2.4 Basis and Dimension - ESE 2030 📏

1Reading¶

Material related to this page, as well as additional exercises, can be found in ALA Ch. 2.4 and LAA 4.3-4.5.

2Learning Objectives¶

By the end of this page, you should know:

a basis of a vector space
the dimension of a vector space
a coordinate system for generic vector spaces

3Basis¶

The previous section was admittedly quite abstract, but it was necessary to get us to the extremely practical notion of a basis of a vector space. This section is where the magic happens: we will show that any $n$ dimensional vector space doesn’t just look like, but “behaves the same” as $\mathbb{R}^n$ .

Another way of thinking about a basis is we are looking for the smallest collection of vectors that allows us to express any vector $\vv v \in V$ as a linear combination from our collection.

4Coordinate System¶

An important reason for specifying a basis for a vector space $V$ is to impose a coordinate system on $V$ . This section will show that if if dim $(V) = n$ , that is, if the basis has $n$ elements, then the coordinate system makes $V$ behave exactly like $\mathbb{R}^n$ !

Theorem 1

Let $\vv v_1 ,\ldots, \vv v_n$ be a basis for a vector space $V$ . Then, for each $\vv v \in V$ , there exists a unique set of coefficients $c_1, \ldots, c_n$ such that

\vv v = c_1 \vv v_1 + c_2 \vv v_2 + \ldots + c_n \vv v_n

(2)

Proof 1 (Proof of Theorem 1)

Since $\vv v_1 ,\ldots, \vv v_n$ is a basis for $V$ , there exists at least one set of coefficients such that (2) holds. Suppose $\vv v$ also has the representation

\vv v = d_1 \vv v_1 + d_2 \vv v_2 + \ldots + d_n \vv v_n,

(3)

then,

\vv 0 = \vv v - \vv v = (d_1 - c_1)\vv v_1 + (d_2 - c_2)\vv v_2 + \ldots + (d_n - c_n)\vv v_n.

(4)

However, since $\vv v_1 ,\ldots, \vv v_n$ forms a basis, they are linearly dependent, meaning (3) is only satisfied for $d_i - c_i = 0 \Leftrightarrow d_i = c_i$ for $= 1, 2, \ldots, n$ .

Example 4 (Change of Basis)

Consider the standard basis $B = \{ \vv e_1, \vv e_2\}$ for $\mathbb R^2$ . Then we have $\bm x_1 \\x_2 \em = x_1 \vv e_1 + x_2 \vv e_2$ , and so the $B$ -coordinates of $\vv x$ are $x_1$ and $x_2$ , as expected.

What if we instead use he basis $B' = \left\{ \bm 1\\1 \em, \bm 1\\-1 \em \right\}$ ? We need to find coordinates $c_1$ and $c_2$ such that

\begin{align*} c_1\bm 1 \\ 1\em + c_2 \bm 1\\ -1\em = \bm c_1 + c_2 \\ c_1 - c_2 \em = \bm x_1 \\ x_2 \em \end{align*}

(6)

This linear system has a solution (it must!), and it is $c_1 = \frac{x_1 + x_2}{2}, c_1 = \frac{x_1 - x_2}{2}$ . In the coordinate system defined by $B'$ , the coordinates for $\bm x_1\\x_2 \em$ are $\frac 1 2 \bm x_1 + x_2 \\ x_1 - x_2 \em$ .

Moving from basis $B$ to $B'$ is called a change of basis. Can you pose this as finding the solution to $A \vv c = \vv x$ ?

Example 5 (Isomorphic Vector Spaces)

Let $B = \{ 1, x, x^2, x^3 \}$ be the standard basis for $P^{(3)}$ . A typical element $p(x) \in P^{(3)}$ has the form $p(x) = a_0 + a_1x + a_2x^2 + a_3x^3$ .

We can read off the coordinates of $p(x)$ with respect to $B$ , which we encode in the vector $\vv p$ :

\begin{align*} \vv p = \bm a_0 \\ a_1 \\ a_2 \\ a_3 \em \end{align*}

(7)

Notice that $\vv p$ lives in $\mathbb R^4$ ! Notice even further that the coefficients of the sum of polynomials:

\begin{align*} p(x) = a_0 + a_1x + a_2x^2 + a_3x^3, \quad q(x) = b_0 + b_1x + b_2x^2 + b_3x^3\\ p(x) + q(x) = (a_0 + b_0) + (a_1 + b_1)x + (a_2 + b_2)x^2 + (a_3 + b_3)x^3 \end{align*}

(8)

can be obtained by adding their $B$ -coordinate vectors in $\mathbb R^4$ :

\begin{align*} \vv p + \vv q = \bm a_0\\a_1\\a_2\\a_3 \em + \bm b_0\\b_1\\b_2\\b_3 \em = \bm a_2 + b_0 \\ a_1 + b_1 \\ a_2 + b_2 \\ a_3 + b_3 \em \end{align*}

(9)

You should check that the same connection between $cp(x)$ and $c\vv p$ also holds. By setting a suitable coordiante system (via a basis), working with elements of $P^{(3)}$ , which are polynomial functions, can be turned into working with ordinary vectors in $\mathbb{R}^4$ !

The idea of going back and forth between the vector spaces is captured in terms of a vector space isomorphism. We do not yet have all of the tools needed to define this rigorously, but for now, we will interpret it as meaning that every vector space calculation in $V$ is accurately reproduced in $W$ , and vice versa.

In the above example, we used that $P^{(3)}$ and $\mathbb R^4$ are isomorphic, so we can add and scale either the polynomials directly, or work with their coefficient vectors in $\mathbb R^4$ . Even though they are closely related, they are not the same thing. Rather $P^{(3)}$ and $\mathbb R^4$ are different ways of representing polynomials of degree $\leq 3$ , connected via the chosen basis $B = \{ 1, x, x^2 ,x^3\}$ .