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# 11.1: Bases in Rⁿ

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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 113

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 $$\Re^{2}$$.