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The unique list of coefficients a basis gives each vector, why independence is what makes it unique, and the isomorphism onto R to the n that lets a polynomial be row-reduced.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to compute the coordinate vector of a vector in any ordered basis by solving the square system whose columns are the basis vectors, explain why the coordinates exist because the basis spans and are unique because it is independent, take coordinates in spaces whose vectors are not lists of numbers — polynomials, matrices, solutions of a differential equation — and use the coordinate map to turn a question about such a space into a row reduction in $\mathbb{R}^n$, and say why the order of a basis is part of the basis.
A basis is an independent spanning set, and you have already used both halves separately. Spanning says every vector can be written as a combination of the basis vectors; independence says the combination is not a matter of choice. This lesson puts the two halves together and reads off the consequence, which is larger than either half suggests.
An ordered basis $\mathcal{B} = (b_1, \dots, b_n)$ is a basis whose vectors have been listed in a fixed order. The coordinate vector of $v$ is $[v]_{\mathcal{B}} = (c_1, \dots, c_n)$, the unique coefficients with $v = c_1b_1 + \dots + c_nb_n$; it is written as a column. The coordinate map is $v \mapsto [v]_{\mathcal{B}}$. An isomorphism is a linear map that is one-to-one and onto, so it has a linear inverse. The standard basis of $\mathbb{R}^n$ is $(e_1, \dots, e_n)$.
It is worth being blunt about this, because the notation hides it. A list of numbers is what a vector looks like after a basis has been chosen. Change the basis and the same vector has a different list.
Fix an ordered basis $\mathcal{B} = (b_1, \dots, b_n)$ of $V$. Spanning gives at least one way of writing $v = c_1b_1 + \dots + c_nb_n$. Independence gives at most one: if $\sum c_ib_i = \sum d_ib_i$, then $\sum (c_i - d_i)b_i = 0$, and independence forces every $c_i - d_i$ to be zero. So the coefficients exist and are unique, and
$$[v]_{\mathcal{B}} = \begin{pmatrix} c_1 \\ \vdots \\ c_n \end{pmatrix}$$
is well defined. Both halves of the definition of a basis are used, and each does a different job.
The coordinate map $v \mapsto [v]_{\mathcal{B}}$ is linear — $[u + v]_{\mathcal{B}} = [u]_{\mathcal{B}} + [v]_{\mathcal{B}}$ and $[cv]_{\mathcal{B}} = c[v]_{\mathcal{B}}$, both straight from collecting coefficients — and it is one-to-one and onto, so it is an isomorphism $V \to \mathbb{R}^n$.
That is the payoff, and it is enormous. Any $n$-dimensional real vector space — polynomials, matrices, solutions of a differential equation — is $\mathbb{R}^n$ with the labels changed. Independence, span, dimension and rank can all be decided in $V$ by taking coordinates and row-reducing in $\mathbb{R}^n$.
Another way: picture
Draw a plane with the usual square grid, then draw a second grid over it made from two slanted vectors. One arrow, one point of the plane — but two addresses, one per grid. The arrow did not move when you drew the second grid, and the two lists of numbers are both correct descriptions of it. The standard basis is simply the grid that is so familiar it stops looking like a choice.
Another way: steps
The construction never mentioned $\mathbb{R}^n$, so it applies anywhere.
Polynomials. $P_2$, the polynomials of degree at most $2$, has basis $(1, x, x^2)$, and $[a + bx + cx^2]_{\mathcal{B}} = (a, b, c)$. Coordinates are just the coefficients here, which makes $P_2$ look trivial — until the basis is changed to $(1,\; 1 + x,\; (1+x)^2)$, when the same polynomial gets a quite different list and finding it means solving a system.
Matrices. The $2 \times 2$ real matrices form a four-dimensional space with basis $(E_{11}, E_{12}, E_{21}, E_{22})$, and the coordinate vector of a matrix is its entries read in that order. So a question about $2 \times 2$ matrices — are these three independent? what do they span? — is a question about vectors in $\mathbb{R}^4$, and row reduction answers it.
Solutions of a differential equation. The solutions of $y'' + y = 0$ form a two-dimensional space with basis $(\cos x, \sin x)$, and a solution's coordinates are the two constants in $A\cos x + B\sin x$.
In each case the same sentence applies: choose a basis, take coordinates, and the problem becomes one you can row-reduce.
A basis as a set is enough to say what dimension a space has. It is not enough to give coordinates, because a coordinate vector is a list and a list has a first entry.
With $\mathcal{B} = ((1, 1), (1, -1))$ the vector $(3, 1)$ has coordinates $(2, 1)$. With the same two vectors written the other way round, $\mathcal{C} = ((1, -1), (1, 1))$, the same vector has coordinates $(1, 2)$. Nothing about the space changed; the labelling did.
This is why textbooks say ordered basis the first few times and then quietly drop the adjective. It matters most when a matrix is being built: the columns of $[T]$ are indexed by the domain's basis in its order, and the rows by the codomain's. Reorder either and the matrix's rows or columns are permuted.
