# Basis of a Vector Space

## Vector Space

Let's introduce some definitions. A set is called closed with respect to some operation if for any elements of the set the result of applying this operation belongs to the given set. A set of vectors that is closed under linear operations is called a vector space (or linear space).

For example, the set of all vectors parallel to a given plane is a two-dimensional vector space. In fact, if the vectors a, b are parallel to the plane π, then any linear combination of these vectors is also parallel to this plane:

$\mathbf{a} \parallel \pi, \mathbf{b} \parallel \pi \implies \lambda\mathbf{a} + \mu\mathbf{b} \parallel \pi, \text{ where } \lambda,\mu \in \mathbb{R}.$

Similarly, the set of all vectors parallel to a given straight line forms a one-dimensional vector space. The zero vector space consists only of the zero vector. A three-dimensional vector space includes the set of all vectors of the space.

More formally, for a set to be a vector space it must satisfy the 8 axioms listed below.

Let V be a non-empty set over a scalar field F. The elements of the set V are vectors a, b, c, ..., and the elements of the field F are numbers λ, μ, ν, ... Two binary operations are defined for the set V over field F:

$\forall \mathbf{a}, \mathbf{b} \in V \;\exists\, \mathbf{c} \in V: \mathbf{a} + \mathbf{b} = \mathbf{c};$
• Scalar multiplication:
$\forall \mathbf{a} \in V,\;\forall \lambda \in F \;\exists\, \mathbf{b} \in V: \lambda\mathbf{a} = \mathbf{b}.$

The set V is a vector space over F, if the following axioms are satisfied:

1. $$\mathbf{a} + \mathbf{b} = \mathbf{b} + \mathbf{a}\;$$ (commutativity)
2. $$\left({\mathbf{a} + \mathbf{b}}\right) + \mathbf{c} = \mathbf{a} + \left({\mathbf{b} + \mathbf{c}}\right)\;$$ (associativity)
3. $$\mathbf{a} + \mathbf{0} = \mathbf{a}\;$$ (existence of identity element)
4. $$\mathbf{a} + \left({-\mathbf{a}}\right) = \mathbf{0}\;$$ (existence of inverse element)
5. $$1 \cdot \mathbf{a} = \mathbf{a}$$
6. $$\lambda\left({\mu\mathbf{a}}\right) = \left({\lambda\mu}\right)\mathbf{a}$$
7. $$\lambda\left({\mathbf{a} + \mathbf{b}}\right) = \lambda\mathbf{a} + \lambda\mathbf{b}$$
8. $$\left({\lambda + \mu}\right)\mathbf{a} = \lambda\mathbf{a} + \mu\mathbf{a}$$

Here a, b, c, 0 are vectors and λ, μ, 1 are numbers.

## Examples of Vector Spaces

A key example of a vector space is the set of all free vectors with respect to the usual operations of adding vectors and multiplying a vector by a real number. However, vector space is a rather abstract concept. It can contain not only vectors but also other objects. For example, the following sets are also vector spaces:

• $${M_{n \times m}} -$$ the set of matrices of dimension $$n \times m$$ with respect to the operation of matrix addition and multiplication by an element of the field of numbers F form a vector space.
• $${P_n\left({x}\right)} -$$ the set of polynomials with real coefficients of degree not higher than n form a vector space with respect to the operation of addition of polynomials and multiplication of a polynomial by a real number.
• The set of all functions of the form $$f : \mathbb{R} \to \mathbb{R}$$ is a vector space; the addition of vectors and their multiplication by numbers are given by the formulas
$\left({f + g}\right)\left({x}\right) = f\left({x}\right) + g\left({x}\right),\;\;\left({\alpha \cdot f}\right)\left({x}\right) = \alpha\cdot f\left({x}\right).$
where $$f\left({x}\right), g\left({x}\right)$$ are functions, $$\alpha, x$$ are real numbers.
• The set of all convergent sequences $$\{u_n\},$$ $$\{v_n\}$$ with the defined operations of addition and multiplication by a number is a vector space. The sum of sequences and the product of a sequence by a number are given by the formulas
$\{u_n + v_n\} = \{u_n\} + \{v_n\},\;\;\{\alpha \cdot u_n\} = \alpha \cdot \{u_n\}.$

From the above examples it can be seen that the elements of a vector space can be of completely different nature: geometric vectors, matrices, polynomials, functions, sequences, and so on.

