Coordinate vectors — change of basis: Exercises

Quick symbol reminder, so nothing appears before it is named:

Level 1 — Recognition

Exercise 1.1

In $R^{2}$ take $B = {b_{1}, b_{2}}$ with $b_{1} = [20]$ , $b_{2} = [03]$ . A vector has coordinate vector $[v]_{B} = [5 - 2]$ . What is the plain vector $v$ (in the standard basis)?

Recall Solution

WHAT: the coordinate vector is a recipe. It says: take $5$ copies of $b_{1}$ and $- 2$ copies of $b_{2}$ . WHY: by the very definition $v = c_{1} b_{1} + c_{2} b_{2}$ . $v = 5 [20] + (- 2) [03] = [100] + [0 - 6] = [10 - 6] .$ Equivalently $v = B [v]_{B}$ with $B = [2003]$ . Same answer: $v = [10 - 6]$ .

Exercise 1.2

For the same $B$ , you are given the plain vector $w = [89]$ . Find $[w]_{B}$ by inspection.

Recall Solution

WHAT: we want how many $b_{1}$ 's and $b_{2}$ 's rebuild $w$ . WHY inspection works here: $b_{1}$ only touches the $x$ -slot, $b_{2}$ only the $y$ -slot (a diagonal basis). So they don't interfere.

$x$ : $2 c_{1} = 8 \Rightarrow c_{1} = 4$ .
$y$ : $3 c_{2} = 9 \Rightarrow c_{2} = 3$ . $[w]_{B} = [43] .$ Check: $4 b_{1} + 3 b_{2} = [89]$ . ✓

Level 2 — Application

Exercise 2.1

$B = {b_{1}, b_{2}}$ with $b_{1} = [12]$ , $b_{2} = [31]$ . Find $[v]_{B}$ for $v = [55]$ .

Recall Solution

WHAT: solve $B [v]_{B} = v$ , i.e. $[v]_{B} = B^{- 1} v$ . WHY $B^{- 1}$ and not $B$ : $B$ turns recipe→plain; we have plain and want recipe, so we run it backwards with the inverse. $B$ is invertible because its columns are linearly independent (they're not multiples of each other). $B = [1231], det B = 1 \cdot 1 - 3 \cdot 2 = - 5.$ $B^{- 1} = \frac{1}{- 5} [1 - 2 - 3 1] = [- \frac{1}{5} \frac{2}{5} \frac{3}{5} - \frac{1}{5}] .$ $[v]_{B} = B^{- 1} [55] = [- 1 + 3 2 - 1] = [21] .$ Check: $2 b_{1} + 1 b_{2} = [24] + [31] = [55]$ . ✓ $[v]_{B} = [21]$ .

The figure below shows $v$ reached by walking $2$ steps of $b_{1}$ then $1$ step of $b_{2}$ .

Figure — Coordinate vectors — change of basis

Exercise 2.2

Two non-standard bases: $B$ as in 2.1, and $C = {c_{1}, c_{2}}$ with $c_{1} = [10]$ , $c_{2} = [11]$ . Build $C \leftarrow B P$ , then use it on the vector from 2.1 (which had $[v]_{B} = [21]$ ).

Recall Solution

WHAT: use $C \leftarrow B P = C^{- 1} B$ . WHY this order: read right to left — $B$ turns $B$ -numbers into standard numbers, then $C^{- 1}$ turns standard into $C$ -numbers. The inner "standard" cancels, leaving $B \to C$ . $C = [1011], C^{- 1} = [10 - 1 1] .$ $C \leftarrow B P = C^{- 1} B = [10 - 1 1] [1231] = [1 - 2 2 3 - 1 1] = [- 1 2 21] .$ Now apply to $[v]_{B} = [21]$ : $[v]_{C} = [- 1 2 21] [21] = [- 2 + 2 4 + 1] = [05] .$ Sanity check with the plain vector $v = [55]$ : $0 c_{1} + 5 c_{2} = 5 [11] = [55]$ . ✓ $[v]_{C} = [05]$ .

Level 3 — Analysis

Exercise 3.1 (polynomials)

In $P_{2}$ (degree $\leq 2$ polynomials) let $B = {1, t, t^{2}}$ and $C = {1, 1 - t, (1 - t)^{2}}$ . Find $[p]_{C}$ for $p (t) = 3 - 4 t + t^{2}$ .

