# 4.8 Bases

## 4.8.1 Basis definition

###### Definition 4.8.1.

A sequence $\mathbf{v}_{1},\ldots,\mathbf{v}_{n}$ of elements of a vector space $V$ is a basis for $V$ if and only if

1. 1.

it is linearly independent, and

2. 2.

it is a spanning sequence for $V$.

Importantly, bases are sequences not sets. This is because the order of a basis matters to some of the definitions we will make later, like the matrix of a linear map.

## 4.8.2 The standard basis for $\mathbb{F}^{n}$

The most important example is the standard basis of $\mathbb{F}^{n}$ (no matter which field $\mathbb{F}$ is). Let $\mathbf{e}_{i}$ be the column vector in $\mathbb{F}^{n}$ with a 1 in position $i$ and 0s elsewhere. When $n=3$, for example, we have

 $\mathbf{e}_{1}=\begin{pmatrix}1\\ 0\\ 0\end{pmatrix},\mathbf{e}_{2}=\begin{pmatrix}0\\ 1\\ 0\end{pmatrix},\mathbf{e}_{3}=\begin{pmatrix}0\\ 0\\ 1\end{pmatrix}.$

Then $\mathbf{e}_{1},\ldots,\mathbf{e}_{n}$ is a basis of $\mathbb{F}^{n}$, called the standard basis. To check this, we must verify the two parts of the definition of basis.

1. 1.

(Linear independence). Suppose $\sum_{i=1}^{n}\lambda_{i}\mathbf{e}_{i}=\mathbf{0}$. To verify linear independence we have to prove all the $\lambda_{i}$ are zero. Using the definition of the $\mathbf{e}_{i}$ we get $\begin{pmatrix}\lambda_{1}\\ \vdots\\ \lambda_{n}\end{pmatrix}=\begin{pmatrix}0\\ \vdots\\ 0\end{pmatrix}$. So $\lambda_{i}=0$ for all $i$ as required.

2. 2.

(Spanning) We have to show that any element of $\mathbb{F}^{n}$ is a linear combination of $\mathbf{e}_{1},\ldots,\mathbf{e}_{n}$. Let $\mathbf{v}=\begin{pmatrix}v_{1}\\ \vdots\\ v_{n}\end{pmatrix}\in\mathbb{F}^{n}$. Then $\mathbf{v}=\sum_{i=1}^{n}\mathbf{v}_{i}\mathbf{e}_{i}$, so $\mathbf{v}$ is a linear combination of the $\mathbf{e}_{i}$ as required.

## 4.8.3 More basis examples

###### Example 4.8.1.

$\mathbb{R}_{\leqslant 3}[x]$ consists of all polynomials of degree at most 3 in the variable $x$. It has a basis $1,x,x^{2},x^{3}$, because

• (linear independence) if $a+bx+cx^{2}+dx^{3}$ is the zero polynomial, that is if it is zero for every value of $x$, then $a=b=c=d=0$. This is because a polynomial of degree $m$ has at most $m$ roots.

• (spanning) every polynomial of degree at most 3 has the form $a+bx+cx^{2}+dx^{3}$ for some $a,b,c,d$, and so is a linear combination of $1,x,x^{2},x^{3}$.

###### Example 4.8.2.

Let $V=M_{m\times n}(\mathbb{F})$ be the $\mathbb{F}$-vector space of all $m\times n$ matrices. Let $E_{ij}$ be the matrix which as a 1 in position $i,j$ and zeroes elsewhere. Then

 $E_{11},E_{21},\ldots,E_{m1},E_{12},E_{22},\ldots E_{mn}$

is a basis for $V$. This can be proved in exactly the same way as we proved that the standard basis of $\mathbb{F}^{n}$ really was a basis.

## 4.8.4 What is a basis good for?

###### Lemma 4.8.1.

If $\mathbf{v}_{1},\ldots,\mathbf{v}_{n}$ is a basis of $V$, every $\mathbf{v}\in V$ can be written uniquely as $\sum_{i=1}^{n}\lambda_{i}\mathbf{v}_{i}$ for some scalars $\lambda_{i}$.

###### Proof.

Every $\mathbf{v}\in V$ can be written this way because the $\mathbf{v}_{i}$ are a basis and hence a spanning sequence for $V$. The problem is to prove that every $\mathbf{v}\in V$ can be written like this in only one way.

