Coordinate vectors — change of basis: Worked examples

This page is the workout room for Coordinate vectors — change of basis. The parent note built the machinery: coordinate vectors, the matrix $P$ whose columns are $[b_{j}]_{C}$ , and the $R^{n}$ shortcut $P_{C \leftarrow B} = C^{- 1} B$ . Here we hit every kind of case the topic can throw at you, one example per "cell".

Before symbols do anything, remember the picture: a vector is an arrow that never moves; a basis is a pair of rulers; coordinates are the numbers you read off those rulers. Change of basis = swapping rulers, arrow frozen.

Definition Reading the symbol

$C \leftarrow B P$ The stacked notation $C \leftarrow B P$ is just the single letter $P$ with a little label written underneath it (that is all the LaTeX \underset does — it parks text below the symbol; it is not a fraction, not an exponent, not a subscript multiplication). If your reader ever finds it fiddly, mentally read it as $P_{C \leftarrow B}$ — same object, label pushed to the side. The label tells you the matrix's job.

Read the arrow $C \leftarrow B$ as "from $B$ to $C$ " (the arrow points at the destination). This matrix consumes a coordinate vector written in the old basis $B$ and produces the same vector's coordinates in the new basis $C$ : $[v]_{C} = C \leftarrow B P [v]_{B} .$ Likewise $B \leftarrow C P$ goes the other way, and (parent note) the two are inverses. Also, whenever we write a bare capital like $B$ , we mean the matrix whose columns are the basis vectors $b_{1}, \dots, b_{n}$ stacked side by side.

The scenario matrix

Every change-of-basis problem is one of these cells. We will cover all of them.

Cell	What makes it special	Example
A Standard → non-standard	read entries, then apply $B^{- 1}$	Ex 1
B Non-standard → standard	just multiply by $B$	Ex 2
C Non-standard → non-standard	full $C^{- 1} B$	Ex 3
D Negative / mixed-sign coordinates	numbers can go below zero	Ex 4
E Round-trip (inverse check)	$P_{B \leftarrow C} P_{C \leftarrow B} = I$	Ex 5
F Non- $R^{n}$ space (polynomials)	no "entries" to read — solve a system	Ex 6
G Degenerate input (the zero vector)	limiting/edge case	Ex 7
H Real-world word problem	rotated map / two observers	Ex 8
I Exam twist (basis reordered)	order of a basis matters	Ex 9

Figure — Coordinate vectors — change of basis

Cell A — standard → non-standard

Worked example Example 1 (Cell A)

Let $B = {b_{1}, b_{2}}$ with $b_{1} = [21]$ , $b_{2} = [- 1 1]$ . The vector $v = [15]$ is given in standard coordinates. Find $[v]_{B}$ .

Forecast: guess before computing — is $v$ more "along $b_{2}$ " (which points up-left) or "along $b_{1}$ "? It leans upward, so expect a healthy $b_{2}$ share.

First name the matrix we'll use: let $B = [21 - 1 1]$ be the matrix whose columns are $b_{1}$ and $b_{2}$ . Now write the meaning of $[v]_{B} = [c_{1} c_{2}]$ : it satisfies $c_{1} b_{1} + c_{2} b_{2} = v$ , which is exactly the matrix product $B [v]_{B} = v$ . Why this step? In the standard basis a vector's coordinates are its entries, so $v$ on the right is already in "standard numbers". $B$ turns $B$ -numbers into standard numbers, so we must undo $B$ .
So $[v]_{B} = B^{- 1} v$ . Here $det B = 2 (1) - (- 1) (1) = 3$ . Why this step? $B$ is invertible exactly because its columns are a basis (independent), so $B^{- 1}$ exists; and computing $det B$ first tells us the inverse exists ( $det \neq = 0$ ) and gives the scaling factor we'll need.
$B^{- 1} = \frac{1}{3} [1 - 1 12]$ . Why this step? For a $2 \times 2$ matrix $[a c b d]$ the inverse is $\frac{1}{a d - b c} [d - c - b a]$ — swap the diagonal, negate the off-diagonal, divide by the determinant.
$[v]_{B} = \frac{1}{3} [1 - 1 12] [15] = \frac{1}{3} [69] = [23]$ . Why this step? Row-times-column: top row $1 (1) + 1 (5) = 6$ , bottom row $- 1 (1) + 2 (5) = 9$ , then the $\frac{1}{3}$ scales both entries. That $\frac{1}{3}$ is exactly why we bothered with $det B$ in step 2.

