Linear-Algebra-Done-Right · Unit 8 · Video 2 · Interactive Practice

Quotient Spaces V/U: Parallel Lines as Vectors

IKey Formulas

Formula	Name	Hypotheses
$v + U = \{v + u : u \in U\}$	Translate of $U$	$v \in V$ , $U$ any subset of $V$
$V/U = \{v + U : v \in V\}$	Quotient space	$U$ a subspace of $V$
$v - w \in U \iff v + U = w + U \iff (v + U) \cap (w + U) \neq \emptyset$	Equal or disjoint	$U$ a subspace of $V$ ; $v, w \in V$
$(v + U) + (w + U) = (v + w) + U$ , $\ \lambda(v + U) = (\lambda v) + U$	Operations on $V/U$	$U$ a subspace of $V$ ; $\lambda \in \mathbf{F}$ ; zero element $0 + U = U$ , inverse $(-v) + U$

Key Insight: An element of $V/U$ is a set with many representatives. Swapping $v$ for another representative $v'$ changes it by $v - v' \in U$ , and because $U$ is closed under addition and scalar multiplication, sums and scalar multiples change only by vectors of $U$ too, so the answer is the same translate.

IISame Line or Disjoint Lines

For the slope-2 line $U$ through the origin, $v + U = w + U$ exactly when $v - w \in U$ ; otherwise the lines never meet.

💡 Challenge: keep $v = (17, 20)$ and drag $w$ to the one point with first coordinate $3$ for which $w + U = v + U$ , then to the one with second coordinate $4$ .

IIIAdding Lines Through Any Representatives

Whichever points represent $L_1 = (17, 20) + U$ and $L_2 = (-14, 0) + U$ , their sum lands on one line.

Operation λ

IVWhy the Subspace Hypothesis Matters

When $U$ is not a subspace, a translate $v + U$ can share points with $U$ without being equal to it.

Subset U

VQuiz Questions

Problem 1 · Which Translate Is It?

Given: $U = \{(x, -3x) : x \in \mathbf{R}\}$ , the line $y = -3x$ in $\mathbf{R}^2$ — find the translate equal to $(2, 1) + U$ .

✅ Correct!

(2, 1) - (0, 7) = (2, -6)

and

-6 = -3 \cdot 2

, so the difference lies in

U

: both translates are the line

y = -3x + 7

❌ Check the sign of the slope.

(2, 1) - (1, -2) = (1, 3)

, whose second coordinate is

+3

times the first. That difference would lie on

y = 3x

, but

U

requires

-3

times the first.

❌ That is the additive inverse.

(-2, -1) + U = -\big((2, 1) + U\big)

, the reflection of that line in the origin. The difference

(2, 1) - (-2, -1) = (4, 2)

is not in

U

, since

2 \neq -12

❌ Not quite. Two translates are equal exactly when

v - w \in U

, that is, when the second coordinate of

v - w

-3

times the first.

Show solution

By the equal-or-disjoint theorem, $(2, 1) + U = w + U$ exactly when $(2, 1) - w \in U$ , i.e. when $(2, 1) - w$ has the form $(x, -3x)$ . Test each option:

$(2, 1) - (1, -2) = (1, 3)$ : $3 \neq -3$ , not in $U$ .
$(2, 1) - (2, -1) = (0, 2)$ : $2 \neq 0$ , not in $U$ .
$(2, 1) - (0, 7) = (2, -6)$ : $-6 = -3 \cdot 2$ , in $U$ . ✓
$(2, 1) - (-2, -1) = (4, 2)$ : $2 \neq -12$ , not in $U$ .

So $(2, 1) + U = (0, 7) + U$ . Geometrically, $(2, 1) + U = \{(2 + x,\ 1 - 3x) : x \in \mathbf{R}\}$ is the line of slope $-3$ through $(2, 1)$ ; taking $x = -2$ gives its point $(0, 7)$ , so it is the line $y = -3x + 7$ . The other three options are lines parallel to it, hence disjoint from it.

Problem 2 · Different Representatives, Same Plane?

Given: the plane $U = \{(x, y, z) \in \mathbf{R}^3 : x + y + z = 0\}$ , $v = (1, 2, 3)$ and $w = (2, 2, 2)$ — decide how $v + U$ and $w + U$ are related.

✅ Correct!

v - w = (-1, 0, 1)

has coordinate sum

0

, so

v - w \in U

and

v + U = w + U

❌ One translate has many representatives. Different vectors can name the same translate; the test is whether

v - w = (-1, 0, 1)

lies in

U

, and its coordinates sum to

0

❌ Translates of a subspace cannot partly overlap. Two translates of

U

are either equal or disjoint, so they never share just a line. These two share a point, so they are equal.

❌ That compares $w$ with $U$ , not with $v$ .

w \notin U

shows only that

w + U \neq U

. To compare

v + U

with

w + U

, test

v - w = (-1, 0, 1)

, which is in

U

Show solution

$U$ is a subspace (it contains $0$ , and sums and scalar multiples of vectors with coordinate sum $0$ again have coordinate sum $0$ ), so the equal-or-disjoint theorem applies:

v - w = (1, 2, 3) - (2, 2, 2) = (-1, 0, 1), \qquad -1 + 0 + 1 = 0,

so $v - w \in U$ and $v + U = w + U$ .