## Basis of a Vector Space

Three linearly independent vectors a, b and c are said to form a basis in space if any vector d can be represented as some linear combination of the vectors a, b and c, that is, if for any vector d there exist real numbers λ, μ, ν such that

$\mathbf{d} = \lambda\mathbf{a} + \mu\mathbf{b} + \nu\mathbf{c}$

This equality is usually called the expansion of the vector d relative to the basis a, b, c and the numbers λ, μ and ν are called the coordinates of the vector d with respect to the basis a, b, c.

Every vector d can be decomposed in a unique way in terms of the basis a, b, c.

A basis for a plane is defined similarly. Two linearly independent vectors a and b lying in a plane π are said to form a basis in this plane if any vector c lying in the plane π can be represented as some linear combination of the vectors a and b, that is, if for any vector c lying in the plane π there exist real numbers λ and μ such that

$\mathbf{c} = \lambda\mathbf{a} + \mu\mathbf{b}$

The following statements are true:

1. Any triple of non-coplanar vectors a, b and c form a basis in space.
2. Any pair of non-collinear vectors a and b lying in a given plane form a basis in this plane.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Check that vectors $$\mathbf{a}\left({-3,1}\right)$$ and $$\mathbf{b}\left({-1,3}\right)$$ form a basis in the plane. Find the coordinates of the vector $$\mathbf{c}\left({-7,5}\right)$$ in this basis.

### Example 2

Check that vectors $$\mathbf{a}\left({3,0,-1}\right),$$ $$\mathbf{b}\left({1,2,-5}\right)$$ and $$\mathbf{c}\left({1,0,-1}\right)$$ form a basis in space. Find the coordinates of the vector $$\mathbf{d}\left({5,0,-3}\right)$$ in this basis.

### Example 3

In the parallelogram $$ABCD,$$ point $$M$$ is the midpoint of the side $$BC$$ and point $$O$$ is the intersection point of the diagonals. Taking $$\mathbf{AB} = \mathbf{a}$$ and $$\mathbf{AD} = \mathbf{b}$$ as the basis vectors, find in this basis the coordinates of the vectors $$\mathbf{BD},$$ $$\mathbf{CO}$$ and $$\mathbf{MD}.$$

### Example 4

Find a basis for the set of $$2 \times 2$$ matrices and determine the coordinates of the matrix

$\left[ \begin{array}{*{20}{r}} -1 & 5\\ 3 & 0 \end{array} \right]$

in this basis.

### Example 1.

Check that vectors $$\mathbf{a}\left({-3,1}\right)$$ and $$\mathbf{b}\left({-1,3}\right)$$ form a basis in the plane. Find the coordinates of the vector $$\mathbf{c}\left({-7,5}\right)$$ in this basis.

Solution.

Two vectors form a basis in the plane if they are not collinear. Since collinear vectors satisfy the relation $$\mathbf{a} = \lambda\mathbf{b},$$ their coordinates must be proportional, that is, the determinant composed of the coordinates of these vectors must be equal to zero. Check it out:

$\left| \begin{array}{*{20}{c}} -3 & 1\\ -1 & 3 \end{array} \right| = -3 \cdot 3 - \left({-1}\right) \cdot 1 = -9 + 1 = -8 \ne 0.$

Hence, the vectors $$\mathbf{a}$$ and $$\mathbf{b}$$ are non-collinear and form a basis. The coordinates of the vector $$\mathbf{c}\left({-7,5}\right)$$ in this basis are determined from the equation