Verify: $2 b_{1} + 3 b_{2} = [42] + [- 3 3] = [15] = v$ ✓. The $b_{2}$ share (3) beats the $b_{1}$ share (2) — forecast holds.

Cell B — non-standard → standard

Worked example Example 2 (Cell B)

Same $B$ as Example 1. Someone hands you $[w]_{B} = [3 - 2]$ . What is $w$ in standard coordinates?

Forecast: the $- 2$ pulls in the $- b_{2}$ direction (down-right), so expect a strongly rightward result.

This direction is the easy one: $w = B [w]_{B}$ , with $B$ the same columns-are-basis matrix from Example 1. Why this step? $[w]_{B}$ literally lists the coefficients $c_{1}, c_{2}$ ; plugging into $c_{1} b_{1} + c_{2} b_{2}$ is just the matrix product $B [w]_{B}$ . No inverse needed — that is what makes going to standard cheap.
$w = [21 - 1 1] [3 - 2] = [6 + 2 3 - 2] = [81]$ . Why this step? Row-times-column again: top $2 (3) + (- 1) (- 2) = 6 + 2 = 8$ , bottom $1 (3) + 1 (- 2) = 3 - 2 = 1$ . No determinant appears because there's no inverse to scale.

Verify: run it back through Example 1's machine: $B^{- 1} w = \frac{1}{3} [1 - 1 12] [81] = \frac{1}{3} [9 - 6] = [3 - 2]$ ✓ — back to the input. Rightward as forecast.

Cell C — non-standard → non-standard

Worked example Example 3 (Cell C)

$B$ as above. Let $C = {c_{1}, c_{2}}$ with $c_{1} = [11]$ , $c_{2} = [01]$ . Build $C \leftarrow B P$ , then convert $[v]_{B} = [23]$ (the $v$ from Ex 1).

Forecast: since $v = [15]$ (Ex 1) and $c_{1}, c_{2}$ span with an easy shear, expect $[v]_{C}$ to have $c_{1}$ -part equal to $v$ 's first entry.

Use $C \leftarrow B P = C^{- 1} B$ , where $C = [1101]$ has columns $c_{1}, c_{2}$ and $B$ is as before. Why this step? The columns of $P$ are $[b_{j}]_{C} = C^{- 1} b_{j}$ ; stacking those columns is exactly $C^{- 1} B$ . Read right-to-left: $B$ takes $B$ →standard, $C^{- 1}$ takes standard→ $C$ ; the "standard" in the middle cancels.
$det C = 1 (1) - 0 (1) = 1$ , so $C^{- 1} = [1 - 1 01]$ . Why this step? Same $2 \times 2$ inverse rule (swap diagonal, negate off-diagonal, divide by $det$ ); here $det = 1$ so no fractions appear.
$C \leftarrow B P = C^{- 1} B = [1 - 1 01] [21 - 1 1] = [2 - 1 - 1 2]$ . Why this step? Multiply column by column: first column $[1 - 1 01] [21] = [2 - 1]$ , second $[1 - 1 01] [- 1 1] = [- 1 2]$ .
$[v]_{C} = C \leftarrow B P [v]_{B} = [2 - 1 - 1 2] [23] = [14]$ . Why this step? This is the payoff — feeding the $B$ -numbers into $P$ spits out the $C$ -numbers: top $2 (2) - 1 (3) = 1$ , bottom $- 1 (2) + 2 (3) = 4$ .

