# Generators of a Vector Space

A **generating system** is a set of vectors S={v_{1},...,v_{m}} in a vector space V $$ S = \{ v_1 , ... , v_m \} \in \ V $$ that allows for the determination of every vector v_{i}∈V in the vector space through the linear combination of the vectors in S with corresponding a_{1},...a_{m} ∈ R scalar coefficients $$ \vec{v}_i=a_1 \vec{v}_1 +...+ a_m \vec{v}_m \ \ \ \forall v_i \in V $$

In essence, the linear combinations of the vectors S={v_{1},...,v_{m}} generate all the vectors in the vector space V.

This is why the vectors in S are known as **generators**.

Therefore, given a set S of vectors in the vector space V

$$ S = \{ v_1, v_2, ... ,v_n \} $$

The smallest linear vector subspace L(S) that contains S is equal to the vector space V.

$$ L(S) = V $$

The number of generating vectors in S can be either finite or infinite, as long as it is always less than or equal to the number of vectors in the vector space V.

$$ m \le i $$

In these notes, I denote the set of generating vectors as S and the vector subspace of S as L(S).

## The Purpose of a Vector Space Generator

A vector space V can consist of an infinite number of vectors. However, it's not necessary to list them all to describe it.

Every vector v ∈ V can be expressed as a linear combination of other vectors in the vector space.

__Example__

$$ v_4 = a_1 v_1 + a_2 v_2 + a_3 v_3 $$ $$ v_5 = a_1 v_1 + a_2 v_2 + a_3 v_3 $$ $$ v_6 = a_1 v_1 + a_2 v_2 + a_3 v_3 $$ $$ \vdots $$

It's sufficient to choose a suitable finite or infinite subset of vectors S={v_{1},..,v_{m}}.

$$ L = \{ v_1 , v_2 , v_3 \} $$

Through the linear combination of the vectors in S with m scalar numbers α_{m}, any other vector in the vector space V can be obtained.

$$ \forall \ v_i \in V \: \: v = a_1 v_1 + a_2 v_2 + a_3 v_3 $$

## An Example of a Vector Space Generator

The entire vector space V = R^{2} over the field of real numbers R can be generated by just two vectors.

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

On the Cartesian plane R^{2}, these two vectors are represented as follows:

To demonstrate that vectors v_{1} and v_{2} are generators of the vector space R^{2}, I write their linear combination with two scalars a_{1} and a_{2} belonging to R.

$$ v = a_1 v_1 + a_2 v_2 \: \: \: \: \: \: with \: \: a_1=1 , a_2=1 $$

$$ v = 1 ( 1 , 0 ) + 1 ( 0 , 1 ) = $$

$$ v = ( 1 , 0 ) + ( 0 , 1 ) $$

$$ v = ( 1 , 1 ) $$

This linear combination generates another vector (v_{3}) in the vector space V, pointing to the coordinates (1,1).

Now, I modify the linear combination using the scalar a_{2}=2.

$$ v = a_1 v_1 + a_2 v_2 \: \: \: \: \: \: with \: \: a_1=1 , a_2=2 $$

$$ v = 1 ( 1 , 0 ) + 2 ( 0 , 1 ) = $$

$$ v = ( 1 , 0 ) + ( 0 , 2 ) $$

$$ v = ( 1 , 2 ) $$

Here, vector v_{2} has doubled in length while v_{1} remains unchanged.

This linear combination generates another distinct vector (v_{4}) in the vector space, pointing to the coordinates (1,2).

In this way, by altering the scalars a_{1} and a_{2} for vectors v_{1} and v_{2}, I can generate every vector v in the vector space V.

Therefore, I can assert that v_{1} and v_{2} form a generating system for the vector space R^{2}.

$$ L_R = \{ v_1 , v_2 \} = R^2$$

And so on.

