Home page for accesible maths 2 Vector spaces 2.D Dimension and bases 2.F Row space and column space

Style control - access keys in brackets

Font (2 3) - + Letter spacing (4 5) - + Word spacing (6 7) - + Line spacing (8 9) - +

2.E Coordinates

You are familiar with writing vectors as linear combinations of the standard basis vectors. For example, if $\vec{v}=(4,5,6)$ , then:

\vec{v}=4\vec{e_{1}}+5\vec{e_{2}}+6\vec{e_{3}}.

Here the scalars $4,5,$ and $6$ are called the coordinates of $\vec{v}$ , with respect to the standard basis. But according to Theorem 2.26, if we were to choose any other basis of our vector space, then we could uniquely express $\vec{v}$ as a linear combination of those, and this would produce a different set of “coordinates”.

Definition 2.42:

If $\mathcal{B}=(\vec{v_{1}},\cdots,\vec{v_{n}})$ is a basis of a vector space $V$ over a field $F$ , and

\vec{v}=\alpha_{1}\vec{v_{1}}+\cdots+\alpha_{n}\vec{v_{n}},

then the sequence $(\alpha_{1},\alpha_{2},\cdots,\alpha_{n})$ of scalars in $F$ are called the coordinates of $\vec{v}$ with respect to the basis $\mathcal{B}$ .

Furthermore, the column vector, which is an $n\times 1$ matrix,

[\vec{v}]_{\mathcal{B}}:=\begin{bmatrix}\alpha_{1}\\ \vdots\\ \alpha_{n}\end{bmatrix}

is called the coordinate matrix of $\vec{v}$ with respect to $\mathcal{B}$ .

When the basis $\mathcal{B}$ is the standard basis of the vector space $F^{n}$ , to avoid cumbersome notation, we may simply write $\vec{v}$ instead of $[\vec{v}]_{\mathcal{B}}$ . Now you might object, since we have been writing vectors horizonally. Sometimes it will be convenient to view $F^{n}$ as column matrices, and this is commonly done in applications, such as statistics; but sometimes it will be convenient to view $F^{n}$ as row vectors, as we have been doing so far. We hope to make it clear whether $\vec{v}$ refers to a row vector or a column vector, when that distinction matters. In any case, it is clear how to switch between the two:

(x_{1},x_{2},\cdots,x_{n})\leftrightarrow\begin{bmatrix}x_{1}\\ x_{2}\\ \vdots\\ x_{n}\end{bmatrix}.

(2.43)

Example 2.44.

Consider the bases $\mathcal{B}=((1,0),(0,1))$ and $\mathcal{C}=((0,1),(3,-1))$ of ${\mathbb{R}}^{2}$ , and let $\vec{v}=(1,1)$ . The coordinate matrices of $\vec{v}$ , with respect to these two different bases are as follows:

$[\vec{v}]_{\mathcal{B}}=\begin{bmatrix}1\\ 1\end{bmatrix}$
$[\vec{v}]_{\mathcal{C}}={\begin{bmatrix}4/3\\ 1/3\end{bmatrix}}$

This is because $(1,1)=1(1,0)+1(0,1)$ and $(1,1)=\frac{4}{3}(0,1)+\frac{1}{3}(3,-1)$ .

Exercise 2.45:

Find the coordinate matrix for the following vectors, with respect to the given bases.

i.

$[(1,1,1)]_{\mathcal{B}}$ , where $\mathcal{B}$ is the standard basis of ${\mathbb{R}}^{3}$ .
ii.

$[(1,2,3)]_{\mathcal{B}}$ , where $\mathcal{B}=((1,0,0),(1,1,0),(1,1,1))$ of ${\mathbb{R}}^{3}$ .
iii.

$[1+x-x^{2}]_{\mathcal{B}}$ , where $\mathcal{B}=(1+x,1-x,x+x^{2})$ of $\mathcal{P}_{2}({\mathbb{C}})$ .

[End of Exercise]

The above exercises are specific cases of the following question: If I have the coordinates of a vector in one basis, how do I find its coordinates in another basis?

You should be able to answer the above exercises by solving a system of equations in the coefficient variables; or possibly by guessing wisely. In Section 4.G we will develop a more systematic method using change of basis matrices.