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# Comprehensive Guide on on Dot Product in Linear Algebra

Linear Algebra
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Vectors
schedule Jan 8, 2023
Last updated
local_offer Linear Algebra
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# What is dot product?

When we are dealing with real numbers, there is only one interpretation of product - multiplication. However, in the world of vectors, there are two types of products - dot product and cross product. The dot product appears everywhere in machine learning and is therefore a must-know topic. In this guide, we will explore the dot product in depth!

Definition.

# Dot product

Suppose we have the following two vectors $\boldsymbol{v}$ and $\boldsymbol{w}$:

$$\boldsymbol{v}=\begin{pmatrix} v_1\\ v_2\\ v_3\\ \end{pmatrix},\;\;\;\; \boldsymbol{w}=\begin{pmatrix} w_1\\ w_2\\ w_3\\ \end{pmatrix}$$

The dot product is defined as:

$$\boldsymbol{v}\cdot{\boldsymbol{w}}= v_1w_1+ v_2w_2+ v_3w_3$$

Note that $\boldsymbol{v}$ and $\boldsymbol{w}$ are in $\mathbb{R}^3$, but the dot product returns a scalar. The dot product is also sometimes referred to as the scalar product or the inner product.

Example.

## Computing the dot product between two vectors

Compute the dot product $\boldsymbol{v}\cdot{\boldsymbol{w}}$ where $\boldsymbol{v}$ and $\boldsymbol{w}$ are defined as:

$$\boldsymbol{v}= \begin{pmatrix} 1\\2 \end{pmatrix},\;\;\;\; \boldsymbol{w}= \begin{pmatrix} 3\\4 \end{pmatrix}$$

Solution. The dot product $\boldsymbol{v}\cdot\boldsymbol{w}$ is:

\begin{align*} \boldsymbol{v}\cdot{\boldsymbol{w}}&= (1)(3)+(2)(4)\\ &=11 \end{align*}
Theorem.

# Using dot product to compute the angle between two vectors

Suppose we have the following two vectors $\boldsymbol{v}$ and $\boldsymbol{w}$:

Let the angle between $\boldsymbol{v}$ and $\boldsymbol{w}$ be $\theta$. The dot product of $\boldsymbol{v}$ and $\boldsymbol{w}$ is:

$$\boldsymbol{v}\cdot{\boldsymbol{w}}= \Vert{\boldsymbol{v}}\Vert\Vert{\boldsymbol{w}}\Vert \cos{(\theta)}$$

Here, $\Vert{\boldsymbol{v}}\Vert$ is the magnitude of $\boldsymbol{v}$.

Proof. Suppose we have the following:

Here, the vector $\boldsymbol{w}-\boldsymbol{v}$ points from $A$ to $B$ because of the geometric interpretation of vector additionlink.

The lengths of the sides are:

Using the cosine rule from elementary trigonometry, we have that:

$$$$\label{eq:xF6GwOYW3PvcsraNGkP} \Vert{\boldsymbol{w}-\boldsymbol{v}}\Vert^2 =\Vert\boldsymbol{v}\Vert^2+ \Vert\boldsymbol{w}\Vert^2-2\Vert{\boldsymbol{v}}\Vert \Vert{\boldsymbol{w}}\Vert\cos(\theta)$$$$

Let's assume $\boldsymbol{w}$ and $\boldsymbol{v}$ are both in $\mathbb{R}^3$, though the proof can easily be generalized to $\mathbb{R}^n$. The vector $\boldsymbol{w}-\boldsymbol{v}$ is:

$$\boldsymbol{w}-\boldsymbol{v} =\begin{pmatrix} w_1-v_1\\ w_2-v_2\\ w_3-v_3\\ \end{pmatrix}$$

The square of the magnitude of $\boldsymbol{w}-\boldsymbol{v}$ is:

\begin{align*} \Vert{\boldsymbol{w}-\boldsymbol{v}}\Vert^2 &= \Big(\;\sqrt{(w_1-v_1)^2+ (w_2-v_2)^2+ (w_3-v_3)^2}\;\Big)^2\\ &= (w_1-v_1)^2+ (w_2-v_2)^2+ (w_3-v_3)^2 \end{align*}

