# 2.8: Bases as Coordinate Systems

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##### Objectives
1. Learn to view a basis as a coordinate system on a subspace.
2. Recipes: compute the $$\mathcal{B}$$-coordinates of a vector, compute the usual coordinates of a vector from its $$\mathcal{B}$$-coordinates.
3. Picture: the $$\mathcal{B}$$-coordinates of a vector using its location on a nonstandard coordinate grid.
4. Vocabulary word: $$\mathcal{B}$$-coordinates.

In this section, we interpret a basis of a subspace $$V$$ as a coordinate system on $$V$$, and we learn how to write a vector in $$V$$ in that coordinate system.

##### Fact $$\PageIndex{1}$$

If $$\mathcal{B}=\{v_1,\: v_2,\cdots ,v_m\}$$ is a basis for a subspace $$V$$, then any vector $$x$$ in $$V$$ can be written as a linear combination

$x=c_1v_1+c_2v_2+\cdots +c_mv_m\nonumber$

in exactly one way.

Proof

Recall that to say $$\mathcal{B}$$ is a basis for $$V$$ means that $$\mathcal{B}$$ spans $$V$$ and $$\mathcal{B}$$ is linearly independent. Since $$\mathcal{B}$$ spans $$V$$, we can write any $$x$$ in $$V$$ as a linear combination of $$v_1,\: v_2,\cdots ,v_m$$. For uniqueness, suppose that we had two such expressions:

\begin{aligned}x&=c_1v_1+c_2v_2+\cdots +c_mv_m \\ x&=c_1'v_1+c_2'v_2+\cdots +c_m'v_m.\end{aligned}

Subtracting the first equation from the second yields

$0=x-x=(c_1-c_1')v_1 +(c_2-c_2')v_2+\cdots +(c_m-c_m')v_m.\nonumber$

Since $$\mathcal{B}$$ is linearly independent, the only solution to the above equation is the trivial solution: all the coefficients must be zero. It follows that $$c_i−c_i'$$ for all $$i$$, which proves that $$c_1=c_1',\: c_2=c_2',\cdots ,c_m=c_m'$$.

##### Example $$\PageIndex{1}$$

Consider the standard bases of $$\mathbb{R}^{3}$$ from Example 2.7.3 in Section 2.7.

$e_1=\left(\begin{array}{c}1\\0\\0\end{array}\right),\quad e_2=\left(\begin{array}{c}0\\1\\0\end{array}\right),\quad e_3=\left(\begin{array}{c}0\\0\\1\end{array}\right).\nonumber$

According to the above fact, Fact $$\PageIndex{1}$$, every vector in $$\mathbb{R}^{3}$$ can be written as a linear combination of $$e_1$$, $$e_2$$, $$e_3$$, with unique coefficients. For example,

$v=\left(\begin{array}{c}\color{Red}{3}\\ \color{blue}{5} \\ \color{Green}{-2}\end{array}\right)\color{black}{=}\color{red}{3}\color{black}{\left(\begin{array}{c}1\\0\\0\end{array}\right)+}\color{blue}{5}\color{black}{\left(\begin{array}{c}0\\1\\0\end{array}\right)}\color{Green}{-2}\color{black}{\left(\begin{array}{c}0\\0\\1\end{array}\right)=}\color{Red}{3}\color{black}{e_1+}\color{blue}{5}\color{black}{e_2}\color{Green}{-2}\color{black}{e_3.}\nonumber$

In this case, the coordinates of $$v$$ are exactly the coefficients of $$e_1$$, $$e_2$$, $$e_3$$.

What exactly are coordinates, anyway? One way to think of coordinates is that they give directions for how to get to a certain point from the origin. In the above example, the linear combination $$3e_1+5e_2−2e_3$$ can be thought of as the following list of instructions: start at the origin, travel $$3$$ units north, then travel $$5$$ units east, then $$2$$ units down.

##### Definition $$\PageIndex{1}$$: $$\mathcal{B}$$-Coordinate Vector

Let $$\mathcal{B}=\{v_1,\: v_2,\cdots ,v_m\}$$ be a basis of a subspace $$V$$, and let

$x=c_1v_1+c_2v_2+\cdots+c_mv_m\nonumber$

be a vector in $$V$$. The coefficients $$c_1,\:c_2,\cdots,c_m$$ are the coordinates of $$x$$ with respect to $$\mathcal{B}$$. The $$\mathcal{B}$$-coordinate vector of $$x$$ is the vector

