_1 + q_2)\br_2 + \cdots + (p_n + q_n)\br_n & - \\
- & \br_2 & - \\
~ & \vdots & ~ \\
- & \br_n & - \\
\end{bmatrix} =
\det\begin{bmatrix}
- & (p_1 + q_1)\bx & - \\
- & \br_2 & - \\
~ & \vdots & ~ \\
- & \br_n & - \\
\end{bmatrix}.
$$

By writing

$$
\Delta = \det\begin{bmatrix}
- & \bx & - \\
- & \br_2 & - \\
~ & \vdots & ~ \\
- & \br_n & - \\
\end{bmatrix},
$$

we have

$$
\det(M) = (p_1 + q_1)\Delta = p_1\Delta + q_1\Delta = \det(A) + \det(B).
$$

This completes the proof.

[^extend]: Every independent set $\alpha$ can be extended to a basis $\beta$ such that $\alpha\subseteq\beta$. Review [Hefferon](https://hefferon.net/linearalgebra/) Two.III for the definition and properties of a basis, in particular, Corollary 2.12.

[^representation]: If $\beta$ is a basis of the vector space $V$, then every vector in $V$ can be uniquely written as a linear combination of $\beta$. Review [Hefferon](https://hefferon.net/linearalgebra/) Two.III for the definition and properties of a basis, in particular, Theorem 1.12.

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