Similarly, $\Vert{\boldsymbol{v}}\Vert^2$ and $\Vert{\boldsymbol{w}}\Vert^2$ are:

\begin{align*} \Vert{\boldsymbol{v}}\Vert^2 &= v_1^2+v^2_2+v^3_3\\ \Vert{\boldsymbol{w}}\Vert^2 &= w_1^2+w^2_2+w^3_3 \end{align*}

Substituting these squared magnitudes into \eqref{eq:xF6GwOYW3PvcsraNGkP} gives:

\begin{align*} (w_1-v_1)^2+ (w_2-v_2)^2+ (w_3-v_3)^2&= v^2_1+v^2_2+v^2_3+ w^2_1+w^2_2+w^2_3 -2 \Vert{\boldsymbol{v}}\Vert\Vert{\boldsymbol{w}}\Vert\cos{(\theta)} \\ v_1w_1+v_2w_2+v_3w_3&=\Vert{\boldsymbol{v}}\Vert\Vert{\boldsymbol{w}}\Vert\cos{(\theta)}\\ \boldsymbol{v}\cdot{\boldsymbol{w}}&=\Vert{\boldsymbol{v}}\Vert\Vert{\boldsymbol{w}}\Vert\cos{(\theta)} \end{align*}

This completes the proof.

Example.

## Computing the angle between two vectors

Compute the angle between the following two vectors:

$$\boldsymbol{v}= \begin{pmatrix} 2\\ 3 \end{pmatrix},\;\;\;\; \boldsymbol{w}= \begin{pmatrix} 4\\ 1 \end{pmatrix}$$

Solution. We know from theoremlink that:

$$\boldsymbol{v}\cdot{\boldsymbol{w}}= \Vert{\boldsymbol{v}}\Vert\Vert{\boldsymbol{w}}\Vert \cos{(\theta)}$$

Rearranging the above to make $\theta$ the subject:

\begin{align*} \theta&= \arccos{\Big(\frac{\boldsymbol{v}\cdot{\boldsymbol{w}}} {\Vert{\boldsymbol{v}}\Vert\Vert{\boldsymbol{w}}\Vert}\Big)}\\ &=\mathrm{arccos}{\Big(\frac{(2)(4)+(3)(1)} {(2^2+3^2)^{1/2}(4^2+1^2)^{1/2}}\Big)}\\ &\approx42.5 \end{align*}

Therefore, the angle between vectors $\boldsymbol{v}$ and $\boldsymbol{w}$ is around $42.5$ degrees. This is confirmed visually below:

# Properties of dot products

Theorem.

## Dot product of two perpendicular vectors

If $\boldsymbol{v}$ and $\boldsymbol{w}$ are vectors that are perpendicular to each other, then:

$$\boldsymbol{v}\cdot{\boldsymbol{w}}=0$$

Proof. If $\boldsymbol{v}$ and $\boldsymbol{w}$ are perpendicular, then we have the following scenario:

From theoremlink, we have that:

\begin{align*} \boldsymbol{v}\cdot{\boldsymbol{w}}&= \Vert{\boldsymbol{v}}\Vert\Vert{\boldsymbol{w}}\Vert \cos{(\theta)}\\ &= \Vert{\boldsymbol{v}}\Vert\Vert{\boldsymbol{w}}\Vert \cos{(90)}\\ &= 0 \end{align*}

This completes the proof.

Theorem.

## Commutative property of dot product

The commutative property of the dot product states that the ordering of the dot product between two vectors does not matter:

$$\boldsymbol{v}\cdot\boldsymbol{w}= \boldsymbol{w}\cdot\boldsymbol{v}$$

Proof. We will prove the case for $\mathbb{R}^3$ here, but the proof can easily be generalized for $\mathbb{R}^n$. Let vectors $\boldsymbol{v}$ and $\boldsymbol{w}$ be:

$$\boldsymbol{v}= \begin{pmatrix} v_1\\v_2\\v_3\\ \end{pmatrix},\;\;\;\; \boldsymbol{w}= \begin{pmatrix} w_1\\w_2\\w_3\\ \end{pmatrix}$$

Starting from the left-hand side:

\begin{align*} \boldsymbol{v}\cdot\boldsymbol{w}&= v_1w_1+v_2w_2+v_3w_3\\ &=w_1v_1+w_2v_2+w_3v_3\\ &=\boldsymbol{w}\cdot\boldsymbol{v} \end{align*}

This completes the proof.

Theorem.

