Linear Coordinate System

A linear coordinate system is a framework used to represent vectors in a vector space using a set of basis vectors. Each vector in the space can be expressed as a linear combination of these basis vectors.

Definition

Let $V$ be a vector space, and $B$ a basis for $V$.

$$V=\{\overrightarrow{b_1},\overrightarrow{b_2},\dots ,\overrightarrow{b_n}\}$$

Any $\vec{x}\in V$ can be written in a unique way as

$$\vec{x}=c_1\overrightarrow{b_1}+c_2\overrightarrow{b_2}+\dots +c_n\overrightarrow{b_n}$$

where $c_1,c_2,\dots ,c_n$ are scalars called the coordinates of $\vec{x}$ with respect to the basis $B$.

Example

Setup

$$V={\mathbb{R}}^2,B=\left\{\left[\begin{array}{c}1\\ 2\end{array}\right],\left[\begin{array}{c}3\\ 0\end{array}\right]\right\}$$

Find the coordinate of $\vec{v}=\left[\begin{array}{c}7\\ -3\end{array}\right]$ in the basis $B$

$$\left[\begin{array}{c}7\\ -3\end{array}\right]=c_1\left[\begin{array}{c}1\\ 2\end{array}\right]+c_2\left[\begin{array}{c}3\\ 0\end{array}\right]$$ $$\{\begin{array}{l}{a}_1+3a_2=7\\ 2a_1=-3\end{array}$$

By observation or by a augmented matrix, we find

$$a_1=-\frac{3}{2},a_2=\frac{17}{6}$$

Therefore, the coordinate of $\overrightarrow{v_1}$ in basis $B$ is

$${\left[\begin{array}{c}\vec{x}\end{array}\right]}_B=\left[\begin{array}{c}-\frac{3}{2}\\ \frac{17}{6}\end{array}\right]$$

Finding a Basis

If $A\in m\times n$ matrix, we can write the matrix equation as:

$$A\vec{x}=\vec{b}$$

And recognize the columns of $A$ as separate vectors:

$$A=\left[\begin{array}{cccc}\overrightarrow{a_1}& \overrightarrow{a_2}& \dots & \overrightarrow{a_n}\end{array}\right]$$

Thus, the equation can be rewritten as:

$$\vec{x}=\left[\begin{array}{c}{x}_1\\ x_2\\ ⋮\\ x_n\end{array}\right]⟹A\vec{x}=x_1\overrightarrow{a_1}+x_2\overrightarrow{a_2}+\dots +x_n\overrightarrow{a_n}=\vec{b}$$

We then define $\{\overrightarrow{a_1},\overrightarrow{a_2},\dots ,\overrightarrow{a_n}\}=B$ as a basis for the vector space spanned by the columns of $A$.

Reaffirming:

• $\vec{x}$ is the coordinate vector of $\vec{b}$ in the basis $B$
• $A$ is a matrix where its columns are the vectors of basis $B$

We can then represent the vector $\vec{b}$ as

$$\vec{b}=A{\left[\begin{array}{c}\vec{b}\end{array}\right]}_B$$

Because the columns of $A$ (by definition of a basis) are linearly independent, the matrix $A$ is invertible.

Multiplying by the inverse yields

$$A^{-1}\cdot \vec{b}={\left[\begin{array}{c}\vec{b}\end{array}\right]}_B$$

Thus, we can find the coordinate of a vector in a basis using the inverse of the matrix representing the basis $A^{-1}$.

This matrix $A$ is commonly called $P_B$.

Example

Coninuing from previous

$$P_B=\left[\begin{array}{cc}1& 3\\ 2& 0\end{array}\right]$$

computing the inverse matrix yields

$$P_B^{-1}=\left[\begin{array}{cc}0& \frac{1}{2}\\ \frac{1}{3}& -\frac{1}{6}\end{array}\right]$$

We then define $\vec{x}$ as

$$\vec{x}=\left[\begin{array}{c}7\\ -3\end{array}\right]$$

Finding the coordinate of $\vec{x}$ in basis $B$:

$$P_B^{-1}\cdot \vec{x}=\left[\begin{array}{cc}0& \frac{1}{2}\\ \frac{1}{3}& -\frac{1}{6}\end{array}\right]\cdot \left[\begin{array}{c}7\\ -3\end{array}\right]=\left[\begin{array}{c}-\frac{3}{2}\\ \frac{17}{6}\end{array}\right]$$

Concluding Example

Setup

$$\overrightarrow{v_1}=\left[\begin{array}{c}3\\ 6\\ 2\end{array}\right],\overrightarrow{v_2}=\left[\begin{array}{c}-1\\ 0\\ 1\end{array}\right]\in {\mathbb{R}}^3$$

$H=\mathrm{span}\{\overrightarrow{v_1},\overrightarrow{v_2}\}$ is a subspace of ${\mathbb{R}}^3$.

$B=\{\overrightarrow{v_1},\overrightarrow{v_2}\}$ is a basis of $H$.

Question

Given $\vec{x}=\left[\begin{array}{c}3\\ 12\\ 7\end{array}\right]$, determine if $\vec{x}\in H$ and if so, find its coordinates in $B$.

Solution

If $\vec{x}\in H$ then $\vec{x}$ is a linear combination of the vectors in basis $B$.

$$\vec{x}=c_1\overrightarrow{v_1}+c_2\overrightarrow{v_2}$$

breaking into vector components

$$\{\begin{array}{l}3c_1-c_2=3\\ 6c_1=12\\ 2c_1+c_2=7\end{array}$$

an augmented matrix can be formed as follows:

$$\left[\begin{array}{ccc}3& -1& 3\\ 6& 0& 12\\ 2& 1& 7\end{array}\right]$$

performing row reduction yields

$$\left[\begin{array}{ccc}1& 0& 2\\ 0& 1& 3\\ 0& 0& 0\end{array}\right]$$

therefore the coordinate of the vector $\vec{x}$ in $B$ is:

$$c_1=2,c_2=3\Rightarrow {\left[\begin{array}{c}\vec{x}\end{array}\right]}_B=\left[\begin{array}{c}2\\ 3\end{array}\right]$$