# Basis

A **basis** of a vector space is a set of vectors in that space that can be used as coordinates for it. The two conditions such a set must satisfy in order to be considered a basis are

- the set must
*span*the vector space; - the set must be
*linearly independent*.

A set that satisfies these two conditions has the property that each vector may be expressed as a finite sum of multiples of basis elements in precisely one way (up to reordering). The "multiples" of the basis elements may be thought of as the *coordinates* of the vector, with respect to that basis.

Bases (the plural of basis) are used to translate the language of linear algebra into the language of matrices. They are solely responsible for the connection between linear transformations and matrices, the usual interpretation of linear transformations. The properties of bases provide the framework for a variety of important properties of vector spaces and linear transformations, like dimension and rank.

The so-called

elementary vectorsdemonstrate the connection between coordinates and bases. In, for instance, four-dimensional Euclidean space $\big($the vector space $\mathbb{R}^4\big),$ the vectors $(1,\,0,\,0,\,0),\ (0,\,1,\,0,\,0),\ (0,\,0,\,1,\,0),\ \text{and}\ (0,\,0,\,0,\,1)$ form a basis for $\mathbb{R}^4$. There is only one way to write any element of $\mathbb{R}^4$ as a sum of real multiples of those four elements; for instance, $\big(7,\,-\sqrt{2},\,\pi,\,0\big) = 7 \cdot (1,\,0,\,0,\,0) - \sqrt{2} \cdot (0,\,1,\,0,\,0) + \pi \cdot (0,\,0,\,1,\,0) + 0 \cdot (0,\,0,\,0,\,1).$

## Span

The span of a set $S$ of vectors seeks to describe the set of all possible vectors that could be reached by performing the usual vector space operations on vectors in $S$. It turns out that this "span" is a vector space itself.

A set $S \subset V$ of vectors (non-uniquely) defines a subspace of $V$ by creating all possible linear combinations of the elements of $S$. So, $\text{Span}(S) = \{a_1 v_1 + \dots + a_n v_n \mid n \in \mathbb{N},\, a_i \in \mathbb{F} \text{ for each } i\}.$

## Linear Independence

Main Article: Linear Independence

The notion of linear independence encapsulates the notion of "redundancy" in the creation of a set's span. If a set $S$ of vectors could span the same space with one less element, then it is known as linearly dependent; if it could not span the same space with one less element, then it is known as linearly independent. Equivalently, the zero-vector $\textbf{0}$ may be expressed as a linear combination of elements of $S$ in precisely one way: with coefficients of zero.

The vector space of subtractive color mixing is created from real multiples of vectors $\color{#D61F06}{\text{red}}$, $\color{#EC7300}{\text{yellow}}$, and $\color{#3D99F6}{\text{blue}}$ with the understanding that $\color{#D61F06}{\text{red}} + \color{#EC7300}{\text{yellow}} + \color{#3D99F6}{\text{blue}} = \textbf{0}$. (When mixing colors, red, yellow, and blue combine to make black.)

For instance, some elements of the vector space are $\sqrt{2} \color{#D61F06}{\text{red}} + \pi \color{#EC7300}{\text{yellow}}\ \text{ and } -3 \color{#D61F06}{\text{red}} + 29 \color{#EC7300}{\text{yellow}} + e^3 \color{#3D99F6}{\text{blue}}.$ However, $3 \color{#D61F06}{\text{red}} + 2 \color{#3D99F6}{\text{blue}} \text{ and } \color{#D61F06}{\text{red}} - 2 \color{#EC7300}{\text{yellow}} \text{ are the same vector},$ since they differ by a multiple of $\textbf{0}$. $($They differ by $2 \cdot ({\color{#D61F06}{\text{red}} + \color{#EC7300}{\text{yellow}} + \color{#3D99F6}{\text{blue}}) = \textbf{0}.})$

What is true of the following sets of vectors in this vector space?

- $\{ \color{#D61F06}{\text{red}} + \color{#EC7300}{\text{yellow}} + \color{#3D99F6}{\text{blue}},\, \color{#D61F06}{\text{red}} + 2 \color{#EC7300}{\text{yellow}} + 4 \color{#3D99F6}{\text{blue}} \}$
- $\{ \color{#D61F06}{\text{red}},\, \color{#EC7300}{\text{yellow}},\, \color{#3D99F6}{\text{blue}} \}$
- $\{ \color{#D61F06}{\text{red}} + \color{#EC7300}{\text{yellow}},\, \color{#EC7300}{\text{yellow}} + \color{#3D99F6}{\text{blue}} \}$

Let $S \subset V$ be a set that spans $V$ and is linearly independent. Then, any vector in $V$ may be expressed as a linear combination of elements of $S$ in only one way.

By definition of span, any vector in $\text{Span}(S) = V$ may be expressed as a linear combination of elements of $S$.

