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16: Basis and Dimension

Last updated

Oct 19, 2022
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- 15.4: Review Problems
- 16.1: Bases in Rⁿ

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In chapter 10, the notions of a linearly independent set of vectors in a vector space $V$ , and of a set of vectors that span $V$ were established: Any set of vectors that span $V$ can be reduced to some minimal collection of linearly independent vectors; such a set is called a \emph{basis} of the subspace $V$ .

Definitions

Let $V$ be a vector space.

Then a set $S$ is a $\textit{basis}$ for $V$ if $S$ is linearly independent and $V = span S$ .
If $S$ is a basis of $V$ and $S$ has only finitely many elements, then we say that $V$ is $\textit{finite-dimensional}$ .
The number of vectors in $S$ is the $\textit{dimension}$ of $V$ .

Suppose $V$ is a $\textit{finite-dimensional}$ vector space, and $S$ and $T$ are two different bases for $V$ . One might worry that $S$ and $T$ have a different number of vectors; then we would have to talk about the dimension of $V$ in terms of the basis $S$ or in terms of the basis $T$ . Luckily this isn't what happens. Later in this chapter, we will show that $S$ and $T$ must have the same number of vectors. This means that the dimension of a vector space is basis-independent. In fact, dimension is a very important characteristic of a vector space.

Example 112

$P_{n}(t)$ (polynomials in $t$ of degree $n$ or less) has a basis $\{1,t,\ldots , t^{n} \}$ , since every vector in this space is a sum

$a^{0}\,1+a^{1}\,t+\cdots +a^{n}\,t^{n}, \qquad a^{i}\in \Re\, ,$

so $P_{n}(t)=span \{1,t,\ldots , t^{n} \}$ . This set of vectors is linearly independent: If the polynomial $p(t)=c^{0}1+c^{1}t+\cdots +c^{n}t^{n}=0$ , then $c^{0}=c^{1}=\cdots =c^{n}=0$ , so $p(t)$ is the zero polynomial. Thus $P_{n}(t)$ is finite dimensional, and $\dim P_{n}(t)=n+1$ .

Theorem

Let $S=\{v_{1}, \ldots, v_{n} \}$ be a basis for a vector space $V$ . Then every vector $w \in V$ can be written $\textit{uniquely}$ as a linear combination of vectors in the basis $S$ :

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

Proof

Since $S$ is a basis for $V$ , then $span S=V$ , and so there exist constants $c^{i}$ such that $w=c^{1}v_{1}+\cdots + c^{n}v_{n}$ .

Suppose there exists a second set of constants $d^{i}$ such that
$w=d^{1}v_{1}+\cdots + d^{n}v_{n}\, .$ Then:
$\begin{eqnarray*} 0_{V}&=&w-w\\ &=&c^{1}v_{1}+\cdots + c^{n}v_{n}-d^{1}v_{1}-\cdots - d^{n}v_{n} \\ &=&(c^{1}-d^{1})v_{1}+\cdots + (c^{n}-d^{n})v_{n}. \\ \end{eqnarray*}$

If it occurs exactly once that $c^{i}\neq d^{i}$ , then the equation reduces to $0=(c^{i}-d^{i})v_{i}$ , which is a contradiction since the vectors $v_{i}$ are assumed to be non-zero.

If we have more than one $i$ for which $c^{i}\neq d^{i}$ , we can use this last equation to write one of the vectors in $S$ as a linear combination of other vectors in $S$ , which contradicts the assumption that $S$ is linearly independent. Then for every $i$ , $c^{i}=d^{i}$ .

This theorem is the one that makes bases so useful--they allow us to convert abstract vectors into column vectors. By ordering the set $S$ we obtain $B=(v_{1},\ldots,v_{n})$ and can write

$w=(v_{1},\ldots,v_{n}) \begin{pmatrix}c^{1}\\ \vdots\\ c^{n}\end{pmatrix}=\begin{pmatrix}c^{1}\\ \vdots\\ c^{n}\end{pmatrix}_{B}\, .$

Remember that in general it makes no sense to drop the subscript $B$ on the column vector on the right--most vector spaces are not made from columns of numbers!

