A sequence ${\mathbf{v}}_{1},\mathrm{\dots},{\mathbf{v}}_{n}$ of elements of a vector space $V$ is a basis for $V$ if and only if
it is linearly independent, and
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.
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
$${\mathbf{e}}_{1}=\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right),{\mathbf{e}}_{2}=\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right),{\mathbf{e}}_{3}=\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right).$$ |
Then ${\mathbf{e}}_{1},\mathrm{\dots},{\mathbf{e}}_{n}$ 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.
(Linear independence). Suppose ${\sum}_{i=1}^{n}{\lambda}_{i}{\mathbf{e}}_{i}=\mathrm{\U0001d7ce}$. To verify linear independence we have to prove all the ${\lambda}_{i}$ are zero. Using the definition of the ${\mathbf{e}}_{i}$ we get $\left(\begin{array}{c}{\lambda}_{1}\\ \mathrm{\vdots}\\ {\lambda}_{n}\end{array}\right)=\left(\begin{array}{c}0\\ \mathrm{\vdots}\\ 0\end{array}\right)$. So ${\lambda}_{i}=0$ for all $i$ as required.
(Spanning) We have to show that any element of ${\mathbb{F}}^{n}$ is a linear combination of ${\mathbf{e}}_{1},\mathrm{\dots},{\mathbf{e}}_{n}$. Let $\mathbf{v}=\left(\begin{array}{c}{v}_{1}\\ \mathrm{\vdots}\\ {v}_{n}\end{array}\right)\in {\mathbb{F}}^{n}$. 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.
${\mathbb{R}}_{\u2a7d3}[x]$ 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 $a+bx+c{x}^{2}+d{x}^{3}$ 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 $a+bx+c{x}^{2}+d{x}^{3}$ for some $a,b,c,d$, and so is a linear combination of $1,x,{x}^{2},{x}^{3}$.
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
$${E}_{11},{E}_{21},\mathrm{\dots},{E}_{m1},{E}_{12},{E}_{22},\mathrm{\dots}{E}_{mn}$$ |
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.
If ${\mathbf{v}}_{\mathrm{1}}\mathrm{,}\mathrm{\dots}\mathrm{,}{\mathbf{v}}_{n}$ is a basis of $V$, every $\mathbf{v}\mathrm{\in}V$ can be written uniquely as ${\mathrm{\sum}}_{i\mathrm{=}\mathrm{1}}^{n}{\lambda}_{i}\mathit{}{\mathbf{v}}_{i}$ for some scalars ${\lambda}_{i}$.
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,
$$\sum _{i=1}^{n}({\lambda}_{i}-{\mu}_{i}){\mathbf{v}}_{i}={\mathrm{\U0001d7ce}}_{V}.$$ |
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 ${\mathbf{v}}_{1},\mathrm{\dots},{\mathbf{v}}_{n}$. ∎
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 ${\mathbf{v}}_{1},\mathrm{\dots},{\mathbf{v}}_{n}$ is a basis of a vector space $V$ and that
$${\lambda}_{1}{v}_{1}+\mathrm{\cdots}+{\lambda}_{n}{v}_{n}={\mu}_{1}{v}_{1}+\mathrm{\cdots}+{\mu}_{n}{v}_{n}.$$ |
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.
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 $\left(\begin{array}{c}0\\ 1\end{array}\right),\left(\begin{array}{c}1\\ 0\end{array}\right)$ is different to $\left(\begin{array}{c}1\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 1\end{array}\right)$ 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
$$\mathbf{u}=\left(\begin{array}{c}1\\ 1\end{array}\right),\mathbf{v}=\left(\begin{array}{c}1\\ -1\end{array}\right).$$ |
Let’s check this. Suppose $a\mathbf{u}+b\mathbf{v}=0$. Then $\left(\begin{array}{c}a+b\\ a-b\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)$, 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 $\left(\begin{array}{c}x\\ y\end{array}\right)\in {\mathbb{R}}^{2}$. We must show there exist $a,b\in \mathbb{R}$ such that $a\mathbf{u}+b\mathbf{v}=\left(\begin{array}{c}x\\ y\end{array}\right)$. 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 $\left(\begin{array}{c}a\\ b\end{array}\right)=a{\mathbf{e}}_{1}+b{\mathbf{e}}_{2}$ 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).