A related habit worth forming: write coordinate vectors as columns. $[v]_{\mathcal{B}}$ is a column because it is about to be multiplied on the left by a matrix, and a row would not fit. The notation is doing work, not decoration.
Reading the entries of a vector as its coordinates in any basis. $(5, 2)$ has coordinates $(5, 2)$ in the standard basis and almost certainly not in the basis the question gave you. Whenever a non-standard basis appears, the entries and the coordinates are different lists, and the whole point of the question is that difference.
Thinking that a change of basis changes the vector. It does not. The vector is the same object throughout; only its description moves. The sentence to keep is: one vector, many addresses.
Losing the order. Swapping two basis vectors swaps two entries of every coordinate vector in the space, and of every row or column of every matrix built from it.
Believing coordinates need the space to be $\mathbb{R}^n$. They need a basis and nothing else. A polynomial, a matrix and a solution of a differential equation all have coordinate vectors, and that is what lets one set of algorithms serve all of them.
Expecting the system to be awkward. The system for coordinates is square, with independent columns, so it is always consistent and never has free variables. If your working produces no solution or a free variable, the arithmetic has gone wrong — or the set you were handed is not a basis.
$\mathcal{B} = ((1, 1), (1, -1))$ and $v = (3, 1)$. Solve $c_1(1, 1) + c_2(1, -1) = (3, 1)$, that is $c_1 + c_2 = 3$ and $c_1 - c_2 = 1$.
Two equations from the two entries.
Adding gives $2c_1 = 4$, so $c_1 = 2$ and then $c_2 = 1$: $[v]_{\mathcal{B}} = (2, 1)$.
Check: $2(1,1) + 1(1,-1) = (3, 1)$.
In the standard basis the same vector has coordinates $(3, 1)$. Two lists, one vector; neither list is more correct than the other.
The basis is part of the answer.
Are $1 + x$, $x + x^2$ and $1 + 2x + x^2$ independent in $P_2$? Take coordinates in $(1, x, x^2)$: $(1,1,0)$, $(0,1,1)$ and $(1,2,1)$.
Coordinates first, always.
Put them in a matrix and reduce. The third is the sum of the first two, so a row of zeros appears and the rank is $2$.
The question is now about $\mathbb{R}^3$.
So the three polynomials are dependent and span a two-dimensional subspace of $P_2$. Nothing was integrated, differentiated or substituted — the isomorphism did the work.
Translate back at the end.
Solve $c_1 + c_2 = 0$ and $c_1 - c_2 = 4$.
Sum and difference of the entries.
Adding: $2c_1 = 4$, so $c_1 = 2$.
Then back-substitute.
$c_2 = -2$, so the coordinate vector is $(2, -2)$ — a negative entry, which is perfectly ordinary and impossible in the standard basis here.
$\mathcal{B} = \{(1, 0),\; (2, 1)\}$ is a basis of $\mathbb{R}^{2}$. Enter, as a column, the coordinate vector of $v = (4, 2)$ in $\mathcal{B}$.
This task has no paper form; do it on a device.
$P_2$ is the space of polynomials of degree at most $2$, with the ordered basis $\{1,\; x,\; x^2\}$. Give the coordinates of each polynomial in that basis.
| Constant coefficient | Coefficient of $x$ | Coefficient of $x^2$ | |
|---|---|---|---|
| $p(x) = 5 + 6x + 6x^2$ | |||
| $p(x) + x^2$ |
$\mathcal{B} = \{(1, 1),\; (1, -1)\}$ is a basis of $\mathbb{R}^{2}$. Match each vector to its coordinate vector in $\mathcal{B}$.
| $(3, 3)$ | $(3, -3)$ | $(9, 0)$ | $(-3, 3)$ | |
|---|---|---|---|---|
| $(6, 0)$ | ||||
| $(0, 6)$ | ||||
| $(9, 9)$ |
$\mathcal{B}$ is a basis of a space of dimension $5$. Which property of $\mathcal{B}$ is what stops a vector having two different coordinate lists?
$\mathcal{B} = \{(1, 3),\; (1, 1)\}$ is a basis of $\mathbb{R}^{2}$. What is the first $\mathcal{B}$-coordinate of $v = (2, 4)$?
Answer:
In $P_2$ with the ordered basis $\{1,\; x,\; x^2\}$, enter as a column the coordinate vector of $p(x) = (2 + x)^2$.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
$\mathcal{B} = \{(1, 0),\; (5, 1)\}$ is a basis of $\mathbb{R}^{2}$. Enter, as a column, the coordinate vector of $v = (6, 5)$ in $\mathcal{B}$.
This task has no paper form; do it on a device.
You can find coordinates in any basis and use them to carry a question into $\mathbb{R}^n$. Say in your own words why a vector cannot have two different coordinate lists in one basis. Next: what happens to those lists, and to a map's matrix, when the basis changes.
10. Your turn: the coordinates of $(0, 4)$ in the basis $((1, 1), (1, -1))$, step 3
Check by recombining.