Next, we would like to establish a method for determining whether a collection of vectors forms a basis for $\Re^{n}$ . But first, we need to show that any two bases for a finite-dimensional vector space has the same number of vectors.

If $S=\{v_{1}, \ldots, v_{n} \}$ is a basis for a vector space $V$ and $T=\{w_{1}, \ldots, w_{m} \}$ is a linearly independent set of vectors in $V$ , then $m\leq n$ .

$indep_span.jpg$

The idea of the proof is to start with the set $S$ and replace vectors in $S$ one at a time with vectors from $T$ , such that after each replacement we still have a basis for $V$ .

Proof

Since $S$ spans $V$ , then the set $\{w_{1}, v_{1}, \ldots, v_{n} \}$ is linearly dependent. Then we can write $w_{1}$ as a linear combination of the $v_{i}$ ; using that equation, we can express one of the $v_{i}$ in terms of $w_{1}$ and the remaining $v_{j}$ with $j\neq i$ . Then we can discard one of the $v_{i}$ from this set to obtain a linearly independent set that still spans $V$ . Now we need to prove that $S_{1}$ is a basis; we must show that $S_{1}$ is linearly independent and that $S_{1}$ spans $V$ .

The set $S_{1}=\{w_{1}, v_{1}, \ldots, v_{i-1}, v_{i+1},\ldots, v_{n} \}$ is linearly independent: By the previous theorem, there was a unique way to express $w_{1}$ in terms of the set $S$ . Now, to obtain a contradiction, suppose there is some $k$ and constants $c^{i}$ such that

$v_{k} = c^{0}w_{1}+c^{1}v_{1}+\cdots + c^{i-1}v_{i-1} + c^{i+1}v_{i+1} + \cdots + c^{n}v_{n}.$

Then replacing $w_{1}$ with its expression in terms of the collection $S$ gives a way to express the vector $v_{k}$ as a linear combination of the vectors in $S$ , which contradicts the linear independence of $S$ . On the other hand, we cannot express $w_{1}$ as a linear combination of the vectors in $\{v_{j} | j\neq i\}$ , since the expression of $w_{1}$ in terms of $S$ was unique, and had a non-zero coefficient for the vector $v_{i}$ . Then no vector in $S_{1}$ can be expressed as a combination of other vectors in $S_{1}$ , which demonstrates that $S_{1}$ is linearly independent.

The set $S_{1}$ spans $V$ : For any $u\in V$ , we can express $u$ as a linear combination of vectors in $S$ . But we can express $v_{i}$ as a linear combination of vectors in the collection $S_{1}$ ; rewriting $v_{i}$ as such allows us to express $u$ as a linear combination of the vectors in $S_{1}$ . Thus $S_{1}$ is a basis of $V$ with $n$ vectors.

We can now iterate this process, replacing one of the $v_{i}$ in $S_{1}$ with $w_{2}$ , and so on. If $m\leq n$ , this process ends with the set $S_{m}=\{w_{1},\ldots, w_{m}$ , $v_{i_1},\ldots,v_{i_{n-m}} \}$ , which is fine.

Otherwise, we have $m>n$ , and the set $S_{n}=\{w_{1},\ldots, w_{n} \}$ is a basis for $V$ . But we still have some vector $w_{n+1}$ in $T$ that is not in $S_{n}$ . Since $S_{n}$ is a basis, we can write $w_{n+1}$ as a combination of the vectors in $S_{n}$ , which contradicts the linear independence of the set $T$ . Then it must be the case that $m\leq n$ , as desired.

$\square$

For a finite-dimensional vector space $V$ , any two bases for $V$ have the same number of vectors.

Let $S$ and $T$ be two bases for $V$ . Then both are linearly independent sets that span $V$ . Suppose $S$ has $n$ vectors and $T$ has $m$ vectors. Then by the previous lemma, we have that $m\leq n$ . But (exchanging the roles of $S$ and $T$ in application of the lemma) we also see that $n\leq m$ . Then $m=n$ , as desired.

Contributor

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

Thumbnail: A linear combination of one basis set of vectors (purple) obtains new vectors (red). If they are linearly independent, these form a new basis set. The linear combinations relating the first set to the other extend to a linear transformation, called the change of basis. (CC0; Maschen via Wikipedia)

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