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11: Basis and Dimension

Last updated

Jul 27, 2023
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- 10.4: Review Problems
- 11.1: Bases in Rⁿ

( \newcommand{\kernel}{\mathrm{null}\,}\)

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 $basis$ for $V$ if $S$ is linearly independent and $V = s p a n S$ .
If $S$ is a basis of $V$ and $S$ has only finitely many elements, then we say that $V$ is $finite-dimensional$ .
The number of vectors in $S$ is the $dimension$ of $V$ .

Suppose $V$ is a $finite-dimensional$ 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 $11.1$ :

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

$\begin{matrix} (11.1) & a^{0} 1 + a^{1} t + \dots + a^{n} t^{n}, a^{i} \in ℜ, \end{matrix}$

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

Theorem

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

$\begin{matrix} (11.2) & w = c^{1} v_{1} + \dots + c^{n} v_{n} . \end{matrix}$

Proof

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

Suppose there exists a second set of constants $d^{i}$ such that
$\begin{matrix} (11.3) & w = d^{1} v_{1} + \dots + d^{n} v_{n} . \end{matrix}$ Then:
$\begin{array}{rcl} 0_{V} & = & w - w \\ = & c^{1} v_{1} + \dots + c^{n} v_{n} - d^{1} v_{1} - \dots - d^{n} v_{n} \\ = & (c^{1} - d^{1}) v_{1} + \dots + (c^{n} - d^{n}) v_{n} . \end{array}$

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 $i$ , $c^{i} = d^{i}$ .

Remark

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 $B = (v_{1}, \dots, v_{n})$ and can write

$\begin{matrix} (11.4) & w = (v_{1}, \dots, v_{n}) (\begin{matrix} c^{1} \\ ⋮ \\ c^{n} \end{matrix}) = {(\begin{matrix} c^{1} \\ ⋮ \\ c^{n} \end{matrix})}_{B} . \end{matrix}$

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 $ℜ^{n}$ . But first, we need to show that any two bases for a finite-dimensional vector space has the same number of vectors.

Lemma

If $S = {v_{1}, \dots, v_{n}}$ is a basis for a vector space $V$ and $T = {w_{1}, \dots, w_{m}}$ 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 ${w_{1}, v_{1}, \dots, v_{n}}$ 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 $S_{1} = {w_{1}, v_{1}, \dots, v_{i - 1}, v_{i + 1}, \dots, v_{n}}$ 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

$\begin{matrix} (11.5) & v_{k} = c^{0} w_{1} + c^{1} v_{1} + \dots + c^{i - 1} v_{i - 1} + c^{i + 1} v_{i + 1} + \dots + c^{n} v_{n} . \end{matrix}$

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 $S_{m} = {w_{1}, \dots, w_{m}$ , $v_{i_{1}}, \dots, v_{i_{n - m}}}$ , which is fine.

Otherwise, we have $m > n$ , and the set $S_{n} = {w_{1}, \dots, w_{n}}$ 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.

$◻$

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

Proof
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)

Remark

Lemma

Contributor

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