Verify: independently, $[v]_{C} = C^{- 1} v = [1 - 1 01] [15] = [14]$ ✓. And $1 c_{1} + 4 c_{2} = [11] + [04] = [15] = v$ ✓. First entry matched — forecast holds.

Cell D — negative / mixed-sign coordinates

Worked example Example 4 (Cell D)

With $B$ from Ex 1, express $u = [- 4 - 1]$ in $B$ -coordinates. Watch the signs.

Forecast: $u$ points down-left, roughly opposite $b_{1}$ . Expect a negative $c_{1}$ .

$[u]_{B} = B^{- 1} u = \frac{1}{3} [1 - 1 12] [- 4 - 1]$ . Why this step? Cell-A machinery again — standard→ $B$ means apply $B^{- 1}$ ; nothing about negatives breaks it — coordinates are free to be any real number, positive or negative.
$= \frac{1}{3} [- 4 - 1 4 - 2] = \frac{1}{3} [- 5 2] = [- \frac{5}{3} \frac{2}{3}]$ . Why this step? Row-times-column: top $1 (- 4) + 1 (- 1) = - 5$ , bottom $- 1 (- 4) + 2 (- 1) = 4 - 2 = 2$ , then scale by $\frac{1}{3}$ . The fraction is fine — coordinates need not be integers.

Verify: $- \frac{5}{3} b_{1} + \frac{2}{3} b_{2} = [- \frac{10}{3} - \frac{5}{3}] + [- \frac{2}{3} \frac{2}{3}] = [- 4 - 1] = u$ ✓. $c_{1} < 0$ as forecast — coordinates need not be positive, and needn't be integers.

Cell E — the round-trip (inverse check)

Worked example Example 5 (Cell E)

Using $B, C$ from Ex 3, confirm $B \leftarrow C P = (C \leftarrow B P)^{- 1}$ and use it to go back.

Forecast: going $C \to B \to C$ must return you unchanged, so the two matrices multiply to the identity $I = [1001]$ .

Compute $B \leftarrow C P = B^{- 1} C$ . Why this step? Swap the roles of $B$ and $C$ in the shortcut $C^{- 1} B$ . Read right-to-left: $C$ sends $C$ →standard, $B^{- 1}$ sends standard→ $B$ .
$B^{- 1} C = \frac{1}{3} [1 - 1 12] [1101] = \frac{1}{3} [2112]$ . Why this step? Column-by-column: first column $\frac{1}{3} [1 - 1 12] [11] = \frac{1}{3} [21]$ , second $\frac{1}{3} [1 - 1 12] [01] = \frac{1}{3} [12]$ .
Check the product with $C \leftarrow B P = [2 - 1 - 1 2]$ from Ex 3: $\frac{1}{3} [2112] [2 - 1 - 1 2] = \frac{1}{3} [3003] = [1001] = I .$ Why this step? This is the whole reason "back-and-forth = nothing"; matrices that undo each other are inverses. The top-left entry is $\frac{1}{3} (2 \cdot 2 + 1 \cdot (- 1)) = \frac{1}{3} (3) = 1$ , and the off-diagonals cancel to $0$ .
Sanity round-trip on $[v]_{C} = [14]$ : $B \leftarrow C P [14] = \frac{1}{3} [2 + 4 1 + 8] = [23] = [v]_{B}$ ✓. Why this step? It confirms the inverse actually returns the Ex-3 starting numbers, not just abstractly but on our concrete vector.

Verify: we recovered exactly $[v]_{B} = [23]$ from Ex 3. The circle closes.

Cell F — a space with no "entries" (polynomials)

Worked example Example 6 (Cell F)

In $V = P_{2}$ (polynomials up to degree 2), $B = {1, t, t^{2}}$ and $C = {1, 1 + t, (1 + t)^{2}}$ . Find $[q]_{C}$ for $q (t) = 3 - 4 t + 2 t^{2}$ .