$\mathbf{c} = \lambda\mathbf{a} + \mu\mathbf{b}.$

Write it in coordinate form

$\left\{ \begin{array}{*{20}{l}} -7 = \left({-3}\right)\lambda - \mu\\ 5 = \lambda + 3\mu \end{array} \right., \Rightarrow \left\{ \begin{array}{*{20}{l}} 7 = 3\lambda + \mu\\ 5 = \lambda + 3\mu \end{array} \right., \Rightarrow \left\{ \begin{array}{*{20}{l}} \mu = 7 - 3\lambda\\ 5 = \lambda + 3\left({7 - 3\lambda}\right) \end{array} \right., \Rightarrow \left\{ \begin{array}{*{20}{l}} \mu = 7 - 3\lambda\\ 5 = \lambda + 21 - 9\lambda \end{array} \right., \Rightarrow \left\{ \begin{array}{*{20}{l}} \mu = 1\\ \lambda = 2 \end{array} \right..$

So the vector $$\mathbf{c}$$ in basis $$\left({\mathbf{a, b}}\right)$$ has the coordinates $$\left({\lambda,\mu}\right) = \left({2,1}\right).$$

### Example 2.

Check that vectors $$\mathbf{a}\left({3,0,-1}\right),$$ $$\mathbf{b}\left({1,2,-5}\right)$$ and $$\mathbf{c}\left({1,0,-1}\right)$$ form a basis in space. Find the coordinates of the vector $$\mathbf{d}\left({5,0,-3}\right)$$ in this basis.

Solution.

For three vectors $$\mathbf{a},$$ $$\mathbf{b}$$ and $$\mathbf{c}$$ to form a basis in space, they must be non-coplanar. Recall that the conditions for complanarity are

$\mathbf{c} = \lambda\mathbf{a} + \mu\mathbf{b}.$

If you make a matrix of three coplanar vectors, its determinant will be equal to zero since the third vector $$\mathbf{c}$$ linearly depends on the vectors $$\mathbf{a}$$ and $$\mathbf{b}.$$ Accordingly, the matrix of three non-coplanar vectors has a non-zero determinant. Let's check what we have in our case:

$\left| \begin{array}{*{20}{c}} 3 & 0 & -1\\ 1 & 2 & -5\\ 1 & 0 & -1 \end{array} \right| = 2 \cdot \left| \begin{array}{*{20}{c}} 3 & -1\\ 1 & -1 \end{array} \right| = 2 \cdot \left({-2}\right) = -4 \ne 0.$

Therefore, the vectors $$\mathbf{a},$$ $$\mathbf{b},$$ $$\mathbf{c}$$ are linearly independent and form a basis in space. Find the coordinates of the vector $$\mathbf{d}$$ in this basis. Write the expansion of this vector in the basis:

$\mathbf{d} = \lambda\mathbf{a} + \mu\mathbf{b} + \nu\mathbf{c}.$

Go to the coordinate form:

$\left\{ \begin{array}{*{20}{l}} 5 = 3\lambda + \mu + \nu\\ 0 = 2\mu\\ -3 = -\lambda - 5\mu - \nu \end{array} \right., \Rightarrow \left\{ \begin{array}{*{20}{l}} 5 = 3\lambda + \nu\\ \mu = 0\\ 3 = \lambda + \nu \end{array} \right., \Rightarrow \left\{ \begin{array}{*{20}{l}} 2 = 2\lambda\\ \mu = 0\\ \nu = 3 - \lambda \end{array} \right., \Rightarrow \left\{ \begin{array}{*{20}{l}} \lambda = 1\\ \mu = 0\\ \nu = 2 \end{array} \right..$

Therefore, the coordinates of the vector $$\mathbf{d}$$ in this basis are $$\left({\lambda,\mu,\nu}\right) = \left({1,0,2}\right).$$

### Example 3.

In the parallelogram $$ABCD,$$ point $$M$$ is the midpoint of the side $$BC$$ and point $$O$$ is the intersection point of the diagonals. Taking $$\mathbf{AB} = \mathbf{a}$$ and $$\mathbf{AD} = \mathbf{b}$$ as the basis vectors, find in this basis the coordinates of the vectors $$\mathbf{BD},$$ $$\mathbf{CO}$$ and $$\mathbf{MD}.$$

Solution.