**Note**. A single vector v_{1} alone is not a generator of the vector space V=R^{2} as it only generates vectors along the x-axis. It is merely a generator of the subspace {x}. Similarly, vector v_{2}, taken individually, does not generate R^{2} as it only produces vectors along the y-axis, thus being a generator of the subspace {y}.

**Example 2**

A vector space generator can also consist of just one vector.

For instance, in a vector space V over the field K=R of real numbers, consider any non-zero vector v.

$$ \vec{v} \in V $$

Visually speaking

The vector v must be non-zero.

$$ \vec{v} \in \vec{0} $$

The set generated by vector L(v) comprises the linear combination of vector v with scalars k from K

$$ L(v) = \{ k \cdot \vec{v} \} $$

The vectors generated share the same direction as the generating vector, meaning they lie on the same line.

What varies is the direction and magnitude (length) of each vector.

**Note**. The generating vector must be non-zero, because any scalar multiplied by a magnitude of zero always results in zero.

**Example 3**

Generally, two non-zero vectors v_{1}, v_{2} generate a vector subspace

$$ \vec{v_1} \ne 0 $$ $$ \vec{v_2} \ne 0 $$

The subspace generated depends on the linear dependence or independence between the two vectors.

- If the two vectors are parallel $$ \vec{v_1} = k \cdot \vec{v_2} $$, then the vector subspace generated by the two vectors L(v
_{1},v_{2}) lies on the line containing the two vectors v_{1}, v_{2}.

- If the vectors are not parallel $$ \vec{v_1} \ne k \cdot \vec{v_2} $$, then the vector subspace generated by the two vectors is the plane containing both vectors v
_{1}and v_{2}.

## How to Determine if Vectors are Generators

To find out if a set of vectors forms a generating system for the vector space V, we need to solve the equation system representing their linear combination.

**A Practical Example**

Consider two vectors v_{1}=(1,0) and v_{2}=(2,1) in the vector space V=R^{2}.

$$ v_1 = (1,1) $$ $$ v_2 = (2,1) $$

Our goal is to determine if these vectors are generators of the vector space V.

The linear combination of vectors v_{1} and v_{2} can be expressed as:

$$ a_1 v_1 + a_2 v_2 $$

$$ a_1 (1,1) + a_2 (2,1) $$

$$ ( a_1 , a_1 ) + (2 a_2 , a_2 ) $$

$$ ( a_1 + 2 a_2 , a_1 + a_2 ) $$

These last two expressions represent the target coordinates (x,y) of the vector generated by the linear combination of the two vectors.

$$ (x,y) = ( a_1 + 2 a_2 , a_1 + a_2 ) $$

We can rewrite this vector equality as a system of equations:

$$ \begin{cases} a_1 + 2 a_2 = x \\ a_1 + a_2 = y \end{cases}$$

If this system has solutions for a_{1} and a_{2} for every possible value of (x,y), then vectors v_{1} and v_{2} are indeed generators of the vector space V=R^{2}.

Conversely, if there are no solutions, then these vectors are not generators of V=R^{2}.

**Note**. To demonstrate that the system has solutions (either one or infinitely many), we can apply the Rouché-Capelli theorem. According to this theorem, a system of linear equations has solutions if the rank of the coefficient matrix A is equal to the rank of the augmented matrix A|B. In such cases, the system has ∞^{n-r} solutions, where r is the rank and n is the number of unknown variables.

In this case, the system has solutions.

Therefore, v_{1} and v_{2} are generators of the vector space R^{2}.

**Verification**

As an additional check, let's compute the solutions of the system.

$$ \begin{cases} a_1 + 2 a_2 = x \\ a_1 = y - a_2 \end{cases}$$

$$ \begin{cases} y - a_2 + 2 a_2 = x \\ a_1 = y - a_2 \end{cases}$$

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

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

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

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

These equations are solved for any combination of x and y values, i.e., over all points on the Cartesian plane.

Thus, v_{1} and v_{2} are indeed generators of the vector space R^{2}.