Recall Solution

WHAT: find $a, b, c$ with $a \cdot 1 + b (1 - t) + c (1 - t)^{2} = 3 - 4 t + t^{2}$ . WHY match top-down: the highest-degree basis vector $(1 - t)^{2}$ is the only one carrying $t^{2}$ , so its coefficient is forced first; then $t$ ; then the constant. This is triangular — one unknown per step. Expand $(1 - t)^{2} = 1 - 2 t + t^{2}$ . Collect: $a + b + c (const), (- b - 2 c) t, c t^{2} .$

$t^{2}$ : $c = 1$ .
$t$ : $- b - 2 c = - 4 \Rightarrow - b - 2 = - 4 \Rightarrow b = 2$ .
const: $a + b + c = 3 \Rightarrow a + 2 + 1 = 3 \Rightarrow a = 0$ . $[p]_{C} = 021 .$ Check: $0 \cdot 1 + 2 (1 - t) + 1 (1 - t)^{2} = 2 - 2 t + 1 - 2 t + t^{2} = 3 - 4 t + t^{2}$ . ✓

Exercise 3.2

Still in $P_{2}$ . Build the full matrix $C \leftarrow B P$ (columns = $[b_{j}]_{C}$ ), and confirm it reproduces the answer of 3.1 when applied to $[p]_{B} = 3 - 4 1$ .

Recall Solution

WHAT: each column is an old basis vector written in $C$ . So express $1, t, t^{2}$ using ${1, 1 - t, (1 - t)^{2}}$ . WHY: by the parent formula, column $j$ of $C \leftarrow B P$ is $[b_{j}]_{C}$ .

$1 = 1 \cdot 1$ , so $[1]_{C} = 100$ .
$t$ : solve $a + b (1 - t) + c (1 - t)^{2} = t$ . Top-down: $t^{2}$ gives $c = 0$ ; $t$ gives $- b = 1 \Rightarrow b = - 1$ ; const $a + b = 0 \Rightarrow a = 1$ . So $[t]_{C} = 1 - 1 0$ .
$t^{2}$ : $t^{2} = (1 - t)^{2} - 1 + 2 t$ . Using the previous line for $2 t$ : $2 t = 2 (1 - t) - 2 + ...$ — cleaner to solve directly. Set $a + b (1 - t) + c (1 - t)^{2} = t^{2}$ : $t^{2}$ gives $c = 1$ ; $t$ gives $- b - 2 c = 0 \Rightarrow b = - 2$ ; const $a + b + c = 0 \Rightarrow a = 1$ . So $[t^{2}]_{C} = 1 - 2 1$ . $C \leftarrow B P = 100 1 - 1 0 1 - 2 1 .$ Apply to $[p]_{B} = 3 - 4 1$ : $100 1 - 1 0 1 - 2 1 3 - 4 1 = 3 - 4 + 1 4 - 2 1 = 021 .$ Matches 3.1. ✓ Notice the matrix is upper-triangular — exactly because $C$ 's $j$ -th vector has degree $j - 1$ , so low-degree $b$ 's never need high-degree $c$ 's.

Level 4 — Synthesis

Exercise 4.1 (chained changes, three bases)

In $R^{2}$ : $B$ with columns $[11], [1 - 1]$ ; $C$ with columns $[10], [12]$ ; $D$ with columns $[20], [01]$ . Show that $D \leftarrow C P C \leftarrow B P = D \leftarrow B P$ , and compute the product.

Recall Solution

WHAT: change $B \to C$ , then $C \to D$ , and check it equals going $B \to D$ directly. WHY it must hold: translating dictionaries compose. Numerically each is $(new)^{- 1} (old)$ , and the middle basis cancels: $(D^{- 1} C) (C^{- 1} B) = D^{- 1} B$ . $B = [11 1 - 1], C = [1012], D = [2001] .$ $C^{- 1} = [10 - \frac{1}{2} \frac{1}{2}], D^{- 1} = [\frac{1}{2} 0 01] .$ $C \leftarrow B P = C^{- 1} B = [10 - \frac{1}{2} \frac{1}{2}] [11 1 - 1] = [\frac{1}{2} \frac{1}{2} \frac{3}{2} - \frac{1}{2}] .$ $D \leftarrow C P = D^{- 1} C = [\frac{1}{2} 0 01] [1012] = [\frac{1}{2} 0 \frac{1}{2} 2] .$ Product: $D \leftarrow C P C \leftarrow B P = [\frac{1}{2} 0 \frac{1}{2} 2] [\frac{1}{2} \frac{1}{2} \frac{3}{2} - \frac{1}{2}] = [\frac{1}{2} 1 \frac{1}{2} - 1] .$ Direct route: $D \leftarrow B P = D^{- 1} B = [\frac{1}{2} 0 01] [11 1 - 1] = [\frac{1}{2} 1 \frac{1}{2} - 1] .$ Identical. ✓ $D \leftarrow B P = [\frac{1}{2} 1 \frac{1}{2} - 1]$ .

