# The Basis of a Vector Space

In linear algebra, a basis is a minimal generating system of the vector space V.

**Definition**

In a vector space V over the field K, a set of vectors B={v_{1},...,v_{n}} that generates V is a basis of V if it consists of linearly independent vectors.

$$ B={v_1,...,v_n} $$

It is also known as a free generating system.

## Characteristics of a Basis, Coordinates, and Dimension

There are two essential conditions for a vector space basis:

**Generators of a vector space**

The set of vectors must be a generating system.**Note**. This means that every vector in the vector space V must be represented by the linear combination of the basis vectors. This is the case in any generating system L_{R}.**Linear Independence**

The vectors must be linearly independent. Each vector v_{n}in the vector basis cannot be written as a linear combination of the other basis vectors {v_{1},..,v_{n}}.**Difference between a Basis and a Generating System**. The characteristic of linear independence of all vectors in the basis B distinguishes bases from generating systems. In a generating system, vectors can also be linearly dependent.

## Coordinates or Weights of Vectors

In a vector basis B, each vector in the vector space V is defined by a unique linear combination, meaning there exists a **unique n-tuple of scalars (a _{1},...,a_{n})** known as coordinates or weights.

$$ v = a_1 v_1 + ... + a_n v_n $$

**Note**. This characteristic of vector bases is demonstrated by the theorem of the uniqueness of vector representation through a basis.

Therefore, the system of linear equations admits **only one solution**.

Thus, determining a basis is equivalent to establishing that a linear system has a unique solution.

## Dimension of the Basis

The number of elements { v_{1},...,v_{n} } in the basis is called the **dimension**. $$ dim ( n ) $$ $$ with \:\: n ∈ Z ≥ 0 $$

The dimension of a basis is denoted by *dim* followed by the cardinality n of the basis elements.

A basis can have either a finite or an infinite number of elements.

**Finite Dimension**. A basis has a finite dimension of n if it consists of n vectors.**Infinite Dimension**. A basis has an infinite dimension if it consists of an infinite set of vectors.

**Zero Dimension**. Only the trivial space {0_{v}} has a dimension of zero. The trivial space consists only of the zero vect/math/the-dimension-theorem-of-a-vector-space-basisor, i.e., a linearly dependent vector. Having no linearly independent vectors within, the trivial space does not have vector bases. For this reason, the trivial space {0_{v}} is the only space with a zero dimension.

__Example__

The basis B consists of two vectors.

$$ B = \{ v_1 , v_2 \} $$

Therefore, the basis has a dimension of 2.

## Examples and Exercises on Vector Bases

**Example 1**

In the vector space V=R^{2} over the field K=R, a generating system is formed by the vectors v_{1} and v_{2}.

$$ v_1=(1,0) $$ $$v_2=(0,1) $$

To determine if they form a basis, I need to check the linear independence of the vectors.

Their linear combination is

$$ v = a_1 v_1 + a_2 v_2 $$ $$ v = a_1 (1,0) + a_2 (0,1) $$ $$ v = (a_1,0) + (0,a_2) $$ $$ v = (a_1,a_2) $$

Expressed in x,y coordinates, this becomes

$$ (x,y) = (a_1,a_2) $$

Which corresponds to the linear system

$$ \begin{cases} a_1=x \\ a_2=y \end{cases} $$

This linear system has an obvious solution. For any (x,y) coordinates in the plane, there are two scalar parameters a_{1} and a_{2} that identify it.

Therefore, the two vectors v_{1} and v_{2} are __linearly independent__.

This constitutes a **basis of the vector space** V.

Also, analyzing the system in matrix form, it's immediately apparent that the rank of the matrix is the maximum for the linear system.

$$ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $$

In the matrix, the complementary minor with a non-zero determinant is of order 2.

$$ \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 1+0 = 1 \ne 0 $$

The rank equals the number of columns in the matrix (2), that is, the number of unknown variables in the linear system.

This confirms the linear independence of the vectors.

**Example 2**

Staying within the vector space V=R^{2} over the field K=R, let's analyze another pair of vectors.

$$ v_1=(1,1) $$ $$v_2=(-1,1) $$

To determine if they form a basis, I need to check their linear independence.

Their linear combination is

$$ v = a_1 v_1 + a_2 v_2 $$ $$ v = a_1 (1,1) + a_2 (-1,1) $$ $$ v = (a_1,a_1) + (-a_2,a_2) $$ $$ v = (a_1-a_2,a_1+a_2) $$

Expressed in x,y coordinates, this becomes

$$ (x,y) = (a_1-a_2,a_1+a_2) $$

Which corresponds to the linear system

$$ \begin{cases} a_1-a_2=x \\ a_1+a_2=y \end{cases} $$

When expressed in matrix form, the system has a rank of two.

$$ \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix} $$

The complementary minor with a non-zero determinant is of order 2 (rank = 2).

$$ \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} = 1+1 = 2 \ne 0 $$

The rank is equal to the number of columns in the matrix.

**Note**. Linear independence of vectors can be verified by observing the rank of the matrix. For more information, click here.

Therefore, __the vectors ____v _{1} and v_{2} are linearly independent__ and the system has at least one solution.