## Position of scalars can be changed

If $\boldsymbol{a}$ and $\boldsymbol{b}$ are vectors in $\mathbb{R}^n$ and $\lambda$ is some constant scalar, then:

$$(\lambda{\boldsymbol{v}})\cdot{\boldsymbol{w}}= \boldsymbol{v}\cdot(\lambda{\boldsymbol{w}}) =\lambda(\boldsymbol{v}\cdot\boldsymbol{w})$$

Proof. We will prove the case for $\mathbb{R}^3$ here, but the proof can easily be generalized for $\mathbb{R}^n$. Let vectors $\boldsymbol{v}$ and $\boldsymbol{w}$ be:

$$\boldsymbol{v}= \begin{pmatrix} v_1\\v_2\\v_3\\ \end{pmatrix},\;\;\;\; \boldsymbol{w}= \begin{pmatrix} w_1\\w_2\\w_3\\ \end{pmatrix}$$

Starting from the left-hand side:

\begin{align*} (\lambda\boldsymbol{v})\cdot{\boldsymbol{w}}&= \begin{pmatrix} \lambda{v_1}\\\lambda{v_2}\\\lambda{v_3}\\ \end{pmatrix}\cdot \begin{pmatrix} w_1\\w_2\\w_3\\ \end{pmatrix}\\ &=\lambda{v_1w_1}+\lambda{v_2w_2}+\lambda{v_3w_3} \end{align*}

To prove the first equality:

\begin{align*} (\lambda\boldsymbol{v})\cdot{\boldsymbol{w}} &=\lambda{v_1w_1}+\lambda{v_2w_2}+\lambda{v_3w_3}\\ &=v_1(\lambda{w_1})+v_2(\lambda{w_2})+v_3(\lambda{w_3})\\ &=\boldsymbol{v}\cdot(\lambda{\boldsymbol{w}}) \end{align*}

To prove the second equality:

\begin{align*} (\lambda\boldsymbol{v})\cdot{\boldsymbol{w}} &=\lambda{v_1w_1}+\lambda{v_2w_2}+\lambda{v_3w_3}\\ &=\lambda(v_1w_1)+\lambda(v_2w_2)+\lambda(v_3w_3)\\ &=\lambda(\boldsymbol{v}\cdot\boldsymbol{w}) \end{align*}

This completes the proof.

Theorem.

# Dot Product Distributes Over Vector Addition

If $\boldsymbol{u}$, $\boldsymbol{v}$ and $\boldsymbol{w}$ are vectors in $\mathbb{R}^n$, then:

\begin{align*} (\boldsymbol{u}+\boldsymbol{v})\cdot\boldsymbol{w}= \boldsymbol{u}\cdot\boldsymbol{w}+\boldsymbol{v}\cdot\boldsymbol{w} \end{align*}

Proof. We will prove the case for $\mathbb{R}^3$ here, but the proof can easily be generalized for $\mathbb{R}^n$. Let vectors $\boldsymbol{u}$, $\boldsymbol{v}$ and $\boldsymbol{w}$ be as follows:

$$\boldsymbol{u}= \begin{pmatrix} u_1\\u_2\\u_3\\ \end{pmatrix},\;\;\;\; \boldsymbol{v}= \begin{pmatrix} v_1\\v_2\\v_3\\ \end{pmatrix},\;\;\;\; \boldsymbol{w}= \begin{pmatrix} w_1\\w_2\\w_3\\ \end{pmatrix}$$

The vector $\boldsymbol{u}+\boldsymbol{v}$ is:

$$\boldsymbol{u}+\boldsymbol{v}= \begin{pmatrix} u_1+v_1\\u_2+v_2\\u_3+v_3\\ \end{pmatrix}$$

The left-hand side $(\boldsymbol{u}+\boldsymbol{v})\cdot\boldsymbol{w}$ is therefore:

\begin{align*} (\boldsymbol{u}+\boldsymbol{v})\cdot\boldsymbol{w}&= \begin{pmatrix} u_1+v_1\\u_2+v_2\\u_3+v_3\\ \end{pmatrix}\cdot \begin{pmatrix} w_1\\w_2\\w_3 \end{pmatrix}\\&= w_1u_1+w_1v_1+ w_2u_2+w_2v_2+ w_3u_3+w_3v_3 \\&=(w_1u_1+w_2u_2+w_3u_3)+(w_1v_1+w_2v_2+w_3v_3) \\&=\boldsymbol{u}\cdot{\boldsymbol{w}}+\boldsymbol{v}\cdot\boldsymbol{w} \end{align*}

This completes the proof.