Exercise 4.2 (transformation in two bases — similarity)

A linear map $T : R^{2} \to R^{2}$ has standard matrix $A = [2013]$ . Let $B$ have columns $b_{1} = [10]$ , $b_{2} = [11]$ . Find the matrix of $T$ in the $B$ -basis, i.e. $[T]_{B} = B^{- 1} A B$ .

Recall Solution

WHAT: to apply $T$ using $B$ -numbers, first convert $B \to$ standard ( $B$ ), apply $A$ , then convert standard $\to B$ ( $B^{- 1}$ ). Reading right to left: $B^{- 1} A B$ . WHY this is a similarity: $[T]_{B}$ and $A$ describe the same map through different rulers, so they are similar via the change-of-basis matrix $B$ . $B = [1011], B^{- 1} = [10 - 1 1] .$ $A B = [2013] [1011] = [2033] .$ $[T]_{B} = B^{- 1} (A B) = [10 - 1 1] [2033] = [2003] .$ $[T]_{B} = [2003]$ — diagonal! The new basis happens to be the eigenvector basis, so $T$ just stretches each axis. (Trace $= 5$ and determinant $= 6$ are preserved from $A$ , as similarity guarantees.)

Level 5 — Mastery

Exercise 5.1 (prove the shortcut)

Prove that in $R^{n}$ , if $B$ and $C$ hold the bases $B, C$ as columns, then $C \leftarrow B P = C^{- 1} B$ . Then explain in one line why $C$ is guaranteed invertible.

Recall Solution

WHAT/WHY, step by step.

For any $v \in R^{n}$ , its plain (standard) form is recovered from $B$ -coordinates by $v = B [v]_{B}$ . Why: the columns of $B$ are the $b_{j}$ , so $B [v]_{B} = \sum c_{j} b_{j} = v$ , the definition of a coordinate vector.
Likewise $v = C [v]_{C}$ .
Set them equal: $C [v]_{C} = B [v]_{B}$ . Why we can: both express the same invariant arrow $v$ .
Multiply on the left by $C^{- 1}$ : $[v]_{C} = C^{- 1} B [v]_{B}$ .
This holds for every $v$ , so the matrix that sends $[v]_{B} \mapsto [v]_{C}$ is exactly $C^{- 1} B$ . That matrix is $C \leftarrow B P$ by definition. $■$

Why $C$ is invertible: its columns are a basis, hence linearly independent, hence $C$ has full rank, hence invertible. Same for $B$ .

Exercise 5.2 (degenerate / edge cases)

(a) What is $B \leftarrow B P$ (same basis both sides)? (b) If $C = E$ (the standard basis), what does $C \leftarrow B P$ reduce to? (c) A student proposes using the vectors $u_{1} = [12], u_{2} = [24]$ as a "basis." Why does the whole change-of-basis machinery break?

Recall Solution

(a) Same ruler in and out changes nothing, so $B \leftarrow B P = I$ . Formula-check: $B^{- 1} B = I$ . ✓

(b) For the standard basis $E = I$ , so $C^{- 1} = I$ and $E \leftarrow B P = I^{- 1} B = B$ . Meaning: to read $B$ -numbers as ordinary numbers you just multiply by $B$ . This is the "standard-basis shortcut" from the parent note.

(c) $u_{2} = 2 u_{1}$ , so the two vectors are linearly dependent — they only span a line, not all of $R^{2}$ . Consequences that each break the machinery:

Not a basis: vectors off that line have no expansion at all, so $[v]_{U}$ may not exist.
No uniqueness: vectors on the line have infinitely many expansions, so the coordinate vector isn't well-defined.
$U$ not invertible: $det [1224] = 4 - 4 = 0$ , so $U^{- 1}$ (needed for $C^{- 1} B$ ) doesn't exist.

All three are the same fact seen three ways: a valid basis must be linearly independent. $■$

Wrap-up recall

Recall One-line answers (cover them)

Plain vector from coordinates? ::: $v = B [v]_{B}$ .
Coordinates from plain vector? ::: $[v]_{B} = B^{- 1} v$ .
$B \to C$ matrix in $R^{n}$ ? ::: $C^{- 1} B$ .
Chain $B \to C \to D$ ? ::: multiply $D \leftarrow C P C \leftarrow B P = D \leftarrow B P$ .
Map $T$ in new basis? ::: $[T]_{B} = B^{- 1} A B$ (a similarity).
Why must a basis be independent? ::: else coordinates fail to exist or fail to be unique, and $B$ isn't invertible.