search
Search
Login
Map of Data Science
menu
menu
search toc
Thanks for the thanks!
close
Comments
Log in or sign up
Cancel
Post
account_circle
Profile
exit_to_app
Sign out
search
keyboard_voice
close
Searching Tips
Search for a recipe:
"Creating a table in MySQL"
Search for an API documentation: "@append"
Search for code: "!dataframe"
Apply a tag filter: "#python"
Useful Shortcuts
/ to open search panel
Esc to close search panel
to navigate between search results
d to clear all current filters
Enter to expand content preview
icon_star
Doc Search
icon_star
Code Search Beta
SORRY NOTHING FOUND!
mic
Start speaking...
Voice search is only supported in Safari and Chrome.
Navigate to

Guide on Basis Vectors

schedule Mar 5, 2023
Last updated
local_offer
Linear Algebra
Tags
map
Check out the interactive map of data science
Definition.

Basis for a vector space

A basis for a vector space $V$ is a set of vectors that:

Example.

Standard basis for R2

Consider the following vectors:

$$\boldsymbol{e}_1=\begin{pmatrix} 1\\0\\ \end{pmatrix},\;\;\;\;\; \boldsymbol{e}_2=\begin{pmatrix} 0\\1\\ \end{pmatrix}$$

Show that these vectors are basis vectors of $\mathbb{R}^2$.

Solution. We first need to show that these vectors span $\mathbb{R}^2$, which means that every vector in $\mathbb{R}^2$ must be expressible using a linear combination of $\boldsymbol{e}_1$ and $\boldsymbol{e}_2$. Let $\boldsymbol{x}$ be any vector in $\mathbb{R}^2$ defined as:

$$\boldsymbol{x}= \begin{pmatrix} x_1\\x_2 \end{pmatrix}$$

This vector can be expressed using $\boldsymbol{e}_1$ and $\boldsymbol{e}_2$ like so:

$$\begin{align*} \begin{pmatrix} x_1\\x_2 \end{pmatrix} &= x_1\begin{pmatrix} 1\\0 \end{pmatrix}+ x_2\begin{pmatrix} 0\\1 \end{pmatrix}\\ &= x_1\boldsymbol{e}_1+ x_2\boldsymbol{e}_1 \end{align*}$$

This means that any vector $\boldsymbol{x}$ in $\mathbb{R}^2$ can be expressed as a linear combination of $\boldsymbol{e}_1$ and $\boldsymbol{e}_2$, and so $\boldsymbol{e}_1$ and $\boldsymbol{e}_2$ span $\mathbb{R}^2$ by definition.

Next, we must show that $\boldsymbol{e}_1$ and $\boldsymbol{e}_2$ are linearly independent. This is obvious because $\boldsymbol{e}_1$ cannot be expressed as some scalar multiple of $\boldsymbol{e}_2$.

Since $\boldsymbol{e}_1$ and $\boldsymbol{e}_2$ span $\mathbb{R}^2$ and are linearly independent, they are basis vectors of $\mathbb{R}^2$. Since they are considered to be the simplest basis vectors of $\mathbb{R}^2$, they are referred to as the standard basis vectors of $\mathbb{R}^2$.

Example.

Non-standard basis for R2

Consider the following two vectors:

$$\boldsymbol{v}_1=\begin{pmatrix} 1\\2\\ \end{pmatrix},\;\;\;\;\; \boldsymbol{v}_2=\begin{pmatrix} 2\\6\\ \end{pmatrix}$$

Show that these vectors form a basis for $\mathbb{R}^2$.

Solution. Let's first show that $\boldsymbol{v}_1$ and $\boldsymbol{v}_2$ span $\mathbb{R}^2$. Let $\boldsymbol{x}$ be any vector in $\mathbb{R}^2$ defined like so:

$$\boldsymbol{x}= \begin{pmatrix} x_1\\x_2 \end{pmatrix}$$

Our goal is to show that there exist constants $c_1$ and $c_2$ such that:

$$\begin{pmatrix} x_1\\x_2 \end{pmatrix}=c_1\begin{pmatrix} 1\\2\\ \end{pmatrix}+ c_2\begin{pmatrix} 2\\3\\ \end{pmatrix}$$

This corresponds to solving the following linear system:

$$\begin{equation}\label{eq:jlmhCMRThDPFEAkUl56} \begin{pmatrix} x_1\\x_2 \end{pmatrix}= \begin{pmatrix} 1&2\\2&3 \end{pmatrix} \begin{pmatrix}c_1\\c_2 \end{pmatrix} \end{equation}$$

The row echelon form of the coefficient matrix is:

$$\begin{pmatrix} 1&2\\2&3 \end{pmatrix}\sim \begin{pmatrix} 1&2\\0&1 \end{pmatrix}\sim \begin{pmatrix} 1&0\\0&1 \end{pmatrix} $$

By theoremlink, we know that because the row echelon form of the coefficient matrix does not contain a row with all zeros, we conclude that there exist scalars $c_1$ and $c_2$ to give us $\boldsymbol{x}$ in \eqref{eq:jlmhCMRThDPFEAkUl56}.