Express the vector $$\mathbf{BD}$$ in terms of vectors $$\mathbf{AB} = \mathbf{a}$$ and $$\mathbf{AD} = \mathbf{b}.$$ By the vector subtraction rule,

$\mathbf{BD} = \mathbf{AD} - \mathbf{AB} = -\mathbf{a} + \mathbf{b}, \Rightarrow \mathbf{BD} = \left({-1, 1}\right).$

Find now the vector $$\mathbf{CO}.$$ Notice that

$\mathbf{AC} = \mathbf{AB} + \mathbf{AD} = \mathbf{a} + \mathbf{b}.$

Since $$\mathbf{CA} = - \mathbf{AC}$$ and $$\mathbf{CO} = \frac{1}{2}\mathbf{CA},$$ we get

$\mathbf{CO} = \frac{1}{2}\mathbf{CA} = -\frac{1}{2}\mathbf{AC} = -\frac{1}{2}\left({\mathbf{a} + \mathbf{b}}\right) = -\frac{1}{2}\mathbf{a} - \frac{1}{2}\mathbf{b},$

or

$\mathbf{CO} = \left({-\frac{1}{2}, -\frac{1}{2}}\right).$

Calculate the vector $$\mathbf{MD}:$$

$\mathbf{MD} = \mathbf{AD} - \mathbf{AM},$

where

$\mathbf{AM} = \mathbf{AB} + \mathbf{BM} = \mathbf{AB} + \frac{1}{2}\mathbf{AD} = \mathbf{a} + \frac{1}{2}\mathbf{b}.$

Hence

$\mathbf{MD} = \mathbf{AD} - \mathbf{AM} = \mathbf{b} - \left({\mathbf{a} + \frac{1}{2}\mathbf{b}}\right) = -\mathbf{a} - \frac{1}{2}\mathbf{b}, \Rightarrow \mathbf{MD} = \left({-1, -\frac{1}{2}}\right).$

### Example 4.

Find a basis for the set of $$2 \times 2$$ matrices and determine the coordinates of the matrix

$\left[ \begin{array}{*{20}{r}} -1 & 5\\ 3 & 0 \end{array} \right]$

in this basis.

Solution.

As a basis we can take the following matrices

$\left[ \begin{array}{*{20}{r}} 1 & 0\\ 0 & 0 \end{array} \right],\left[ \begin{array}{*{20}{r}} 0 & 1\\ 0 & 0 \end{array} \right],\left[ \begin{array}{*{20}{r}} 0 & 0\\ 1 & 0 \end{array} \right],\left[ \begin{array}{*{20}{r}} 0 & 0\\ 0 & 1 \end{array} \right].$

Then an arbitrary $$2 \times 2$$ matrix is represented in this basis as follows:

$\left[ \begin{array}{*{20}{r}} a_{11} & a_{12}\\ a_{21} & a_{22} \end{array} \right] = a_{11}\left[ \begin{array}{*{20}{r}} 1 & 0\\ 0 & 0 \end{array} \right] + a_{12}\left[ \begin{array}{*{20}{r}} 0 & 1\\ 0 & 0 \end{array} \right] + a_{21}\left[ \begin{array}{*{20}{r}} 0 & 0\\ 1 & 0 \end{array} \right] + a_{22}\left[ \begin{array}{*{20}{r}} 0 & 0\\ 0 & 1 \end{array} \right].$

In our case we have

$\left[ \begin{array}{*{20}{r}} -1 & 5\\ 3 & 0 \end{array} \right] = -1\cdot\left[ \begin{array}{*{20}{r}} 1 & 0\\ 0 & 0 \end{array} \right] + 5\cdot\left[ \begin{array}{*{20}{r}} 0 & 1\\ 0 & 0 \end{array} \right] + 3\cdot\left[ \begin{array}{*{20}{r}} 0 & 0\\ 1 & 0 \end{array} \right].$

This matrix in the given basis has coordinates $$\left({-1,5,3,0}\right).$$