Basis of a vector space

A basis of a vector space V is a set of linearly independent vectors whose span is the space V.

Why we introduce a basis of a vector space

Let V be a vector space and W ⊆ V a subset such that the span span(W) of this set is equal to the space V, that is span(W) = V. Example: let W = {(1, 0), (0, 1), (2, 3)} and V = ℝ².

Clearly span(W) = ℝ². The set W has only three vectors, yet thanks to the span we can use it to describe the whole infinite set of vectors ℝ². This raises a question: can we remove some more vectors from W and still keep the property span(W) = ℝ²?

If we threw away the last vector (2, 3), for example, we would get the set W₁ = {(1, 0), (0, 1)}. The span of this set is again ℝ². So the question becomes: what is the smallest set of vectors whose span is ℝ²?

If we removed either of the two remaining vectors from W₁, the resulting set, call it W₂, would no longer have ℝ² as its span. For example, the span of the set W₂ = {(1, 0)} would be span(W₂) = {(a, 0)  ∣  a ∈ ℝ}. This is certainly not equal to ℝ², since it is missing the vector (0, 1), for instance.

So the set W₁ = {(1, 0), (0, 1)} is a minimal set of vectors whose span is the space ℝ². Other questions we might ask are:

1. Is there some completely different one-element set whose span is the space ℝ²?
2. Is there a unique smallest set of vectors whose span is the space ℝ²?

As we all know from the tabloids, if a newspaper headline is a question, the answer is no, and this case is no different. There is no one-element set whose span is ℝ². We can prove it like this:

Is there a one-element set whose span is ℝ²?

Suppose there is a set {x} containing one vector x whose linear combinations make up the whole space ℝ². Then there must be real numbers a₁, a₂ such that

$$\begin{array}{rcl}{a}_1\cdot \mathbf{x}& =& (1,0)\\ a_2\cdot \mathbf{x}& =& (0,1)\end{array}$$

Why these particular vectors? It hardly matters which vectors we choose, as long as they are linearly independent.

Write x = (x₁, x₂). Then certainly x₁, x₂ ≠ 0, because if x₁ = 0, we could hardly get the vector (1, 0) as a linear combination of the vector x. The same goes for x₂. We just write out the previous pair of equations in full:

$$\begin{array}{rcl}{a}_1\cdot (x_1,x_2)& =& (1,0)\\ a_2\cdot (x_1,x_2)& =& (0,1)\end{array}$$

Writing these as separate equations:

$$\begin{array}{rcl}{a}_1\cdot x_1& =& 1\\ a_1\cdot x_2& =& 0\\ a_2\cdot x_1& =& 0\\ a_2\cdot x_2& =& 1\end{array}$$

The second and third equations tell us that a₁ = a₂ = 0, because if x₁, x₂ ≠ 0, the only way to get zero in the products a₁ · x₂ and a₂ · x₁ is a₁ = a₂ = 0. But this contradicts the first and fourth equations: if a₁ = 0, then a₁ · x₁ = 0, but the first equation says that a₁ · x₁ = 1.

So no one-element set can have the space ℝ² as its span.

Is there a smallest set of vectors whose span is ℝ²?

We know that there is no one-element set with this property, and we know that the set W = {(1, 0), (0, 1)} has ℝ² as its span. So we ask whether this is the only two-element set whose span is ℝ².

Of course it is not, because another two-element set, W = {(2, 0), (0, 2)}, also has the span ℝ². Any pair of vectors (a, 0), (0, a) with a ∈ ℝ, a ≠ 0 has ℝ² as its span. In fact, any pair of linearly independent vectors has ℝ² as its span. So there are infinitely many two-element sets whose span can generate ℝ².

At the same time, any three-element subset W₃ ⊆ ℝ² consists of dependent vectors. Why? If we have three vectors (a, b), (c, d), (e, f), we can write them in a matrix:

$$\left(\begin{array}{ccc}a& c& e\\ b& d& f\end{array}\right)$$

This matrix has rank at most two, so it must contain a linearly dependent column (vector).

Definition of a basis of a vector space

Let V be a vector space and let B ⊆ V. We say that B is a basis of the vector space V if:

1. span(B) = V
2. B is a linearly independent set of vectors.

A basis is an independent set of vectors whose span is the space V. So a basis is a minimal set of vectors whose span is the space V. If we removed one element from a basis, its span would no longer be the space V. If we added one vector, that vector would be a linear combination of the other vectors.

In the space ℝⁿ, the set of n-tuples (1, 0, …, 0), (0, 1, 0, …, 0), …, (0, 0, …, 0, 1) forms a basis of ℝⁿ, and this basis is called the standard basis. For ℝ³, for example, we get the standard basis (1, 0, 0), (0, 1, 0), (0, 0, 1). The vector (a, b, c) is then obtained as the combination

$$a\cdot (1,0,0)+b\cdot (0,1,0)+c\cdot (0,0,1)$$

If we removed the vector (0, 1, 0), the span of the new set would not be ℝ³. If we added the vector (14, −2, 5), it would be a linear combination of the other vectors with coefficients a = 14, b = −2, c = 5.

Basic properties of a basis of a vector space

• Every vector space V has a basis; for the trivial space {0} – the space that contains a single vector, the zero vector – it is the empty set. Every linearly independent set B₀ ⊆ V is either a basis of V or can be extended to one. If B₀ is not a basis, that is span(B₀) ≠ V, then there is a vector x₀ ∈ V ∖ span(B₀). This vector is not a combination of vectors from B₀. We add it to the set B₀ and get a new set B₁ = B₀ ∪ {x₀}. Again we can say that B₁ is either a basis or can be extended to one. If the space V has a finite basis, then after finitely many steps we get a set of independent vectors B_n that is a basis of V.

• A basis of a space is not unique; there is always more than one.

• All bases of a vector space have the same number of elements (vectors), or are infinite. For example, all bases of the space ℝ² have two vectors, and no set of three elements or of one element is a basis of ℝ².

• If we have a basis B = {e₁, …, e_n} of a space V, then every vector x in V can be written as a linear combination of the vectors from the basis, that is, there are real coefficients a₁, …, a_n such that

  $$\mathbf{x}=a_1\cdot {\mathbf{e}}_1+\dots +a_n\cdot {\mathbf{e}}_n$$

  These coefficients are unique. That is, if for coefficients b₁, …, b_n we also had

  $$\mathbf{x}=b_1\cdot {\mathbf{e}}_1+\dots +b_n\cdot {\mathbf{e}}_n$$

  then this must mean that a₁ = b₁, …, a_n = b_n.