# Relationship between magnitude and dot product

The following relationships between vector magnitude and dot product is useful for future proofs.

Theorem.

## Dot product of a vector with itself is equal to the square of the vector's magnitude

The magnitude of a vector $\boldsymbol{v}$ can be expressed as dot product like so:

$$\Vert{\boldsymbol{v}}\Vert =(\boldsymbol{v}\cdot\boldsymbol{v})^{1/2}$$

Taking the square on both sides gives:

$$\Vert{\boldsymbol{v}}\Vert^2 =\boldsymbol{v}\cdot\boldsymbol{v}$$

Proof. Consider the following vector:

$$\boldsymbol{v}=\begin{pmatrix} v_1\\ v_2\\ \vdots\\ v_n\\ \end{pmatrix}$$

By definition, the magnitude of $\boldsymbol{v}$ is:

\begin{align*} \Vert\boldsymbol{v}\Vert &=(v_1^2+v_2^2+\cdots+v_n^2)^{1/2}\\ &=\Big((v_1)(v_1)+(v_2)(v_2)+\cdots+(v_n)(v_n)\Big)^{1/2}\\ &=(\boldsymbol{v}\cdot\boldsymbol{v})^{(1/2)} \end{align*}

This completes the proof.

Theorem.

## Expressing dot product of two vectors using magnitude (2)

If $\boldsymbol{v}$ and $\boldsymbol{w}$ are vectors, then:

$$\boldsymbol{v}\cdot{\boldsymbol{w}}= \frac{1}{4}\Vert\boldsymbol{v}+\boldsymbol{w}\Vert^2- \frac{1}{4}\Vert\boldsymbol{v}-\boldsymbol{w}\Vert^2$$

Proof. Let's start from the right-hand side:

\begin{align*} \Vert\boldsymbol{v}+\boldsymbol{w}\Vert^2&= (\boldsymbol{v}+\boldsymbol{w})\cdot(\boldsymbol{v}+\boldsymbol{w})\\ &= (\boldsymbol{v}\cdot\boldsymbol{v})+ 2(\boldsymbol{v}\cdot\boldsymbol{w})+(\boldsymbol{w}\cdot\boldsymbol{w})\\ &= \Vert\boldsymbol{v}\Vert^2+ 2(\boldsymbol{v}\cdot\boldsymbol{w})+ \Vert\boldsymbol{w}\Vert^2\\ \end{align*}

Note that the first step follows from theoremlink.

Similarly, we have that:

\begin{align*} \Vert\boldsymbol{v}-\boldsymbol{w}\Vert^2&= (\boldsymbol{v}-\boldsymbol{w})\cdot(\boldsymbol{v}-\boldsymbol{w})\\ &= (\boldsymbol{v}\cdot\boldsymbol{v})- 2(\boldsymbol{v}\cdot\boldsymbol{w})+(\boldsymbol{w}\cdot\boldsymbol{w})\\ &= \Vert\boldsymbol{v}\Vert^2- 2(\boldsymbol{v}\cdot\boldsymbol{w})+ \Vert\boldsymbol{w}\Vert^2\\ \end{align*}

Therefore:

\begin{align*} \frac{1}{4}\Vert\boldsymbol{v}+\boldsymbol{w}\Vert^2- \frac{1}{4}\Vert\boldsymbol{v}-\boldsymbol{w}\Vert^2&= \frac{1}{4}( \Vert\boldsymbol{v}\Vert^2+ 2(\boldsymbol{v}\cdot\boldsymbol{w})+ \Vert\boldsymbol{w}\Vert^2 )-\frac{1}{4} (\Vert\boldsymbol{v}\Vert^2- 2(\boldsymbol{v}\cdot\boldsymbol{w})+ \Vert\boldsymbol{w}\Vert^2)\\ &=\frac{1}{2}(\boldsymbol{v}\cdot\boldsymbol{w})+ \frac{1}{2}(\boldsymbol{v}\cdot\boldsymbol{w})\\ &=\boldsymbol{v}\cdot\boldsymbol{w} \end{align*}

This completes the proof.

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