Basis of a Vector Space

# Basis of a Vector Space

We will now look at a new definition regarding vector spaces.

 Definition: A set of vectors $\{ v_1, v_2, ..., v_n \}$ is said to be a Basis of the $\mathbb{F}$-vector space $V$ if both $V = \mathrm{span} (v_1, v_2, ..., v_n)$ and $\{v_1, v_2, ..., v_n \}$ is a linearly independent set. From the definition, to prove that a set of vectors $\{ v_1, v_2, ..., v_n \}$ from $V$ is a basis of $V$, we must show that this set of vectors is a spanning set of $V$ and that this set of vectors is linearly independent.

For example, consider the vector space $\mathbb{R}^3$ which we are familiar with already, and recall the standard unit vectors $\vec{i} = (1, 0, 0)$, $\vec{j} = (0, 1, 0)$ and $\vec{k} = (0, 0, 1)$. The set of vectors $\{ \vec{i}, \vec{j}, \vec{k} \}$ form a basis of $\mathbb{R}^n$. First notice that any vector $(x, y, z) \in \mathbb{R}^n$ can be written in the form $(x, y, z) = a_1(1, 0, 0) + a_2(0,1,0) + a_3(0, 0, 1) = (a_1, a_2, a_3)$ (by choosing $a_1 = x$, $a_2 = y$, and $a_3 = z$) for all $a_1, a_2, a_3 \in \mathbb{F}$ and so $V = \mathrm{span} (\vec{i}, \vec{j}, \vec{k})$. Furthermore, this set of vectors is linearly independent since $a_1(1,0,0) + a_2(0,1,0) + a_3(0,0,1) = (a_1, a_2, a_3) = (0, 0, 0)$ is only true for $a_1 = a_2 = a_3 = 0$.

For another example, consider the set of polynomials with degree less than or equal to $n$. The set $\{ 1, x, x^2, ..., x^n \}$ forms a basis as you should verify.

We will now look at a very important theorem which defines whether a set of vectors is a basis of a finite-dimensional vector space or not.

 Theorem 1: A set of vectors $B = \{ v_1, v_2, ..., v_n \}$ from the vector space $V$ is a basis if and only if each vector $v \in V$ can be written uniquely as a linearly combination of the vectors in $B$, that is $v = a_1v_1 + a_2v_2 + ... + a_nv_n$.
• Proof: $\Rightarrow$ Let $B = \{ v_1, v_2, ..., v_n \}$ be a basis of the vector space $V$, and let $v \in \mathbb{V}$. We know that by definition $B$ is also a spanning set, and so $v = a_1v_1 + a_2v_2 + ... + a_nv_n$ where $a_i \in \mathbb{F}$. Now suppose also that $v = b_1v_1 + b_2v_2 + ... + b_nv_n$. In other words, suppose that $v$ has two represents as a linear combination from the vectors in $B$. The it follows from subtracting these equations that:
(1)
\begin{align} \quad 0 = v - v = (a_1v_1 + a_2v_2 + ... + a_nv_n) - (b_1v_1 + b_2v_2 + ... + b_nv_n) \\ 0 = (a_1 - b_1)v_1 + (a_2 - b_2)v_2 + ... + (a_n - b_n)v_n \end{align}
• Now recall that by definition the basis $B$ is also a linearly independent set which implies that the representation of $0 = (a_1 - b_1)v_1 + (a_2 - b_2)v_2 + ... + (a_n - b_n)v_n$ is unique, and so $a_1 - b_1 = 0$, $a_2 - b_2 = 0$, …, $a_n - b_n = 0$ which implies that $a_1 = b_1$ and $a_2 = b_2$, and …, $a_n = b_n$. Therefore $v \in V$ has a unique representation as a linear combination of the basis vectors.
• $\Leftarrow$ Instead, suppose that every vector $v \in V$ could be unique written as a linear combination of the vectors in $B$, that is $v = a_1v_1 + a_2v_2 + ... + a_nv_n$ with $a_i \in \mathbb{F}$. By definition, the set of vectors $\{ v_1, v_2, ..., v_n \}$ would be a spanning set of $V$.
• Also, since $0 \in V$ we note that $0 = a_1v_1 + a_2v_2 + ... + a_nv_n$ has a unique representation which must trivially be that $a_1 = a_2 = ... = a_n = 0$, and so the set of vectors $\{a_1, a_2, ..., a_n \}$ is also linearly independent.
• Since $\{ v_1, v_2, ..., v_n \}$ is a spanning set of $V$ and is linearly independent, then by definition $\{ v_1, v_2, ..., v_n \}$ is a basis of $V$. $\blacksquare$

## Example 1

Consider the vector space $M_{22}$ of all $2 \times 2$ matrices whose entries are in $\mathbb{F}$. Find a basis of $M_{22}$.

The simplest basis for $M_{22}$ is the step of vectors $\left \{ \begin{bmatrix} 1 & 0\\ 0 & 0\end{bmatrix}, \begin{bmatrix} 0 & 1\\ 0 & 0\end{bmatrix}, \begin{bmatrix} 0 & 0\\ 1 & 0\end{bmatrix}, \begin{bmatrix} 0 & 0\\ 0 & 1\end{bmatrix} \right \}$.

