Types: N/A
Examples: N/A
Constructions: N/A
Generalizations: N/A

Properties: N/A
Sufficiencies: N/A
Questions: N/A

Algebraic Properties of Matrix-Column-Vector Multiplication

Let matrix $A \in R^{m \times n}$ , let vectors $\vec{x}, \vec{y} \in R^{n \times 1}$ , and let scalars $α, β \in R$ . Then, each of the following laws holds true:

Distributivity: $A (\vec{x} + \vec{y}) = A \vec{x} + A \vec{y}$
Scalar Multiplication: $A (α \vec{x}) = α (A \vec{x})$
Linearity: $A (α \vec{x} + β \vec{y}) = α A \vec{x} + β A \vec{y}$

Remark. When trying to prove theorems, we want to transform our theorem into the general form of a conditional statement (read if $P$ , then $Q$ ).

\underset{antecedent}{P} ⟹ \underset{consequent}{Q}

We first assume the antecedent is true.
Then, the consequent must follow from that assumption.
If something is true, then the other thing must also be true.
$⧫$

Distributivity

For distributivity, we can say

If A \in R^{m \times n} and \vec{x}, \vec{y} \in R^{n \times 1}, then A (\vec{x} + \vec{y}) = A \vec{x} + A \vec{y}

Concrete Example

Before proving the general case, let's create a concrete example (not a general proof).

Distributivity

Let $m = 3$ and $n = 4$ . Let $A \in R^{m \times n}$ and $\vec{x}, \vec{y} \in R^{n \times 1}$ .

With our antecedent in mind, we write our matrix

[A]_{m \times n} = {[\begin{matrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \end{matrix}]}_{3 \times 4}

We can also write our vectors

\vec{x} = {[\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{matrix}]}_{4 \times 1}, \vec{y} = {[\begin{matrix} y_{1} \\ y_{2} \\ y_{3} \\ y_{4} \end{matrix}]}_{4 \times 1}

With this information at hand, let's compute our desired result. We want to show

\underset{expression 1}{\underset{⏟}{A (\vec{x} \cdot \vec{y})}} = \underset{expression 2}{\underset{⏟}{A \vec{x} + A \vec{y}}}

Let's analyze each expression.

\begin{aligned} Expression 1: [A]_{m \times n} \cdot ([\vec{x}]_{1 \times n} + [\vec{y}]_{1 \times n}) \\ Expression 2: [A]_{m \times n} \cdot [\vec{x}]_{n \times 1} + [A]_{m \times n} \cdot [\vec{y}]_{n \times 1} \end{aligned}

Some theorems/definitions that could be useful are:

Forward Direction

Let's start with expression 1 and work our way towards expression 2.

\begin{aligned} \underset{expression 1}{\underset{⏟}{A \cdot (\vec{x} + \vec{y})}} & = \underset{expression 1.1}{\underset{⏟}{[\begin{array}{c} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \end{array}] ([\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}] + [\begin{array}{c} y_{1} \\ y_{2} \\ y_{3} \\ y_{4} \end{array}])}} \\ = \underset{expression 1.2}{\underset{⏟}{{[\begin{array}{c} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{44} \end{array}]}_{m \times n} \cdot {[\begin{array}{c} x_{1} + y_{1} \\ x_{2} + y_{2} \\ x_{3} + y_{3} \\ x_{4} + y_{4} \end{array}]}_{n \times 1}}} \\ = \sum_{k = 1}^{n} (x_{k} + y_{k}) \cdot A (:, k) \\ = \underset{k = 1}{\underset{⏟}{(x_{1} + y_{1}) [\begin{array}{c} a_{11} \\ a_{21} \\ a_{31} \end{array}]}} + \underset{k = 2}{\underset{⏟}{(x_{2} + y_{2}) [\begin{array}{c} a_{12} \\ a_{22} \\ a_{32} \end{array}]}} + \underset{k = 3}{\underset{⏟}{(x_{3} + y_{3}) [\begin{array}{c} a_{13} \\ a_{23} \\ a_{33} \end{array}]}} + \underset{k = 4}{\underset{⏟}{(x_{4} + y_{4}) [\begin{array}{c} a_{14} \\ a_{24} \\ a_{34} \end{array}]}} \end{aligned}

Notice how our sum is composed of 4.1 Scalar-Vector Multiplication products. Consider

\begin{aligned} (x_{1} + y_{1}) [\begin{array}{c} a_{11} \\ a_{21} \\ a_{31} \end{array}] & = [\begin{array}{c} (x_{1} + y_{1}) a_{11} \\ (x_{1} + y_{1}) a_{21} \\ (x_{1} + y_{1}) a_{31} \end{array}] \\ = [\begin{array}{c} x_{1} a_{11} + y_{1} a_{11} \\ x_{1} a_{21} + y_{1} a_{21} \\ x_{1} a_{31} + y_{1} a_{31} \end{array}] \\ = [\begin{array}{c} x_{1} a_{11} \\ x_{1} a_{21} \\ x_{1} a_{31} \end{array}] + [\begin{array}{c} y_{1} a_{11} \\ y_{1} a_{21} \\ y_{1} a_{31} \end{array}] \\ = x_{1} [\begin{array}{c} a_{11} \\ a_{21} \\ a_{31} \end{array}] + y_{1} [\begin{array}{c} a_{11} \\ a_{21} \\ a_{31} \end{array}] \end{aligned}

Conjecture

$(α + β) \cdot \vec{a} = α \vec{a} + β \vec{a}$

Note: For conjectures, we either prove in general or quote a result (for math heads we prove every result ourselves).