Next, we must show that $\boldsymbol{v}_1$ and $\boldsymbol{v}_2$ are linearly independent. This is easy because the vectors clearly cannot be expressed as a scalar multiple of the other.

Because vectors $\boldsymbol{v}_1$ and $\boldsymbol{v}_2$ span $\mathbb{R}^2$ and are linearly independent, they are indeed a basis for $\mathbb{R}^2$. This example demonstrates that basis for $\mathbb{R}^2$ is not unique - in fact, there are infinitely many pairs of vectors that span $\mathbb{R}^2$ and are linearly independent. Typically, we use the standard basis vectorslink because they are easy to work with.

Example.

Vectors that do not form a basis for R2

Consider the following two vectors:

$$\boldsymbol{v}_1=\begin{pmatrix} 1\\2\\ \end{pmatrix},\;\;\;\;\; \boldsymbol{v}_2=\begin{pmatrix} 3\\6\\ \end{pmatrix}$$

Show that these vectors do not form a basis for $\mathbb{R}^2$.

Solution. We can easily see that $\boldsymbol{v}_2=3\boldsymbol{v}_1$. This means that:

  • $\boldsymbol{v}_1$ and $\boldsymbol{v}_2$ are linearly dependent.

  • since $\boldsymbol{v}_2$ can be expressed as a linear combination of $\boldsymbol{v}_1$, and so the spanning set traces out a line instead of the entire $\mathbb{R}^2$.

Note that both criteria are not satisfied in this case but only one is sufficient to claim that the vectors do not form a basis for $\mathbb{R}^2$.

Let's now look at examples where one criterion is satisfied while the other is not.

Example.

Case when set is not linearly independent but spans vector space

Consider the following vectors:

$$\boldsymbol{v}_1=\begin{pmatrix} 1\\2\\ \end{pmatrix},\;\;\;\;\; \boldsymbol{v}_2=\begin{pmatrix} 2\\4\\ \end{pmatrix},\;\;\;\;\; \boldsymbol{v}_3=\begin{pmatrix} 4\\4\\ \end{pmatrix}$$

Show that these vectors do not form a basis for $\mathbb{R}^2$.

Solution. We see that $\boldsymbol{v}_2=2\boldsymbol{v}_1$ so the set of vectors is linearly dependent. This already means that the vectors do not form a basis for $\mathbb{R}^2$. Let's check whether or not the set spans $\mathbb{R}^2$. Let's focus on $\boldsymbol{v}_2$ and $\boldsymbol{v}_3$ and obtain their row echelon form:

$$\begin{pmatrix} 2&4\\4&4 \end{pmatrix}\sim \begin{pmatrix} 1&2\\1&1 \end{pmatrix} \sim \begin{pmatrix} 1&2\\0&1 \end{pmatrix} $$

By theoremlink, because the row echelon form of the coefficient matrix does not have a row with all zeros, any vector in $\mathbb{R}^2$ can be expressed as a linear combination of $\boldsymbol{v}_2$ and $\boldsymbol{v}_3$. Therefore, the set of vectors spans $\mathbb{R}^2$.

Example.

Case when set is linearly independent but does not span vector space

Consider the following vector:

$$\boldsymbol{v}=\begin{pmatrix} 1\\2\\ \end{pmatrix}$$

Show that this vector does not form a basis for $\mathbb{R}^2$.

Solution. Since we only have one vector, this vector must be linearly independent. However, $\boldsymbol{v}$ clearly does not span $\mathbb{R}^2$ because some vectors in $\mathbb{R}^2$ cannot be expressed as a scalar multiple of $\boldsymbol{v}$. Therefore, $\boldsymbol{v}$ does not form a basis for $\mathbb{R}^2$.

We have already covered the standard basis of $\mathbb{R}^2$ in the earlier sectionlink. We can also generalize the idea to $\mathbb{R}^n$.

Definition.

Standard basis of Rn

The standard basis vectors of $\mathbb{R}^n$ is defined as a set containing the following $n$ vectors:

$$\left\{ \begin{pmatrix} 1\\0\\\vdots\\0 \end{pmatrix}, \begin{pmatrix} 0\\1\\\vdots\\0 \end{pmatrix},\;\cdots\;, \begin{pmatrix} 0\\0\\\vdots\\1 \end{pmatrix} \right\}$$
robocat
Published by Isshin Inada
Edited by 0 others
Did you find this page useful?
thumb_up
thumb_down
Comment
Citation
Ask a question or leave a feedback...