This constitutes a **basis of the vector space**.

To verify this, let's try solving the system of linear equations using the substitution method.

$$ \begin{cases} a_1-a_2=x \\ a_1+a_2=y \end{cases} $$

$$ \begin{cases} a_1-a_2=x \\ a_1=y-a_2 \end{cases} $$

$$ \begin{cases} (y-a_2)-a_2=x \\ a_1=y-a_2 \end{cases} $$

$$ \begin{cases} a_2=(y-x)/2 \\ a_1=y-a_2 \end{cases} $$

$$ \begin{cases} a_2=(y-x)/2 \\ a_1=y-(y-x)/2 \end{cases} $$

$$ \begin{cases} a_2=(y-x)/2 \\ a_1=-x/2 \end{cases} $$

The system has a solution for any coordinate (x,y) on the plane.

The linear independence of vectors v_{1} and v_{2} is confirmed.

## How Many Bases Exist in a Vector Space

There isn't just one basis in a vector space.

Every real vector space contains an infinite number of bases.

If there are various bases in a vector space, every vector in the space can be represented differently depending on the chosen basis.

**The Case of the Trivial Vector Space.**

The trivial vector space {0_{v}} is an exception to the rule.

The trivial vector space {0_{v}} has no basis.

__Demonstration__

The trivial vector space lacks a basis because it contains only the null vector.

A null vector is always linearly dependent.

**Note**. A vector is linearly independent if it equals the null vector only when all coefficients of the linear combination are zero. $$ \vec{v} = k_1 \vec{v_1} + ... + k_n \vec{v_n} = \vec{0} $$ In the case of the null vector, however, I can obtain the null vector even if any one of the coefficients is non-zero. For example, k_{1}≠0 and all others k_{2},...k_{n}=0. $$ \vec{0} = k_1 \vec{v_1} = \vec{0} $$ Therefore, the null vector can never be linearly independent. Hence, it is always linearly dependent.

Thus, the trivial vector space has no linearly independent vectors within it and cannot have a basis.

## The Canonical Basis

A basis is called canonical if each vector v_{i} has all elements at zero except for the i-th element.

In every vector space K^{n}, there always exists a canonical basis.

**Example**

In a vector space R^{4} over the field R, I have four vectors.

$$ v_1 = (1,0,0,0) $$ $$ v_2 = (0,1,0,0) $$ $$ v_3 = (0,0,1,0) $$ $$ v_4 = (0,0,0,1) $$

This is an identity matrix because its diagonal is composed of 1s while the other elements are 0s.

$$ \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} $$

It is immediately evident that the vectors are linearly independent because the rank of the matrix is 4, equaling the number of columns.

The linear combination of vectors is as follows:

$$ v = a_1 v_1 + a_2 v_2 + a_3 v_3 + a_4 v_4 $$ $$ v = a_1 (1,0,0,0) + a_2 (0,1,0,0) + a_3 (0,0,1,0) + a_4 (0,0,0,1) $$ $$ v=(a_1,0,0,0)+(0,a_2,0,0)+(0,0,a_3,0)+(0,0,0,a_4) $$ $$ v=(a_1,a_2,a_3,a_4) $$

Therefore,

$$ (x,y,z,w)=(a_1,a_2,a_3,a_4) $$

can also be expressed as a system:

$$ \begin{cases} a_1 = x \\ a_2=y \\ a_3=z \\ a_4 = w \end{cases} $$

It's a basis because the system admits a single solution.

## Theorems on Vector Bases

Main theorems of a vector space basis:

**Theorem of Unique Vector Representation in a Basis**

Every vector v in the vector space V is representable by a vector basis B through a unique combination of scalar numbers a_{1},...,a_{n}.

**Theorem of Linear Dependence of Every Vector Relative to the Basis**

All vectors in a vector space V over the field are linearly dependent relative to the vectors of the vector basis B.

**Theorem of the Dimension of a Basis**

If a vector space over the field K has a basis B with a finite number of elements (dimension), then every other basis B' of V has the same number of elements, i.e., the same dimension as B.

__Corollary__

The number of elements in a basis depends not on the choice of the basis but on the vector space itself.

**Theorem of Completing the Basis**

If a basis has k<n elements, it is an incomplete basis and can always be completed by adding the missing n-k linearly independent vectors.

**Other Theorems on Bases**

- In a finitely generated vector space V of dimension n, from any set of generators {v
_{1},v_{2},..,v_{s}} with s>n, it is always possible to obtain a basis B of the vector space by eliminating linearly dependent vectors from the set of generators (proof). - In a finitely generated vector space V of dimension n, from any set of linearly independent vectors {v
_{1},v_{2},..,v_{p}} with p<n, it is always possible to obtain a basis B of the vector space by adding linearly independent vectors to the set of vectors (proof). - In a vector space V with a known dimension dim(V)=n, if {v
_{1},v_{2},..,v_{n}} are a set of generators of V, then they are also a basis of V (proof) - In a vector space V with a known dimension dim(V)=n, if {v
_{1},v_{2},..,v_{n}} are a set of linearly independent vectors, then they are also a basis of V (proof)