Forecast: $(1 + t)^{2} = 1 + 2 t + t^{2}$ , so its $t^{2}$ -term is 1. Only $c_{3}$ carries $t^{2}$ , so expect its coordinate to equal $q$ 's $t^{2}$ -coefficient, namely $2$ .

There are no "entries to read" — a polynomial isn't a column of numbers until you pick $B$ . So solve a system: $a \cdot 1 + b (1 + t) + c (1 + t)^{2} = 3 - 4 t + 2 t^{2}$ . Why this step? Coordinates are defined as the unique coefficients in the basis expansion; finding them is literally solving that linear equation. Uniqueness is guaranteed by Linear independence of $C$ .
Match the highest power first: $t^{2}$ appears only in $c (1 + t)^{2}$ , giving $c = 2$ . Why this step? Working top-down means each new equation has just one new unknown — a triangular system solves instantly.
$t$ -term: from $b (1 + t) + c (1 + 2 t)$ the $t$ -coefficient is $b + 2 c = - 4$ , so $b = - 4 - 4 = - 8$ . Why this step? With $c$ already pinned to $2$ , this equation has only $b$ left to find.
constant term: $a + b + c = 3$ , so $a = 3 - (- 8) - 2 = 9$ . Why this step? The constant equation is last because it involves all three — but now $b, c$ are known, so $a$ pops out.
$[q]_{C} = 9 - 8 2$ .

Verify: $9 (1) - 8 (1 + t) + 2 (1 + 2 t + t^{2}) = 9 - 8 - 8 t + 2 + 4 t + 2 t^{2} = 3 - 4 t + 2 t^{2} = q$ ✓. Top coordinate $= 2$ as forecast; the middle coordinate went strongly negative — Cell D behaviour showing up again.

Cell G — the degenerate input (zero vector)

Worked example Example 7 (Cell G — edge case)

What is $[0]_{B}$ for any basis $B$ , and what does the change-of-basis matrix do to it?

Forecast: the zero vector is the origin — no rulers, no measuring — so guess "all zeros" no matter what $B$ is.

Seek $c_{1} b_{1} + \dots + c_{n} b_{n} = 0$ . Why this step? That's the defining equation of $[0]_{B}$ .
By Linear independence, the only solution is $c_{1} = \dots = c_{n} = 0$ . Why this step? "Linearly independent" means the sole combination giving $0$ is the all-zero one. This is the same uniqueness that makes coordinates well-defined at all.
So $[0]_{B} = 0$ , and $C \leftarrow B P 0 = 0$ for every $P$ . Why this step? Any matrix times the zero column is the zero column — the origin is fixed by every linear map.

Verify: with $C \leftarrow B P = [2 - 1 - 1 2]$ from Ex 3, $P [00] = [00]$ ✓. The origin is the one point every observer agrees on.

Cell H — real-world word problem

Worked example Example 8 (Cell H)

Two surveyors mark a buried chest. Ann uses East–North axes $E = {e_{1}, e_{2}}$ . Bob's map is rotated $9 0^{\circ}$ counter-clockwise, so his rulers are $f_{1} = [01]$ (points North) and $f_{2} = [- 1 0]$ (points West). Ann reads the chest at $[v]_{E} = [34]$ (3 East, 4 North). What does Bob write?

Forecast: rotating the axes $+ 9 0^{\circ}$ makes a vector's numbers rotate $- 9 0^{\circ}$ . A point "3 East, 4 North" should become, in Bob's frame, "4 along- $f_{1}$ , 3 along- $f_{2}$ " (with a sign to sort out). Guess something like $[4 - 3]$ .