$[x]_{\mathcal{B}}=\left(\begin{array}{c}c_1 \\c_2\\ \vdots \\ c_m\end{array}\right)\quad\text{ in }\mathbb{R}^{m}.\nonumber$

If we change the basis, then we can still give instructions for how to get to the point $$(3,\:5,\:−2)$$, but the instructions will be different. Say for example we take the basis

$v_1=e_1+e_2=\left(\begin{array}{c}1\\1\\0\end{array}\right),\quad v_2=e_2=\left(\begin{array}{c}0\\1\\0\end{array}\right),\quad v_3=e_3=\left(\begin{array}{c}0\\0\\1\end{array}\right).\nonumber$

We can write $$(3,\:5,\:−2)$$ in this basis as $$3v_1+2v_2−2v_3$$. In other words: start at the origin, travel northeast $$3$$ times as far as $$v_1$$, then $$2$$ units east, then $$2$$ units down. In this situation, we can say that “$$3$$ is the $$v_1$$-coordinate of $$(3,\:5,\:−2)$$, $$2$$ is the $$v_2$$-coordinate of $$(3,\:5,\:−2)$$, and $$−2$$ is the $$v_3$$-coordinate of $$(3,\:5,\:−2)$$.”

##### Note $$\PageIndex{1}$$

The above definition, Definition $$\PageIndex{1}$$, gives a way of using $$\mathbb{R}_m$$ to label the points of a subspace of dimension $$m$$: a point is simply labeled by its $$\mathcal{B}$$-coordinate vector. For instance, if we choose a basis for a plane, we can label the points of that plane with the points of $$\mathbb{R}^2$$.

##### Example $$\PageIndex{2}$$: A Nonstandard Coordinate System of $$\mathbb{R}^2$$)

Define

$v_1=\left(\begin{array}{c}1\\1\end{array}\right),\quad v_2=\left(\begin{array}{c}1\\-1\end{array}\right),\quad\mathcal{B}=\{v_1,\:v_2\}.\nonumber$

1. Verify that $$\mathcal{B}$$ is a basis for $$\mathbb{R}^2$$.
2. If $$[w]_{\mathcal{B}}=\left(\begin{array}{c}1\\2\end{array}\right)$$, then what is $$w$$?
3. Find the $$\mathcal{B}$$-coordinates of $$v=\left(\begin{array}{c}5\\3\end{array}\right)$$.
###### Solution
1. By the basis theorem in Section 2.7, Theorem 2.7.3, any two linearly independent vectors form a basis for $$\mathbb{R}^2$$. Clearly $$v_1,\:v_2$$ are not multiples of each other, so they are linearly independent.
2. To say $$[w]_{\mathcal{B}}=\left(\begin{array}{c}1\\2\end{array}\right)$$ means that $$1$$ is the $$v_1$$-coordinate of $$w$$, and that $$2$$ is the $$v_2$$-coordinate:
$w=v_1+2v_2=\left(\begin{array}{c}1\\1\end{array}\right)+2\left(\begin{array}{c}1\\-1\end{array}\right)=\left(\begin{array}{c}3\\-1\end{array}\right).\nonumber$
3. We have to solve the vector equation $$v=c_1v_1+c_2v_2$$ in the unknowns $$c_1,\: c_2$$. We form an augmented matrix and row reduce:
$\left(\begin{array}{cc|c} 1&1&5 \\ 1&-1&3\end{array}\right) \quad\xrightarrow{\text{RREF}}\quad\left(\begin{array}{cc|c} 1&0&4 \\ 0&1&1\end{array}\right).\nonumber$
We have $$c_1=4$$ and $$c_2=1$$, so $$v=4v_1+v_2$$ and $$[v]_{\mathcal{B}}=\left(\begin{array}{c}4\\1\end{array}\right)$$.

In the following picture, we indicate the coordinate system defined by $$\mathcal{B}$$ by drawing lines parallel to the “$$v_1$$-axis” and “$$v_2$$-axis”. Using this grid it is easy to see that the $$\mathcal{B}$$-coordinates of $$v$$ are $$\left(\begin{array}{c}5\\1\end{array}\right)$$, and that the $$\mathcal{B}$$-coordinates of $$w$$ are $$\left(\begin{array}{c}1\\2\end{array}\right)$$.

Figure $$\PageIndex{1}$$

This picture could be the grid of streets in Palo Alto, California. Residents of Palo Alto refer to northwest as “north” and to northeast as “east”. There is a reason for this: the old road to San Francisco is called El Camino Real, and that road runs from the southeast to the northwest in Palo Alto. So when a Palo Alto resident says “go south two blocks and east one block”, they are giving directions from the origin to the Whole Foods at $$w$$.

