A vector space is a nonempty set V = { u, v, w, ……}, whose elements are vectors. Some important operations are carried out with them, among which the following stand out:

- Sum between two vectors u + v resulting z, which belongs to the set V.

- multiplication of a real number α by a vector v: α v giving another vector and belonging to V.

Artistic vision of a vector space. Source: Pixabay

To denote a vector we use bold (v is a vector), and for scalars or numbers Greek letters (α is a number).

Axioms and properties

For a vector space to be given, the following eight axioms must hold:

1-commutability: u + v = v + u

2-Transitivity: (u + v) + w = u + (v + w)

3-Existence of the null vector 0 such that 0 + v = v

4-Existence of the opposite: the opposite of v is (- v), since v + (- v) = 0

5-Distributivity of the product with respect to the vector sum: α (u + v) = α u + α v

6-Distributivity of the product with respect to the scalar sum: (α + β) v = α v + β v

7-Associativity of the scalar product: α (β v) = (α β) v

8-The number 1 is the neutral element since: 1 v = v

Examples of vector spaces

Example 1

Vectors in the (R²) plane are an example of a vector space. A vector in the plane is a geometric object that has magnitude and direction. It is represented by an oriented segment that belongs to said plane and with a size proportional to its magnitude.

The sum of two vectors in the plane can be defined as the geometric translation operation of the second vector after the first. The result of the sum is the oriented segment that starts from the origin of the first and reaches the tip of the second.

In the figure it can be seen that the sum in R² is commutative.

Figure 2. Vectors in the plane form vector space. Source: self made.

The product of a number α and a vector is also defined. If the number is positive, the direction of the original vector is kept and the size is α times the original vector. If the number is negative, the direction is the opposite, and the size of the resulting vector is the absolute value of the number.

The vector opposite any vector v is - v = (- 1) v.

The null vector is a point in the R² plane, and the number zero times a vector gives the null vector.

All that has been said is illustrated in Figure 2.

Example 2

The set P of all polynomials of degree less than or equal to two, including degree zero, form a set that satisfies all the axioms of a vector space.

Let the polynomial P (x) = a x² + bx + cy Q (x) = d x² + ex + f

The sum of two polynomials is defined: P (x) + Q (x) = (a + d) x² + (b + e) ​​x + (c + f)

The sum of polynomials belonging to the set P is commutative and transitive.

The null polynomial belonging to the set P is one that has all its coefficients equal to zero:

0 (x) = 0 x² + 0 x + 0

The sum of a scalar α by a polynomial is defined as: α P (x) = α ∙ a x² + α ∙ bx + α ∙ c

The opposite polynomial of P (x) is -P (x) = (-1) P (x).

From all of the above it follows that the set P of all polynomials of degree less than or equal to two is a vector space.

Example 3

The set M of all matrices of m rows xn columns whose elements are real numbers form a real vector space, with respect to the operations of addition of matrices and product of a number by a matrix.

Example 4

The set F of continuous functions of real variable, form a vector space, since it is possible to define the sum of two functions, the multiplication of a scalar by a function, the null function and the symmetric function. They also fulfill the axioms that characterize a vector space.

Base and dimension of a vector space

Base

The basis of a vector space is defined as a set of linearly independent vectors such that any vector of that vector space can be generated from a linear combination of them.

Linearly combining two or more vectors consists of multiplying the vectors by some scalar and then adding them vectorially.

For example, in the vector space of vectors in three dimensions formed by R³, the canonical basis defined by the unit vectors (of magnitude 1) i, j, k is used.

Where i = (1, 0, 0); j = (0, 1, 0); k = (0, 0, 1). These are the Cartesian or canonical vectors.

Any vector V belonging to R³ is written as V = a i + b j + c k, which is a linear combination of the base vectors i, j, k. A scalar or numbers a, b, c are known as Cartesian components of V.

It is also said that the base vectors of a vector space form a generator set of the vector space.

Dimension

The dimension of a vector space is the cardinal number of a vector basis for that space; that is, the number of vectors that make up said base.

This cardinal is the maximum number of linearly independent vectors of that vector space, and at the same time the minimum number of vectors that form a generator set of that space.

The bases of a vector space are not unique, but all the bases of the same vector space have the same dimension.

Vector subspace

A vector subspace S of a vector space V is a subset of V in which the same operations are defined as in V and fulfills all vector space axioms. Therefore, the subspace S will also be a vector space.

Example of vector subspace are the vectors that belong to the XY plane. This subspace is a subset of a vector space of dimensionality greater than the set of vectors belonging to the three-dimensional space XYZ.