To identify the plane: every element of $v + U$ is $(1 + x,\ 2 + y,\ 3 + z)$ with $x + y + z = 0$ , so its coordinates sum to $6$ . Conversely, if $p$ has coordinate sum $6$ , then $p - v$ has coordinate sum $0$ , so $p = v + (p - v) \in v + U$ . Hence $v + U$ is the plane $x + y + z = 6$ , parallel to $U$ , and the same argument with $w$ gives the same plane. Since $(2, 2, 2) \notin U$ , this plane is not $U$ itself, so it misses the origin.

Problem 3 · Arithmetic in $\mathbf{R}^2/U$

Given: $U = \{(x, -3x) : x \in \mathbf{R}\}$ in $\mathbf{R}^2$ ; since $(1, 4) - (0, 7) = (1, -3) \in U$ , the translate $(1, 4) + U$ is also $(0, 7) + U$ — compute $2\big((1, 4) + U\big) + \big((-2, 0) + U\big)$ .

With the representatives $(0, 7)$ and $(-2, 0)$ , which vector do the rules produce?

Which line is the resulting element of $\mathbf{R}^2/U$ ?

✅ Correct! With

(1, 4)

the rules give

2(1, 4) + (-2, 0) = (0, 8)

; with

(0, 7)

they give

(-2, 14)

. The difference

(2, -6)

lies in

U

, so both name the line

y = -3x + 8

❌ The scalar was dropped.

2\big((0, 7) + U\big) = (0, 14) + U

, so the representative is

(0, 14) + (-2, 0)

, not

(0, 7) + (-2, 0)

❌ The scalar multiplies only the first translate.

(-2, 0) + U

is added once:

(0, 14) + (-2, 0) = (-2, 14)

❌ Apply the rules to representatives.

\lambda(v + U) = (\lambda v) + U

and

(v + U) + (w + U) = (v + w) + U

, so compute

2(0, 7) + (-2, 0)

❌ The scalar was dropped. The intercepts are

7

and

-6

, and doubling a line doubles its intercept:

2 \cdot 7 + (-6) = 8

, not

7 + (-6)

❌ Check the second intercept. The line

(-2, 0) + U

has equation

y = -3(x + 2) = -3x - 6

, so its intercept is

-6

, not

6

❌ The operations are well defined. Different representatives give results that differ by a vector of

U

, hence the same translate:

(0, 8) - (-2, 14) = (2, -6) \in U

❌ Find the line through the representative. The result is the line of slope

-3

through

(-2, 14)

Show solution

Step 1: representatives. With $(0, 7)$ for the first translate,

2\big((0, 7) + U\big) + \big((-2, 0) + U\big) = \big(2(0, 7) + (-2, 0)\big) + U = (-2, 14) + U.

With $(1, 4)$ instead, the same rules give $\big(2(1, 4) + (-2, 0)\big) + U = (0, 8) + U$ .

Step 2: same translate. $(0, 8) - (-2, 14) = (2, -6)$ and $-6 = -3 \cdot 2$ , so the difference is in $U$ and $(0, 8) + U = (-2, 14) + U$ .

Step 3: the line. The line of slope $-3$ through $(0, 8)$ is $y = -3x + 8$ ; check $(-2, 14)$ : $-3(-2) + 8 = 14$ ✓.

Intercept check. Each line of slope $-3$ is $y = -3x + c$ with $c$ the second coordinate plus $3$ times the first: $(1, 4) + U$ has $c = 7$ and $(-2, 0) + U$ has $c = -6$ . Then $2 \cdot 7 + (-6) = 8$ ✓.

Problem 4 · An Additive Inverse in $\mathbf{R}^3/U$

Given: the line $U = \{(t, t, 0) : t \in \mathbf{R}\}$ through the origin in $\mathbf{R}^3$ — find the additive inverse of $(2, 3, 1) + U$ in $\mathbf{R}^3/U$ .

✅ Correct!

(2, 3, 1) + (-5, -6, -1) = (-3, -3, 0) \in U

, so the sum of the two translates is

(-3, -3, 0) + U = U

, the zero element. It is

(-2, -3, -1) + U

written with another representative.

❌ $U$ is the zero element, not the inverse.

\big((2, 3, 1) + U\big) + U = (2, 3, 1) + U

, which is not

U

because

(2, 3, 1) \notin U

❌ That is the same translate.

(2, 3, 1) - (1, 2, 1) = (1, 1, 0) \in U

, so

(1, 2, 1) + U = (2, 3, 1) + U

; adding it to itself gives

(4, 6, 2) + U \neq U

❌ Negate every coordinate.

(2, 3, 1) + (-2, -3, 1) = (0, 0, 2)

, which is not of the form

(t, t, 0)

Show solution

The zero element of $\mathbf{R}^3/U$ is $0 + U = U$ , and $-(v + U) = (-v) + U$ . With $v = (2, 3, 1)$ the inverse is $(-2, -3, -1) + U$ , a line parallel to $U$ .

None of the options is written with the representative $(-2, -3, -1)$ , so test each against it:

(-5, -6, -1) - (-2, -3, -1) = (-3, -3, 0) \in U,

so $(-5, -6, -1) + U = (-2, -3, -1) + U$ is the inverse. Directly:

\big((2, 3, 1) + U\big) + \big((-5, -6, -1) + U\big) = (-3, -3, 0) + U = U.

For the others: $(1, 2, 1) + U$ equals $(2, 3, 1) + U$ itself; $(-2, -3, 1) - (-2, -3, -1) = (0, 0, 2) \notin U$ ; and $U$ is the zero element, the inverse only of itself.

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