Figure $$\PageIndex{2}$$: A picture of the basis $$\mathcal{B}=\{v_1,\: v_2\}$$ of $$\mathbb{R}^2$$. The grid indicates the coordinate system defined by the basis $$\mathcal{B}$$; one set of lines measures the $$v_1$$-coordinate, and the other set measures the $$v_2$$-coordinate. Use the sliders to find the $$\mathcal{B}$$-coordinates of $$w$$.

##### Example $$\PageIndex{3}$$

Let

$v_1=\left(\begin{array}{c}2\\-1\\1\end{array}\right)\quad v_2=\left(\begin{array}{c}1\\0\\-1\end{array}\right).\nonumber$

These form a basis $$\mathcal{B}$$ for a plane $$V=\text{Span}\{v_1,\: v_2\}$$ in $$\mathbb{R}^3$$. We indicate the coordinate system defined by $$\mathcal{B}$$ by drawing lines parallel to the “$$v_1$$-axis” and “$$v_2$$-axis”:

Figure $$\PageIndex{3}$$

We can see from the picture that the $$v_1$$-coordinate of $$\color{red}{u_1}$$ is equal to $$1$$, as is the $$v_2$$-coordinate, so $$[\color{red}{u_1}\color{black}{]_{\mathcal{B}}=\left(\begin{array}{c}1\\1\end{array}\right)}$$. Similarly, we have

$[\color{blue}{u_2}\color{black}{]_{\mathcal{B}}}=\left(\begin{array}{c}-1 \\ \frac{1}{2}\end{array}\right)\quad [\color{Green}{u_3}\color{black}{]_{\mathcal{B}}}=\left(\begin{array}{c}\frac{3}{2} \\ -\frac{1}{2}\end{array}\right)\quad [\color{orange}{u_4}\color{black}{]_{\mathcal{B}}}=\left(\begin{array}{c}0\\ \frac{3}{2}\end{array}\right).\nonumber$

Figure $$\PageIndex{4}$$: Left: the $$\mathcal{B}$$-coordinates of a vector $$x$$. Right: the vector $$x$$. The violet grid on the right is a picture of the coordinate system defined by the basis $$\mathcal{B}$$; one set of lines measures the $$v_1$$-coordinate, and the other set measures the $$v_2$$-coordinate. Drag the heads of the vectors $$x$$ and $$[x]_{\mathcal{B}}$$ to understand the correspondence between $$x$$ and its $$\mathcal{B}$$-coordinate vector.

##### Example $$\PageIndex{4}$$: A Coordinate System on a Plane

Define

$v_1=\left(\begin{array}{c}1\\0\\1\end{array}\right),\: v_2=\left(\begin{array}{c}1\\1\\1\end{array}\right),\quad\mathcal{B}=\{v_1,\: v_2\},\quad V=\text{Span}\{v_1,\: v_2\}.\nonumber$

1. Verify that $$\mathcal{B}$$ is a basis for $$V$$.
2. If $$[w]_{\mathcal{B}}=\left(\begin{array}{c}5\\2\end{array}\right)$$, then what is $$w$$?
3. Find the $$\mathcal{B}$$-coordinates of $$v=\left(\begin{array}{c}5\\3\\5\end{array}\right)$$.
###### Solution
1. We need to verify that $$\mathcal{B}$$ spans $$V$$, and that it is linearly independent. By definition, $$V$$ is the span of $$\mathcal{B}$$; since $$v_1$$ and $$v_2$$ are not multiples of each other, they are linearly independent. This shows in particular that $$V$$ is a plane.
2. To say $$[w]_{\mathcal{B}}=\left(\begin{array}{c}5\\2\end{array}\right)$$ means that $$5$$ is the $$v_1$$-coordinate of $$w$$, and that $$2$$ is the $$v_2$$-coordinate:
$w=5v_1 +2v_2=5\left(\begin{array}{c}1\\0\\1\end{array}\right)+2\left(\begin{array}{c}1\\1\\1\end{array}\right)=\left(\begin{array}{c}7\\2\\7\end{array}\right).\nonumber$
3. We have to solve the vector equation $$v=c_1v_1+c_2v_2$$ in the unknowns $$c_1,\: c_2$$. We form an augmented matrix and row reduce:
$\left(\begin{array}{cc|c} 1&1&5\\0&1&3\\1&1&5\end{array}\right) \quad\xrightarrow{\text{RREF}}\quad \left(\begin{array}{cc|c} 1&0&2\\0&1&3\\0&0&0\end{array}\right).\nonumber$

We have $$c_1=2$$ and $$c_2=3$$, so $$v=2v_1+3v_2$$ and $$[v]_{\mathcal{B}}=\left(\begin{array}{c}2\\3\end{array}\right)$$.

