4.4 Coordinate systems

Big idea

A basis gives every vector exactly one list of weights. Package that list as a column and you have the coordinate vector of the vector relative to that basis. The vector itself did not change; only the description did, and the description depends entirely on which basis you chose.

That is the point that takes getting used to. The column $(5, 1)$ in $\mathbb{R}^2$ looks like a coordinate vector already, and it is: relative to the standard basis. Choose a different basis and the same arrow gets a different column. Coordinates are always relative, and the notation carries the basis so you never lose track of which one.

The payoff is that coordinates flatten any space onto $\mathbb{R}^n$. A polynomial of degree at most two becomes a column of three numbers, and questions about independence or span of polynomials become questions about columns, answered by row reduction. A space you can only describe abstractly gets a working coordinate system, and everything you know about $\mathbb{R}^n$ transfers.

Decoder

If $\mathcal{B}$ is a basis for $V$ and $\mathbf{x}$ is in $V$, the coordinates of $\mathbf{x}$ relative to $\mathcal{B}$ are the unique weights making $\mathbf{x}$ the corresponding linear combination of the basis vectors.

Unpacked: write $\mathbf{x}$ as a mixture of the basis vectors. The numbers you mixed with, in the order the basis was listed, are the coordinates. Stack them into a column and write $[\mathbf{x}]_{\mathcal{B}}$.

The word “unique” is doing real work and comes from independence. Without it, the same vector would have several coordinate columns and the notation would be meaningless. The word “order” matters too: a basis is an indexed list, not just a set, because swapping two basis vectors swaps two coordinates.

Definitions and results

Coordinate vector. Let $\mathcal{B} = \{\mathbf{b}_1, \dots, \mathbf{b}_n\}$ be a basis for $V$ and let $\mathbf{x}$ be in $V$. The unique scalars with $\mathbf{x} = c_1\mathbf{b}_1 + \cdots + c_n\mathbf{b}_n$ are the coordinates of $\mathbf{x}$ relative to $\mathcal{B}$, and the column $[\mathbf{x}]_{\mathcal{B}} = (c_1, \dots, c_n)$ is the coordinate vector.

Change-of-coordinates matrix in $\mathbb{R}^n$. When $V = \mathbb{R}^n$, let $P_{\mathcal{B}}$ be the matrix whose columns are the basis vectors $\mathbf{b}_1, \dots, \mathbf{b}_n$ written in standard coordinates. Then

$$ \mathbf{x} = P_{\mathcal{B}}\,[\mathbf{x}]_{\mathcal{B}} $$

so $P_{\mathcal{B}}$ converts $\mathcal{B}$-coordinates into standard ones. Since its columns are a basis, $P_{\mathcal{B}}$ is invertible, and $[\mathbf{x}]_{\mathcal{B}} = P_{\mathcal{B}}^{-1}\mathbf{x}$ converts the other way.

Which direction is which. Multiplying by $P_{\mathcal{B}}$ goes from $\mathcal{B}$ to standard, because the product is literally the combination of basis vectors with the given weights. Going from standard to $\mathcal{B}$ needs the inverse, or equivalently solving a system. Mixing up the direction is the standard error here, and the fix is to recompute one example you already know the answer to.

Finding coordinates by solving. In practice you can skip the inverse. Row reduce the augmented matrix whose columns are the basis vectors followed by $\mathbf{x}$; the solution is $[\mathbf{x}]_{\mathcal{B}}$.

The coordinate mapping. The map sending $\mathbf{x}$ to $[\mathbf{x}]_{\mathcal{B}}$ is a linear transformation from $V$ to $\mathbb{R}^n$ that is one-to-one and onto. Coordinates of a sum are the sum of the coordinates, and coordinates of a scalar multiple are the scaled coordinates.

Isomorphism. Because the coordinate mapping is linear and reversible, $V$ and $\mathbb{R}^n$ have the same structure; they are isomorphic. Any statement about sums and scalar multiples holds in $V$ exactly when it holds for the coordinate columns. So a set of polynomials is independent exactly when their coordinate columns are independent, and spans exactly when the columns span.

What the isomorphism buys. Take any finite-dimensional space, pick a basis, and every question about spans, independence, bases and dimension becomes a question about columns of numbers, answered by row reduction. That is how a claim about polynomials or matrices gets settled without any new theory: translate to coordinates, reduce, translate the answer back. The translation is faithful in both directions, so nothing is lost on the way.

What coordinates do not preserve. Coordinates carry sums and scalar multiples and nothing else. Lengths and angles are not preserved unless the basis is chosen to preserve them, which is a question for a later chapter. A basis can be badly skewed and still be perfectly valid.