Suppose that

 $\sum_{i=1}^{n}\lambda_{i}\mathbf{v}_{i}=\sum_{i=1}^{n}\mu_{i}\mathbf{v}_{i}.$

Then subtracting one side from the other,

 $\sum_{i=1}^{n}(\lambda_{i}-\mu_{i})\mathbf{v}_{i}=\mathbf{0}_{V}.$

Linear independence of the $\mathbf{v}_{i}$ tells us $\lambda_{i}-\mu_{i}=0$ for all $i$, so $\lambda_{i}=\mu_{i}$ for all $i$. We have proved that there is only one expression for $\mathbf{v}$ as a linear combination of the elements of the basis $\mathbf{v}_{1},\ldots,\mathbf{v}_{n}$. ∎

This means that a basis gives a way of giving coordinates to an arbitrary vector space, no matter what the elements look like. Once we fix a basis of $V$, there is a one-one correspondence between the elements of $V$ and the coefficients needed to express them in terms of that basis — you could call these the coordinates of the vector in terms of this basis.

A basis also allows us to compare coefficients. Suppose $\mathbf{v}_{1},\ldots,\mathbf{v}_{n}$ is a basis of a vector space $V$ and that

 $\lambda_{1}v_{1}+\cdots+\lambda_{n}v_{n}=\mu_{1}v_{1}+\cdots+\mu_{n}v_{n}.$

Then the uniqueness result Lemma 4.8.1 tells us we can compare coefficients to get that $\lambda_{1}=\mu_{1}$, $\lambda_{2}=\mu_{2}$, and so on.

## 4.8.5 Multiple bases for the same vector space

A given vector space can have many different bases. This is true in a trivial sense: as we saw before, basis are sequences, the order matters, so $\begin{pmatrix}0\\ 1\end{pmatrix},\begin{pmatrix}1\\ 0\end{pmatrix}$ is different to $\begin{pmatrix}1\\ 0\end{pmatrix},\begin{pmatrix}0\\ 1\end{pmatrix}$ but clearly still a basis of $\mathbb{R}^{2}$. But it is also true in a more interesting way. Take $\mathbb{R}^{2}$, for example: we know $\mathbf{e}_{1},\mathbf{e}_{2}$ is a basis, but so also is

 $\mathbf{u}=\begin{pmatrix}1\\ 1\end{pmatrix},\mathbf{v}=\begin{pmatrix}1\\ -1\end{pmatrix}.$

Let’s check this. Suppose $a\mathbf{u}+b\mathbf{v}=0$. Then $\begin{pmatrix}a+b\\ a-b\end{pmatrix}=\begin{pmatrix}0\\ 0\end{pmatrix}$, so $a+b=0=a-b$ from which it follows $a=b=0$ and $\mathbf{u},\mathbf{v}$ is linearly independent. To show $\mathbf{u},\mathbf{v}$ spans $\mathbb{R}^{2}$, let $\begin{pmatrix}x\\ y\end{pmatrix}\in\mathbb{R}^{2}$. We must show there exist $a,b\in\mathbb{R}$ such that $a\mathbf{u}+b\mathbf{v}=\begin{pmatrix}x\\ y\end{pmatrix}$. The condition $a$ and $b$ must satisfy is $a+b=x,a-b=y$. It is always possible to find such $a$ and $b$: solving the equations you get $a=(x+y)/2,b=(x-y)/2$, so $\mathbf{u},\mathbf{v}$ spans $\mathbb{R}^{2}$.

Here’s why a vector space having several different bases is useful. The expression of an element $\mathbf{v}$ in terms of different bases can tell us different things about $\mathbf{v}$. In other words, different bases give different ways of looking at the elements of the vector space.

Say for example you are representing an image as an element of $\mathbb{R}^{n}$. The smallest possible example is a 2-pixel image which we could represent as an element $\begin{pmatrix}a\\ b\end{pmatrix}=a\mathbf{e}_{1}+b\mathbf{e}_{2}$ in $\mathbb{R}^{2}$, where the first coordinate tells me how bright the first pixel is and the second tells me how bright the second is.

Now consider the alternative basis $\mathbf{e}_{1}+\mathbf{e}_{2},\mathbf{e}_{1}-\mathbf{e}_{2}$. Any image $a\mathbf{e}_{1}+b\mathbf{e}_{2}$ can be re-written in terms of the new basis:

 $a\mathbf{e}_{1}+b\mathbf{e}_{2}=\frac{a+b}{2}(\mathbf{e}_{1}+\mathbf{e}_{2})+% \frac{a-b}{2}(\mathbf{e}_{1}-\mathbf{e}_{2}).$

So the new basis is giving us a different description of the image. It tells us how bright the image is overall (the coefficient $(a+b)/2$ of $\mathbf{e}_{1}+\mathbf{e}_{2}$ is the average brightness of the two pixels, so it measures the overall image brightness) and how different in brightness the two pixels are (the coefficient $(a-b)/2$ of $\mathbf{e}_{1}-\mathbf{e}_{2}$ is a measure of how different the brightnesses $a$ and $b$ of the two pixels are).