Figure $$\PageIndex{5}$$: A picture of the plane $$V$$ and the basis $$\mathcal{B}=\{v_1,\: v_2\}$$. The violet grid is a picture of the coordinate system defined by the basis $$\mathcal{B}$$; one set of lines measures the $$v_1$$-coordinate, and the other set measures the $$v_2$$-coordinate. Use the sliders to find the $$\mathcal{B}$$-coordinates of $$v$$.

##### Example $$\PageIndex{5}$$: A Coordinate System on Another Plane

Define

$v_1=\left(\begin{array}{c}2\\3\\2\end{array}\right),\quad v_2=\left(\begin{array}{c}-1\\1\\1\end{array}\right),\quad v_3=\left(\begin{array}{c}2\\8\\6\end{array}\right),\quad V=\text{Span}\{v_1,\: v_2,\: v_3\}.\nonumber$

1. Find a basis $$\mathcal{B}$$ for $$V$$.
2. Find the $$\mathcal{B}$$-coordinates of $$x=\left(\begin{array}{c}4\\11\\8\end{array}\right).$$
###### Solution
1. We write $$V$$ as the column space of a matrix $$A$$, then row reduce to find the pivot columns, as in Example 2.7.6, in Section 2.7.
$A=\left(\begin{array}{ccc}2&-1&2\\3&1&8\\2&1&6\end{array}\right) \quad\xrightarrow{\text{RREF}}\quad \left(\begin{array}{ccc}1&0&2\\0&1&2\\0&0&0\end{array}\right).\nonumber$
The first two columns are pivot columns, so we can take $$\mathcal{B}=\{v_1,\:v_2\}$$ as our basis for $$V$$.
2. We have to solve the vector equation $$x=c_1v_1+c_2v_2$$. We form an augmented matrix and row reduce:
$\left(\begin{array}{cc|c} 2&-1&4 \\ 3&1&11\\2&1&8\end{array}\right) \quad\xrightarrow{\text{RREF}}\quad \left(\begin{array}{cc|c} 1&0&3\\0&1&2\\0&0&0\end{array}\right).\nonumber$
We have $$c_1=3$$ and $$c_2=2$$, so $$x=3v_1+2v_2$$, and thus $$[x]_{\mathcal{B}}=\left(\begin{array}{c}3\\2\end{array}\right)$$.

Figure $$\PageIndex{6}$$: A picture of the plane $$V$$ and the basis $$\mathcal{B}=\{v_1,\:v_2\}$$. The violet grid is a picture of the coordinate system defined by the basis $$\mathcal{B}$$; one set of lines measures the $$v_1$$-coordinate, and the other set measures the $$v_2$$-coordinate. Use the sliders to find the $$\mathcal{B}$$-coordinates of $$x$$.

##### Recipes: $$\mathcal{B}$$-Coordinates

If $$\mathcal{B}=\{v_1,\: v_2,\cdots ,v_m\}$$ is a basis for a subspace $$V$$ and $$x$$ is in $$V$$, then

$[x]_{\mathcal{B}}=\left(\begin{array}{c}c_1 \\ c_2 \\ \vdots \\ c_m\end{array}\right)\quad\text{means}\quad x=c_1v_1+c_2v_2+\cdots c_mv_m.\nonumber$

Finding the $$\mathcal{B}$$-coordinates of $$x$$ means solving the vector equation

$x=c_1v_1+c_2v_2+\cdots +c_mv_m\nonumber$

in the unknowns $$c_1,\: c_2,\cdots ,c_m$$. This generally means row reducing the augmented matrix

$\left(\begin{array}{cccc|c} |&|&\quad&|&| \\ v_1&v_2&\quad &v_m&x \\ |&|&\quad &|&|\end{array}\right).\nonumber$

##### Remark

Let $$\mathcal{B}=\{v_1,\: v_2,\cdots ,v_m\}$$ be a basis of a subspace $$V$$. Finding the $$\mathcal{B}$$-coordinates of a vector $$x$$ means solving the vector equation

$x=c_1v_1+c_2v_2+\cdots +c_mv_m.\nonumber$

If $$x$$ is not in $$V$$, then this equation has no solution, as $$x$$ is not in $$V=\text{Span}\{v_1,\: v_2,\cdots ,v_m\}$$. In other words, the above equation is inconsistent when $$x$$ is not in $$V$$.

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