First note that this set of vectors spans $M_{22}$, since $a_1 \begin{bmatrix} 1 & 0\\ 0 & 0\end{bmatrix} + a_2 \begin{bmatrix} 0 & 1\\ 0 & 0\end{bmatrix} + a_3 \begin{bmatrix} 0 & 0\\ 1 & 0\end{bmatrix} + a_4 \begin{bmatrix} 0 & 0\\ 0 & 1\end{bmatrix} = \begin{bmatrix} a_1 & a_2\\ a_3 & a_4\end{bmatrix}$, and for any vector $x \in M_{22}$ can be written as a linear combination of this set of vectors, that is $\begin{bmatrix} a_1 & a_2\\ a_3 & a_4\end{bmatrix} = \begin{bmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{bmatrix}$ when $x_{11} = a_1$, $x_{12} = a_2$, $x_{21} = a_3$ and $x_{22} = a_4$ and so $V = \mathrm{span} \left ( \begin{bmatrix} 1 & 0\\ 0 & 0\end{bmatrix}, \begin{bmatrix} 0 & 1\\ 0 & 0\end{bmatrix}, \begin{bmatrix} 0 & 0\\ 1 & 0\end{bmatrix}, \begin{bmatrix} 0 & 0\\ 0 & 1\end{bmatrix} \right )$.

Now we need to show that this set of vectors is linearly independent. Clearly $\begin{bmatrix} a_1 & a_2\\ a_3 & a_4\end{bmatrix} =\begin{bmatrix} 0 & 0\\ 0 & 0\end{bmatrix}$ if and only if $a_1 = a_2 = a_3 = a_4 = 0$, and so this set is linearly independent.

Therefore this set in all is a basis for $M_{22}$.

## Example 2

Consider the vector space of complex numbers $\mathbb{C}$. Find a basis of $\mathbb{C}$.

The simplest basis is the set of vectors $\{ 1, i \}$. We note that $\mathbb{C} = \mathrm{span} (1, i)$ since any complex number $z$ can be written in the form $a + bi$ where $a, b \in \mathbb{F}$. Furthermore, this set of vectors is linearly independent since $a + bi = 0$ if and only if $a = b = 0$.

## Example 3

Consider the vector space of complex numbers $\mathbb{C}$. Prove that the set of vectors $\{ 1 + i, 1 - i \}$ is a basis of $\mathbb{C}$.

To prove this we must show that $\mathbb{C} = \mathrm{span} (1 + i, 1 - i)$ and that $\{ 1 + i, 1 - i \}$ is a linearly independent set of vectors. First, let's chose that this set spans $\mathbb{C}$.

We note that $1 = \frac{1}{2}(1 + i) + \frac{1}{2}(1 - i)$ and that $i = \frac{1}{2} (1 + i ) - \frac{1}{2}(1 - i)$ which we verified in example 2 above to span $\mathbb{C}$ and so $\mathbb{C} = \mathrm{span} (1 + i, 1 - i)$.

Now we will show this set of vectors is linearly independent. We have that:

(2)
\begin{align} a(1 + i) + b(1 - i) = 0 \\ a + ai +b - bi = 0 \\ (a + b) + (a - b)i =0 \end{align}

The vector equation above is only satisfied if $a + b = 0$ $a - b = 0$ and so $a = b = 0$. Therefore $\{ 1 + i, 1 - i \}$ is a basis of $\mathbb{C}$.

## Example 4

Prove that if $B = \{ v_1, v_2, ..., v_n \}$ is a basis of the vector space $V$, then $B' = \{ v_1 - v_2, v_2 - v_3, ..., v_{n-1} - v_n, v_n \}$ is also a basis of $V$.

To show that $B'$ is a basis, we must show that $V = \mathrm{span} (B')$ and that $B'$ is a linearly independent set.

First, since $B$ is a basis of $V$, by the definition of a basis, $B$ spans $V$, that is $\{ v_1, v_2, ..., v_n \}$ spans $V$. We want to show that $B'$ also spans $V$. Let $v \in V$, and consider the following vector equation:

(3)
\begin{align} \quad v = a_1(v_1 - v_2) + a_2(v_2 - v_3) + ... + a_{n-1}(v_{n-1} - v_n) + a_nv_n \\ \quad v = a_1v_1 - a_1v_2 + a_2v_2 - a_2v_3 + ... + a_{n-1}v_{n-1} - a_{n-1}v_n + a_nv_n \\ \quad v = a_1v_1 + (a_2 - a_1)v_2 + (a_3 - a_2)v_3 + ... + (a_n - a_{n-1})v_n \end{align}

Since $v \in \mathrm{span} (B)$ there exists scalars $b_1, b_2, ..., b_n \in \mathbb{F}$ such that:

(4)
\begin{equation} v = b_1v_1 + b_2v_2 + ... + b_nv_n \end{equation}

So if $b_1 = a_1$, $b_2 = a_2 - a_1$, …, $b_n = a_n - a_{n-1}$ then $v$ is a linear combination of the vectors in $B'$ and so $v \in \mathrm{span} (B')$.

Now we need to show that $B'$ is linearly independent. Consider the vector equation:

(5)
\begin{align} \quad 0 = a_1(v_1 - v_2) + a_2(v_2 - v_3) + ... + a_{n-1}(v_{n-1} - v_n) + a_nv_n \\ \quad 0 = a_1v_1 + (a_2 - a_1)v_2 + (a_3 - a_2)v_3 + ... + (a_n - a_{n-1})v_n \end{align}

Since $B = \{ v_1, v_2, ..., v_n \}$ is a spanning set of $V$ and hence linearly independent, this implies that $a_1 = 0$, $a_2 - a_1 = 0$, $a_3 - a_2 = 0$, and so forth, or in other words, $a_1 = a_2 = ... = a_n = 0$. So then $B'$ is also linearly independent. $\blacksquare$