By definition of linear independence, $\textbf{0}$ may be expressed as a linear combination of elements of $S$ in only one way. Now suppose a vector $u \in V$ can be expressed as a linear combination of elements of $S$ in two different ways: $u = a_1 v_1 + a_2 v_2 + \dots + a_n v_n = b_1 v'_1 + b_2 v'_2 + \dots + b_k v'_k.$ But then $\textbf{0} = a_1 v_1 + \dots + a_n v_n - b_1 v'_1 - \dots - b_k v'_k$ and not all terms cancel out on the righthand side (otherwise, the linear combinations would be the same). This contradicts the definition of linear independence.

It follows that any vector in $V$ may be expressed as a linear combination of elements of $S$ in only one way. $_\square$

This theorem reconciles the definition of a basis with its crucial property. It is also necessary to show that there do, in fact, exist bases for arbitrary vector spaces, but that follows from mathematical induction for finite-dimensional vector spaces and Zorn's lemma for infinite-dimensional vector spaces. The properties of linearity provide a strong groundwork for further results, like those regarding the "size" of a vector space.

## Dimension

In order to discuss the "dimension" of a vector space, it is important to realize that this is not a defining concept of vector space. While many examples like $\mathbb{R}^n$ have a very immediate concept of dimension (the number of coordinates, $n$), there are a myriad of others where it is not so clear. And indeed, it is even surprising that certain vectors form a basis!

The real vector space of "Fibonacci-like" sequences contains all sequences $\{a_n\}_{n \ge 0}$ satisfying $a_n + a_{n+1} = a_{n+2}$ for all natural numbers $n$. Two sequences can be added by adding their corresponding terms, and a sequence can be multiplied by a scalar by multiplying all elements in the sequence by a real number. (Note that the resulting sequence in each case still satisfies the Fibonacci property and is therefore still in the vector space.)

The Fibonacci numbers $F = \{0,\, 1,\, 1,\, 2,\, 3,\, 5,\, \ldots\}$ and the Lucas number $L = \{2,\,1,\,3,\,4,\,7,\,\ldots\}$ are both elements of this sequence. The two geometric series $G_+ = \left\{1,\,\varphi,\,\varphi^2,\,\varphi^3,\,\dots\right\}$ and $G_- = \left\{1,\,-\varphi^{-1},\,\varphi^{-2},\,-\varphi^{-3},\,\ldots\right\}$, where $\varphi^2 = \varphi + 1$ and $\varphi > 0$ are also elements of this sequence.

What makes a basis for this vector space?

Suppose $\mathcal{B}_1$ is a basis for the vector space $V$, and suppose $\mathcal{B}_2$ is another basis for $V$. Then $\mathcal{B}_1$ and $\mathcal{B}_2$ have the same cardinality.

If $\mathcal{B}_1$ is finite, then the

dimensionof $V$ is known as the number of elements in $\mathcal{B}_1$, $\text{dim}(V) = | \mathcal{B}_1 |.$

If there exists a basis with a finite number of elements, then an injective linear transformation from the vector space to itself must be a bijection. This can be shown by mathematical induction (on the number of elements in the finite basis) and is necessary for the following proof.

Suppose $\mathcal{B}_1 = \{v_1,\,v_2,\,\dots,\,v_n\}$ is finite and smaller than $\mathcal{B}_2$. Then $\mathcal{B}_2$ contains vectors $u_1,\,u_2,\,\dots,\,u_n,\,u_{n+1}$. Consider the linear transformation defined by $T: v_i \mapsto u_i$ for each integer $i$, $1 \le i \le n$. The idea is to show that this mapping is injective but not surjective, creating a contradiction (with the above fact).

Let $v \in V$ be a vector such that $T(v) = \textbf{0}$. Then, $v = a_1 v_1 + \dots + a_n v_n$ for some scalars $a_i$, so $T(v) = T(a_1 v_1 + \dots + a_n v_n) = a_1 T(v_1) + \dots + a_n T(v_n) = a_1 u_1 + \dots + a_n u_n = \textbf{0}.$ Since the $u_i$'s are linearly independent, it follows that $a_i = 0$ for all $i$ and therefore $v = \textbf{0}$. Then, $T$ must be injective, since its kernel is $\{\textbf{0}\}$.

Now, suppose $v' \in V$ is a vector such that $T(v') = u_{n+1}$. There must be such a vector in order for $T$ to be surjective. Then, $v' = b_1 v_1 + \dots + b_n v_n$ for some scalars $b_i$, so $T(v') = T(b_1 v_1 + \dots + b_n v_n) = b_1 T(v_1) + \dots + b_n T(v_n) = b_1 u_1 + \dots + b_n u_n = u_{n+1}.$ Since the $u_i$'s are linearly independent, there is no possible choice of $b_i$'s that satisfy this equation. Then, $T$ must not be surjective.

However, in a vector space with a finite basis, an injection must be surjective. It follows that $\mathcal{B}_1$ cannot be finite if $\mathcal{B}_2$ is larger than it. In other words, $\mathcal{B}_1$ and $\mathcal{B}_2$ must have the same number of elements. This holds for infinite bases too, but the argument requires more knowledge of infinite sets. (In particular, a $\kappa$-union of finite sets has cardinality $\kappa$.)

## Change of Basis

Main Article: Change of Basis