Another example of a vector subspace S1 of the vector space S formed by all 2 × 2 matrices with real elements is defined below:

On the other hand, S2 defined below, although it is a subset of S, does not form a vector subspace:

Solved exercises

-Exercise 1

Let the vectors V1 = (1, 1, 0); V2 = (0, 2, 1) and V3 = (0, 0, 3) in R³.

a) Show that they are linearly independent.

b) Show that they form a basis in R³, since any triple (x, y, z) can be written as a linear combination of V1, V2, V3.

c) Find the components of the triple V = (-3,5,4) in the base V1, V2, V3.

Solution

The criterion to demonstrate linear independence consists in establishing the following set of equations in α, β and γ

α (1, 1, 0) + β (0, 2, 1) + γ (0, 0, 3) = (0, 0, 0)

In case the only solution to this system is α = β = γ = 0 then the vectors are linearly independent, otherwise they are not.

To obtain the values ​​of α, β and γ we propose the following system of equations:

α ∙ 1 + β ∙ 0 + γ ∙ 0 = 0

α ∙ 1 + β ∙ 2 + γ ∙ 0 = 0

α ∙ 0 + β ∙ 1 + γ ∙ 3 = 0

The first leads to α = 0, the second α = -2 ∙ β but since α = 0 then β = 0. The third equation implies that γ = (- 1/3) β, but since β = 0 then γ = 0.

Answer to

It is concluded that it is a set of linearly independent vectors in R³.

Answer b

Now let's write the triple (x, y, z) as a linear combination of V1, V2, V3.

(x, y, z) = α V1 + β V2 + γ V3 = α (1, 1, 0) + β (0, 2, 1) + γ (0, 0, 3)

α ∙ 1 + β ∙ 0 + γ ∙ 0 = x

α ∙ 1 + β ∙ 2 + γ ∙ 0 = y

α ∙ 0 + β ∙ 1 + γ ∙ 3 = z

Where do you have:

α = x

α + 2 β = y

β + 3 γ = z

The first indicates α = x, the second β = (yx) / 2 and the third γ = (z- y / 2 + x / 2) / 3. In this way we have found the generators of α, β and γ of any triplet of R³

Answer c

Let's move on to find the components of the triple V = (-3,5,4) in the base V1, V2, V3.

We substitute the corresponding values ​​in the expressions found above for the generators.

In this case we have: α = -3; β = (5 - (- 3)) / 2 = 4; γ = (4- 5/2 + (- 3) / 2) / 3 = 0

That is:

(-3,5,4) = -3 (1, 1, 0) + 4 (0, 2, 1) + 0 (0, 0, 3)

By last:

V = -3 V1 + 4 V2 + 0 V3

We conclude that V1, V2, V3 form a basis in the vector space R³ of dimension 3.

-Exercise 2

Express the polynomial P (t) = t² + 4t -3 as a linear combination of P1 (t) = t² -2t + 5, P2 (t) = 2t² -3t and P3 (t) = t + 3.

Solution

P (t) = x P1 (t) + y P2 (t) + z P3 (t)

where the numbers x, y, z are to be determined.

By multiplying and grouping terms with the same degree in t, we obtain:

t² + 4 t -3 = (x + 2y) t² + (-2x -3y + z) t + (5x + 3z)

Which leads us to the following system of equations:

x + 2y = 1

-2x -3y + z = 4

5x + 3z = -3

The solutions of this system of equations are:

x = -3, y = 2, z = 4.

That is:

P (t) = -3 P1 (t) + 2 P2 (t) + 4 P3 (t)

-Exercise 3

Show that the vectors v1 = (1, 0, -1, 2); v2 = (1, 1, 0, 1) and v3 = (2, 1, -1, 1) of R⁴ are linearly independent.

Solution

We linearly combine the three vectors v1, v2, v3 and demand that the combination adds the null element of R⁴

a v1 + b v2 + c v3 = 0

That is to say, a (1, 0, -1, 2) + b (1, 1, 0, 1) + c (2, 1, -1, 1) = (0, 0, 0, 0)

This leads us to the following system of equations:

a + b + 2 c = 0

b + c = 0

-a - c = 0

2 a + b + c = 0

Subtracting the first and fourth we have: -a + c = 0 which implies a = c.

But if we look at the third equation, we have that a = -c. The only way that a = c = (- c) holds is for c to be 0 and therefore a will also be 0.

a = c = 0

If we plug this result into the first equation then we conclude that b = 0.

Finally a = b = c = 0, so that it can be concluded that the vectors v1, v2 and v3 are linearly independent.

References

1. Lipschutz, S. 1993. Linear algebra. Second edition. McGraw-Hill. 167-198.