Bob's basis-as-columns is $F = [01 - 1 0]$ . We want $[v]_{F} = F^{- 1} [v]_{E}$ . Why this step? $F$ turns Bob-numbers into Ann's standard East–North numbers; to go the other way we invert. This is exactly the Cell-A move with $F$ in place of $B$ , because Ann's frame is the standard basis here.
$det F = 0 \cdot 0 - (- 1) (1) = 1$ , so $F^{- 1} = [0 - 1 10]$ . Why this step? Same $2 \times 2$ inverse rule (swap diagonal, negate off-diagonal, divide by $det = 1$ ); a rotation's inverse is the rotation the other way, which this sign pattern shows.
$[v]_{F} = [0 - 1 10] [34] = [4 - 3]$ . Why this step? Row-times-column: top $0 (3) + 1 (4) = 4$ , bottom $- 1 (3) + 0 (4) = - 3$ . This is Bob's reading.

Verify: $4 f_{1} + (- 3) f_{2} = 4 [01] - 3 [- 1 0] = [34]$ = Ann's chest ✓. Bob writes $[4 - 3]$ — matching the forecast. Same chest, different numbers; the rotation matrix just relabels.

Cell I — the exam twist (order matters)

Worked example Example 9 (Cell I)

The exam gives $B = {b_{1}, b_{2}}$ with $b_{1} = [10]$ , $b_{2} = [11]$ , and a reordered basis $B^{'} = {b_{2}, b_{1}}$ — same vectors, swapped order. For $[v]_{B} = [52]$ , find $[v]_{B^{'}}$ .

Forecast: if only the order flips, surely the coordinates just swap: $[25]$ . Let's confirm the machinery agrees.

The change-of-basis matrix $B^{'} \leftarrow B P$ has columns $[b_{1}]_{B^{'}}$ and $[b_{2}]_{B^{'}}$ . Why this step? Same rule as always — columns are the old vectors ( $b_{1}, b_{2}$ in $B$ 's order) written in the new coordinates ( $B^{'}$ ).
In $B^{'}$ , the first ruler is $b_{2}$ and the second is $b_{1}$ . So $b_{1} = 0 \cdot b_{2} + 1 \cdot b_{1} \Rightarrow [b_{1}]_{B^{'}} = [01]$ , and $b_{2} = 1 \cdot b_{2} + 0 \cdot b_{1} \Rightarrow [b_{2}]_{B^{'}} = [10]$ . Why this step? Reordering just renames which coordinate slot each ruler owns; the first slot now belongs to $b_{2}$ , the second to $b_{1}$ .
$B^{'} \leftarrow B P = [0110]$ — the swap matrix. Why this step? Its two columns are exactly the coordinate vectors from step 2, placed side by side.
$[v]_{B^{'}} = [0110] [52] = [25]$ . Why this step? Multiplying by the swap matrix exchanges the two entries: top becomes the old bottom ( $2$ ), bottom becomes the old top ( $5$ ).

Verify: in $B^{'}$ that means $2 b_{2} + 5 b_{1}$ ; in $B$ the original was $5 b_{1} + 2 b_{2}$ — identical sums, same $v$ ✓. Order matters: an ordered basis is part of the data.

Active recall

Recall Cover the answers

Q: To go standard→ $B$ , multiply by? → $B^{- 1}$ (Cell A).
Q: To go $B$ →standard, multiply by? → $B$ (Cell B, no inverse).
Q: $B$ → $C$ matrix in $R^{n}$ ? → $C^{- 1} B$ (Cell C).
Q: Can a coordinate be negative or fractional? → yes (Cell D).
Q: What is $[0]_{B}$ ? → the zero column, always (Cell G).
Q: Does reordering a basis change coordinates? → yes, it permutes them (Cell I).

Which matrix converts standard coordinates to

B

-coordinates in

R^{n}

B^{- 1}

, where

B

has the basis vectors as columns.

Why doesn't going to the standard basis need an inverse?

Because

B [v]_{B}

directly rebuilds

v

from its coefficients.

For polynomials, how do you find coordinates without matrix inversion?

Solve the expansion equation, matching highest powers down to constants.

What are the coordinates of the zero vector in any basis?

All zeros, because independence forbids any non-trivial combination equal to

0

Does swapping the order of basis vectors change the change-of-basis matrix?

Yes — you get a permutation (swap) matrix.