As the word itself suggests, the concept of basis of a vector space contains all the information necessary to rebuild the vector space, starting from very “few” vectors.

ℝ^2 =⟨(1,0),(0,1)⟩=⟨(1,0),(0,1),(1,1)⟩

How can we find a minimal set (a set as small as possible) of generators for the space $ℝ^2$?

If we remove from the set any one vector that is a linear combination of the others, the new set obtained generates the same vector space. We may repeat this step as many times as necessary until all vectors are independent.

Let $V = ⟨𝐯_1, . . . , 𝐯_n⟩ ≠$ { $0$ }. Then there exists a subset of { $𝐯_1, . . . , 𝐯_n$ }, consisting of linearly independent vectors, which generates $V$ .

Let $V$ be a vector space. The set { $𝐯_1, . . . , 𝐯_n$ } is called a basis if:

1. The vectors $𝐯_1, . . . , 𝐯_n$ are linearly independent.
2. The vectors $𝐯_1, . . . , 𝐯_n$ generate $V$.

Henceforth, we will say that a set $X$ is maximal with respect to a certain property if $X$ enjoys that property, but as soon as we add an element to $X$, then $X$ does not enjoy the property anymore.

Also, we will say that a set $X$ is minimal with respect to a certain property if $X$ enjoys that property, but as soon as we remove an element from $X$, then $X$ does not enjoy the property anymore.

Let $𝐯_1, . . . , 𝐯_n$ be vectors in a vector space $V$ .

1. { $𝐯_1, . . . , 𝐯_n$ } is a basis of $V$ if and only if it is a minimal set of generators of $V$.
2. { $𝐯_1, . . . , 𝐯_n$ } is a basis of $V$ if and only if it is a maximal set of linearly independent vectors.

If a vector space $V ≠$ { $0$ } is generated by a finite number of vectors $𝐯_1, . . . , 𝐯_n$, then there exists a basis of $V$.

A vector space, as we already know, can admit different bases, however, as we shall see:

• All the bases have the same number of elements,
• This number is called the dimension of the vector space.

Completion Theorem Let S = {v1, . . . .vm} be a set of linearly independent vectors in a finitely generated vector space V. If B = {w1,...,wn} is a basis of V (we know that there is always at least one) then m ≤ n, and we can always add to S n−m vectors from the basis {w1,...,wn} in order to obtain a basis of V.

All the bases of the same finitely generated vector space have the same number of elements.

The number of elements of a basis of a vector space is called the dimension of the vector space, and it is denoted with dim(V ). When this number is finite, or equivalently, when V is generated by a finite number of vectors, V is said to be finite dimensional.

Canonical bases

• ℝⁿ

  The canonical basis is given by C = { $𝐮̂_1 ,...,𝐮̂_n$ }, where $𝐮̂_i$ is the vector that has 1 in position i and 0 in the other positions.

  For example the canonical basis of ℝ³ is C = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}, and ℝ³ has dimension 3, or we can also say that it is a three-dimensional vector space.

  Dim = n

• ℝ[x]

  The canonical basis is given by C = { $x^n,x^{n−1},...,x^1$ }.

  Dim = n+1

• $𝑴_{𝚖⨯𝚗}$

  $𝑴_{𝚖⨯𝚗} = ｛ 𝑼_{1,1} .,. 𝑼_{m,n} ｝$ where $𝑼_{i,j}$ is 1 at position $(i,j)$ and 0 in the other positions.

  C_{2,2}= \begin{Bmatrix}
  \begin{pmatrix}
   1 & 0 \\
   0 & 0 
  \end{pmatrix} & \begin{pmatrix}
   0 & 1 \\
   0 & 0 
  \end{pmatrix} & \begin{pmatrix}
   0 & 0 \\
   1 & 0 
  \end{pmatrix} & \begin{pmatrix}
   0 & 0 \\
   0 & 1 
  \end{pmatrix} 
  \end{Bmatrix}

Dim = m·n

Let $V$ be a vector space of dimension $n$ and let $W$ be a subspace of $V$ . Then:

• $dim(W ) ≤ dim(V )$
• $dim(W) = dim(V) ⟺ V = W

Let V be a vector space of dimension n, and let { $v_1, . . . , v_n$ } be set of n vectors of V . The following are equivalent:

• { $v_1, . . . , v_n$ } is a basis of $V$;
• $v_1, . . . , v_n$ are linearly independent;
• $v_1, . . . , v_n$ generate $V$

Let $𝓑$ = { $v_1, . . . , v_n$ } be an ordered basis for the vector space $V$ (that is, we fixed an order in the set of vectors numbering them) and let $𝐯 ∈ V$ . Then there exists a unique n−tuple of scalars $(α_1, . . . , α_n)$, such that

𝐯 = α_1v_1 + ⋯ + α_nv_n

The scalars $(α_1, . . . , α_n)$ are called the components of $𝐯 ∈ V$ in the basis $𝓑$ or also the coordinates.

(𝐯)_𝓑 = (α_1, ⋯ , α_n)

Given a matrix $A ∈ M_{m,n}(ℝ)$ the elementary row operations do not change the subspace of $ℝ^n$ generated by the row vectors of $A$.