## Difference Between a Set of Generators and Linear Span

A **linear span** is a set of vectors S_{p} in the vector space V

$$ S_p = \{ v_1,.v_2, ...,v_m \} $$

whose linear combination is a vector in V.

$$ a_1 v_1 + ... + a_m v_m \in V $$

Every linear combination of the vectors in the span S_{p} belongs to the vector space V.

$$ L(S_p) \subseteq V $$

However, it does not imply that every vector in the vector space V can be obtained through a linear combination of the span set S_{p}.

$$ L(S_p) \ne V $$

**Note**. The vectors in the span {v_{1},...,v_{n}} are certainly generators for a subspace L(S_{p}). It does not necessarily mean they are for the entire vector space V.

Conversely, a **set of generators** is a collection S of vectors

$$ S = \{ v_1,.v_2, ...,v_n \} $$

whose linear combination generates all vectors in the vector space V.

$$ L(S) = V $$

Therefore, a set of vectors forming a span is not necessarily a set of generators.

$$ L(S_p) \ne L(S) = V $$

## How to Minimize a Set of Generators

A set of vectors {v_{1},...,v_{m}} forms a generating system, but it's not necessarily the smallest possible set of generators for the vector space V.

Indeed, consider a generating system:

$$ v = a_1 v_1 + ... + a_m v_m $$

If you add another vector v_{m+1} with a scalar a_{m+1} = 0, you end up with another generating system for the same vector space V.

$$ v = a_1 v_1 + ... + a_m v_m + a_{m+1} v_{m+1} $$

$$ with \:\:\: a_{m+1} = 0 $$

Therefore, even though a set of vectors is a generating system for the vector space, it can potentially be further reduced.

**Example**

**How Do You Reduce a Set of Generators?**

To reduce a set of generators, one must eliminate the vectors linearly dependent on others.

__Demonstration__

In a set of generators, the vector v_{m} is linearly dependent on the others.

$$ v = a_1 v_1 + ... + a_m v_m $$

Hence, v_{m} can be rewritten as a linear combination of the other vectors.

$$ v = β _1 v_1 + ... + β _{m-1} v_{m-1} $$

Substituting this linear combination into the generating system:

$$ v = a_1 v_1 + ... + a_m ( β _1 v_1 + ... + β _{m-1} v_{m-1} ) $$

$$ v = a_1 v_1 + ... + a_m β _1 v_1 + ... + a_m β _{m-1} v_{m-1} ) $$

$$ v = v_1 ( a_1 + a_m β _1 ) + ... + v_{m-1} ( a_{m-1} + a_m β _{m-1} ) $$

The coefficients (a_{1}+a_{m}β_{1} ) ... (a_{m-1}+a_{m}β_{m-1} ) are still scalar numbers from the field R.

So, they can be rewritten as γ_{1} , ... , γ_{m-1 }_{∈} R

$$ v = γ _1 v_1 + ... + γ _{m-1} v_{m-1} $$

This shows that even { v_{1}, v_{2} } can be a generating system for the vector space V.

And so on.

**Example **

## Difference Between a Set of Generators and a Basis

A set of generators is called a **basis of the vector space** V if the vectors in the set S $$ S = \{ v_1, v_2, ... , v_n \} $$ are linearly independent.

A basis is also known as a **free system of generators**.

The linear combinations of vectors in the set S={v_{1},v_{2},..,v_{n}} generate all vectors in the vector space V.

$$ \vec{v}_i=a_1 \vec{v}_1 +...+ a_m \vec{v}_n \ \ \ \forall \ v_i \in V $$

The same holds true in generating systems.

However, in the case of a basis, **each vector in the vector space V is generated by one and only one linear combination of the vectors in set S**.

**Note**. Conversely, in a generating system with linearly dependent vectors, any vector in the vector space V can be composed of infinite combinations of vectors in set S.

In these cases, the generating system is called a **basis of the vector space**.