Basis of a vector space

Let \(V\) be a vector space. A minimal set of vectors in \(V\) that spans \(V\) is called a basis for \(V\).

Equivalently, a basis for \(V\) is a set of vectors that

• is linearly independent;

• spans \(V\).

As a result, to check if a set of vectors form a basis for a vector space, one needs to check that it is linearly independent and that it spans the vector space. If at least one of these conditions fail to hold, then it is not a basis.

Examples

1. In \(\mathbb{R}^3\), every vector has the form \(\begin{bmatrix} a\\b\\c\end{bmatrix}\) where \(a,b,c\) are real numbers. Note that \(\mathbb{R}^3\) is spanned by the set \(\left\{\begin{bmatrix} 1\\0\\0\end{bmatrix}, \begin{bmatrix} 0\\1\\0\end{bmatrix}, \begin{bmatrix} 0\\0\\1\end{bmatrix}\right\}\) since \(a\begin{bmatrix} 1\\0\\0\end{bmatrix}+ b\begin{bmatrix} 0\\1\\0\end{bmatrix}+ c\begin{bmatrix} 0\\0\\1\end{bmatrix}= \begin{bmatrix} a\\b\\c\end{bmatrix}\). Clearly, \(a\begin{bmatrix} 1\\0\\0\end{bmatrix}+ b\begin{bmatrix} 0\\1\\0\end{bmatrix}+ c\begin{bmatrix} 0\\0\\1\end{bmatrix}= \begin{bmatrix} 0\\0\\0\end{bmatrix}\) if and only if \(a=b=c=0\). Hence, the set is a linearly independent set that spans \(\mathbb{R}^3\) and is therefore a basis for \(\mathbb{R}^3\). (Note that the set \(\left\{\begin{bmatrix} 1\\0\\0\end{bmatrix}, \begin{bmatrix} 0\\1\\0 \end{bmatrix}, \begin{bmatrix} 0\\0\\1 \end{bmatrix}, \begin{bmatrix} 1\\1\\1 \end{bmatrix} \right\}\) is not a basis for \(\mathbb{R}^3\) even though it spans \(\mathbb{R}^3\) since it is not a linearly independent set.)

2. The set \(\{x^2, x, 1\}\) is a basis for the vector space of polynomials in \(x\) with real coefficients having degree at most \(2\).

Observe that \(\mathbb{R^3}\) has infinitely many vectors yet we managed to have a description of all of them using just three vectors. We can think of a basis as a minimal way to describe a vector space which makes many types of computations over vector spaces feasible.

Dimension of a vector space

Let \(V\) be a vector space not of infinite dimension. An important result in linear algebra is the following:

Every basis for \(V\) has the same number of vectors.

It can be shown that every set of linearly independent vectors in \(V\) has size at most \(\dim(V)\). For example, a set of four vectors in \(\mathbb{R}^3\) cannot be a linearly independent set.

Quick Quiz

Exercises

1. Let \(\mathbb{F}\) denote a field. Give a basis for \(\mathbb{F}^4\).  

   There are many possible answers. One possible answer is \(\left\{ \begin{bmatrix} 1\\ 0 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0\\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0\\ 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0\\ 0 \\ 0 \\ 1 \end{bmatrix} \right \}\).

2. Give a basis for \(\mathbb{R}^{2\times 2}\).  

   There are many possible answers. One possible answer is \(\left\{ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \right \}\).

3. What is the dimension of the vector space of polynomials in \(x\) with real coefficients having degree at most three?  

   The dimension is 4 since every such polynomial is of the form \(ax^3 + bx^2 + cx + d\) where \(a,b,c,d \in \mathbb{R}\). and has for example, \(\{x^3,x^2,x,1\}\) as a basis.