Worked examples

Coordinates in a non-standard basis of $\mathbb{R}^2$

Let $\mathcal{B} = \{\mathbf{b}_1, \mathbf{b}_2\}$ with

$$ \mathbf{b}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \qquad \mathbf{b}_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix} $$

and let $\mathbf{x} = (5, 1)$. Solve $c_1\mathbf{b}_1 + c_2\mathbf{b}_2 = \mathbf{x}$, which reads $c_1 + c_2 = 5$ and $c_1 - c_2 = 1$. Adding gives $2c_1 = 6$, so $c_1 = 3$ and $c_2 = 2$. So $[\mathbf{x}]_{\mathcal{B}} = (3, 2)$.

Check by rebuilding: $3(1,1) + 2(1,-1) = (3 + 2,\; 3 - 2) = (5, 1)$. That is $\mathbf{x}$.

The same computation with the matrix: $P_{\mathcal{B}} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}$ has determinant $-2$, so

$$ P_{\mathcal{B}}^{-1} = \frac{1}{-2}\begin{bmatrix} -1 & -1 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} 1/2 & 1/2 \\ 1/2 & -1/2 \end{bmatrix} $$

and $P_{\mathcal{B}}^{-1}(5, 1) = (5/2 + 1/2,\; 5/2 - 1/2) = (3, 2)$, agreeing with the direct solve.

Going the easy direction

Let $\mathcal{C} = \{(2, 1), (-1, 3)\}$ and suppose $[\mathbf{y}]_{\mathcal{C}} = (2, -1)$. No system is needed; multiply out the combination:

$$ \mathbf{y} = 2\begin{bmatrix} 2 \\ 1 \end{bmatrix} - 1\begin{bmatrix} -1 \\ 3 \end{bmatrix} = \begin{bmatrix} 4 + 1 \\ 2 - 3 \end{bmatrix} = \begin{bmatrix} 5 \\ -1 \end{bmatrix} $$

Check by going back. Solving $c_1(2,1) + c_2(-1,3) = (5,-1)$ gives $2c_1 - c_2 = 5$ and $c_1 + 3c_2 = -1$. From the second, $c_1 = -1 - 3c_2$; substituting gives $-2 - 6c_2 - c_2 = 5$, so $c_2 = -1$ and $c_1 = 2$. That is the coordinate vector we started with.

This is the asymmetry to remember. Coordinates to vector is one multiplication. Vector to coordinates is a system.

Coordinates of a polynomial

Work in $\mathbb{P}_2$ with the basis $\mathcal{B} = \{1,\; t - 1,\; (t-1)^2\}$, and take $p(t) = 3 - 2t + t^2$.

Write $p = a + b(t-1) + c(t-1)^2$ and expand the right-hand side: $a + bt - b + c t^2 - 2ct + c$, which collects to

$$ (a - b + c) + (b - 2c)t + c\,t^2 $$

Match coefficients against $3 - 2t + t^2$. The $t^2$ terms give $c = 1$. The $t$ terms give $b - 2 = -2$, so $b = 0$. The constants give $a - 0 + 1 = 3$, so $a = 2$. Therefore $[p]_{\mathcal{B}} = (2, 0, 1)$.

Check by expanding the answer: $2 + 0 + (t-1)^2 = 2 + t^2 - 2t + 1 = 3 - 2t + t^2$, which is $p$.

Note how different the two descriptions look. Relative to the standard basis $\{1, t, t^2\}$ the same polynomial has coordinates $(3, -2, 1)$. Same polynomial, two columns, and neither is more correct than the other.

Why coordinates answer questions about polynomials

Are $1 + t$, $t + t^2$ and $1 + 2t + t^2$ independent in $\mathbb{P}_2$? Take coordinates relative to $\{1, t, t^2\}$: the columns are $(1,1,0)$, $(0,1,1)$ and $(1,2,1)$. The third is the sum of the first two, so the columns are dependent, so the polynomials are dependent.

Check in the original space: $(1 + t) + (t + t^2) = 1 + 2t + t^2$, which is the third polynomial exactly. Row reduction never had to know what a polynomial is.

Practice

First bases and the unique-representation property that makes coordinates well defined.

Then the coordinates themselves: both directions in $\mathbb{R}^n$ and coordinates in a polynomial basis.

Videos

Watch the running example where the same arrow gets two different coordinate columns because two people picked different basis vectors. That framing is the whole section.

Quiz

Five items on coordinate vectors, the change-of-coordinates matrix, and coordinates in a polynomial basis.