11.1: Bases in Rⁿ

In review question 2, chapter 10 you checked that

$$\Re^{n} = span \left\{ \begin{pmatrix}1\\0\\ \vdots \\ 0\end{pmatrix}, \begin{pmatrix}0\\1\\ \vdots \\ 0\end{pmatrix}, \ldots, \begin{pmatrix}0\\0\\ \vdots \\ 1\end{pmatrix}\right\},$$

and that this set of vectors is linearly independent. (If you didn't do that problem, check this before reading any further!) So this set of vectors is a basis for $\Re^{n}$, and $\dim \Re^{n}=n$. This basis is often called the $\textit{standard}$ or $\textit{canonical basis}$ for $\Re^{n}$. The vector with a one in the $i$th position and zeros everywhere else is written $e_{i}$. (You could also view it as the function $\{1,2,\ldots,n\}\to \mathbb{R}$ where $e_{i}(j)=1$ if $i=j$ and $0$ if $i\neq j$.) It points in the direction of the $i^{th}$ coordinate axis, and has unit length. In multivariable calculus classes, this basis is often written $\{ i, j, k \}$ for $\Re^{3}$.

Note that it is often convenient to order basis elements, so rather than writing a set of vectors, we would write a list. This is called an ordered basis. For example, the canonical ordered basis for $\mathbb{R^{n}}$ is $(e_{1},e_{2},\ldots,e_{n})$. The possibility to reorder basis vectors is not the only way in which bases are non-unique:

Remark (Bases are not unique)

While there exists a unique way to express a vector in terms of any particular basis, bases themselves are far from unique.

For example, both of the sets:

$$\left\{ \begin{pmatrix}1\\0\end{pmatrix}, \begin{pmatrix}0\\1\end{pmatrix} \right\} \textit{ and } \left\{ \begin{pmatrix}1\\1\end{pmatrix}, \begin{pmatrix}1\\-1\end{pmatrix} \right\}$$

are bases for $\Re^{2}$. Rescaling any vector in one of these sets is already enough to show that $\Re^{2}$ has infinitely many bases. But even if we require that all of the basis vectors have unit length, it turns out that there are still infinitely many bases for $\Re^{2}$ (see review question 3).

To see whether a collection of vectors $S=\{v_{1}, \ldots, v_{m} \}$ is a basis for $\Re^{n}$, we have to check that they are linearly independent and that they span $\Re^{n}$. From the previous discussion, we also know that $m$ must equal $n$, so lets assume $S$ has $n$ vectors. If $S$ is linearly independent, then there is no non-trivial solution of the equation

$$0 = x^{1}v_{1}+\cdots + x^{n}v_{n}.$$

Let $M$ be a matrix whose columns are the vectors $v_{i}$ and $X$ the column vector with entries $x^{i}$. Then the above equation is equivalent to requiring that there is a unique solution to

$$MX=0\, .$$

To see if $S$ spans $\Re^{n}$, we take an arbitrary vector $w$ and solve the linear system

$$w=x^{1}v_{1}+\cdots + x^{n}v_{n}$$

in the unknowns $x^{i}$. For this, we need to find a unique solution for the linear system $MX=w$.

Thus, we need to show that $M^{-1}$ exists, so that

$$X=M^{-1}w$$

is the unique solution we desire. Then we see that $S$ is a basis for $V$ if and only if $\det M\neq 0$.

Theorem

Let $S=\{v_{1}, \ldots, v_{m} \}$ be a collection of vectors in $\Re^{n}$. Let $M$ be the matrix whose columns are the vectors in $S$. Then $S$ is a basis for $V$ if and only if $m$ is the dimension of $V$ and

$$\det M \neq 0.$$

Remark

Also observe that $S$ is a basis if and only if ${\rm RREF}(M)=I$.

Example $\PageIndex{1}$:

Let
$$S=\left\{ \begin{pmatrix}1\\0\end{pmatrix}, \begin{pmatrix}0\\1\end{pmatrix} \right\} \textit{ and } T=\left\{ \begin{pmatrix}1\\1\end{pmatrix}, \begin{pmatrix}1\\-1\end{pmatrix} \right\}.$$
Then set $M_{S}=\begin{pmatrix} 1 & 0\\ 0 & 1\\ \end{pmatrix}$. Since $\det M_{S}=1\neq 0$, then $S$ is a basis for $\Re^{2}$.\\
Likewise, set $M_{T}=\begin{pmatrix} 1 & 1\\ 1 & -1\\ \end{pmatrix}$. Since $\det M_{T}=-2\neq 0$, then $T$ is a basis for $\Re^{2}$.

Contributor

• David Cherney, Tom Denton, and Andrew Waldron (UC Davis)