Going back to our original forward direction problem, we can distribute our conjecture across where we stopped previously.

\begin{aligned} = x_{1} [\begin{array}{c} a_{11} \\ a_{21} \\ a_{31} \end{array}] + y_{1} [\begin{array}{c} a_{11} \\ a_{21} \\ a_{31} \end{array}] + x_{1} [\begin{array}{c} a_{12} \\ a_{22} \\ a_{32} \end{array}] + y_{1} [\begin{array}{c} a_{12} \\ a_{22} \\ a_{32} \end{array}] + x_{1} [\begin{array}{c} a_{13} \\ a_{23} \\ a_{33} \end{array}] + y_{1} [\begin{array}{c} a_{13} \\ a_{23} \\ a_{33} \end{array}] + x_{1} [\begin{array}{c} a_{14} \\ a_{24} \\ a_{34} \end{array}] + y_{1} [\begin{array}{c} a_{14} \\ a_{24} \\ a_{34} \end{array}] \end{aligned}

Backward Direction

Now, lets start from expression 2 and work backwards towards expression 1.

\begin{aligned} = A \vec{x} + A \vec{y} \\ = \sum_{k = 1}^{n} x_{k} A (:, k) + \sum_{k = 1}^{n} y_{k} A (:, k) \\ = x_{1} [\begin{array}{c} a_{11} \\ a_{21} \\ a_{31} \end{array}] + y_{1} [\begin{array}{c} a_{11} \\ a_{21} \\ a_{31} \end{array}] + x_{1} [\begin{array}{c} a_{12} \\ a_{22} \\ a_{32} \end{array}] + y_{1} [\begin{array}{c} a_{12} \\ a_{22} \\ a_{32} \end{array}] + x_{1} [\begin{array}{c} a_{13} \\ a_{23} \\ a_{33} \end{array}] + y_{1} [\begin{array}{c} a_{13} \\ a_{23} \\ a_{33} \end{array}] + x_{1} [\begin{array}{c} a_{14} \\ a_{24} \\ a_{34} \end{array}] + y_{1} [\begin{array}{c} a_{14} \\ a_{24} \\ a_{34} \end{array}] \end{aligned}

We are able to write our sum as such because vector addition is commutative. If we generalize our solution, we will see that

If A \in R^{m \times n} and \vec{x}, \vec{y} \in R^{n \times 1}, then A (\vec{x} + \vec{y}) = A \vec{x} + A \vec{y}

Scalar Multiplication

For this theorem, we might state

\underset{P}{\underset{⏟}{If A \in R^{m \times n}, \vec{x} \in R^{n \times 1}, and α \in R,}} then \underset{Q}{\underset{⏟}{A (α \vec{x}) = α (A \vec{x})}}

Some theorems/definitions that could be useful for proving this:

We want to prove that two vectors are equal. We know that two vectors are equal if the each entry is to the corresponding entry of the other vector. A way we can get the entry-by-entry definition of matrix-column-vector multiplication is through the inner product definition. With these definitions in mind, let's formalize our proof.

\begin{proof}
Suppose that $A \in R^{m \times n}$ , $\vec{x} \in R^{n}$ , and $α \in R$ . For any $i = 1, 2, \dots, m$ , we want to show that ${entry}_{(i, 1)} (A (α \vec{x})) = {entry}_{(i, 1)} (α (A \vec{x}))$ . Consider

\begin{aligned} {entry}_{(i, 1)} (A (a \vec{x})) & = (A (i, :))^{T} \cdot (α \vec{x}) \\ [\begin{array}{c} a_{i 1} \\ a_{i 2} \\ ⋮ \\ a_{i n} \end{array}] \cdot [\begin{array}{c} a x_{1} \\ a x_{2} \\ ⋮ \\ a x_{n} \end{array}] & = a_{i 1} (α x_{1}) + a_{i 2} (α x_{2}) + \dots + a_{i n} (α x_{n}) \\ = α (a_{i 1} + a_{i 2} + \dots + a_{i n}) \\ = α ((A (i, :))^{T} \cdot \vec{x}) \\ = {entry}_{(i, 1)} (α (A \vec{x})) \end{aligned}

This is exactly what we wanted to show.

\end{proof}