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Linear Algebra
52 guides
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1. Vectors
2. Matrices
3. Linear equations
4. Matrix determinant
5. Vector space
6. Special matrices
7. Eigenvalues and Eigenvectors
8. Orthogonality
9. Matrix decomposition
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# Guide on Basis Vectors

schedule Mar 5, 2023
Last updated
local_offer
Linear Algebra
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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:

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

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:

$$\left\{ \begin{pmatrix} 1\\0\\\vdots\\0 \end{pmatrix}, \begin{pmatrix} 0\\1\\\vdots\\0 \end{pmatrix},\;\cdots\;, \begin{pmatrix} 0\\0\\\vdots\\1 \end{pmatrix} \right\}$$
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