Linear-Algebra-Done-Right · Unit 8 · Video 3 · Interactive Practice

The Quotient Map and V/null T ≅ range T

IKey Formulas

Formula	Name	What it says
$\pi(v) = v + U$	Quotient map $\pi\colon V \to V/U$	Linear, with $\text{null}\, \pi = U$ and $\text{range}\, \pi = V/U$
$\dim V/U = \dim V - \dim U$	Dimension of a quotient	$V$ finite-dimensional; the fundamental theorem applied to $\pi$
$\widetilde{T}(v + \text{null}\, T) = Tv$	Induced map $\widetilde{T}\colon V/(\text{null}\, T) \to W$	Well defined: $u - v \in \text{null}\, T \Rightarrow Tu = Tv$
$\widetilde{T} \circ \pi = T$ , $\ V/(\text{null}\, T) \cong \text{range}\, T$	Factoring $T$	$\widetilde{T}$ is injective and $\text{range}\, \widetilde{T} = \text{range}\, T$

Key Insight: $T$ may send many vectors to $0$ , but they all lie in one translate, $\text{null}\, T$ , which is the zero element of $V/(\text{null}\, T)$ . Passing to the quotient collapses that translate, so every linear map is an injective map composed with a surjective one: $T = \widetilde{T} \circ \pi$ .

II $\widetilde{T}$ Is Well Defined

For $T(x, y) = y - 2x$ , two representatives of one translate differ by a vector of $\text{null}\, T$ , so they give one output.

IIIThe Quotient Map and $\dim V/U$

$\pi$ kills exactly $U$ and hits all of $\mathbf{R}^3/U$ , so the fundamental theorem splits $\dim \mathbf{R}^3$ between them.

U = span of

IVThe Triangle $T = \widetilde{T} \circ \pi$

Factoring $T$ through $\mathbf{R}^3/(\text{null}\, T)$ makes it injective, and both routes count $3 - \dim \text{null}\, T$ dimensions.

Row 1 Row 2

VQuiz Questions

Problem 1 · The Quotient Map on $\mathcal{P}_4(\mathbf{R})$

Given: $U = \text{span}(1, x^2, x^4)$ , a subspace of $\mathcal{P}_4(\mathbf{R})$ , and the quotient map $\pi\colon \mathcal{P}_4(\mathbf{R}) \to \mathcal{P}_4(\mathbf{R})/U$ — find $\text{null}\, \pi$ and $\dim \mathcal{P}_4(\mathbf{R})/U$ .

What is the null space of the quotient map?

What is the dimension of the quotient?

✅ Correct!

\pi(p) = 0 + U

exactly when

p \in U

, so

\text{null}\, \pi = U

, and

\dim \mathcal{P}_4(\mathbf{R})/U = 5 - 3 = 2

❌ Not quite. The zero element of the quotient is

0 + U = U

, and

p + U = U

for every

p \in U

, not only

p = 0

. For example

\pi(x^2) = x^2 + U = U

❌ Not quite. A null space is a subspace of the domain, here

\mathcal{P}_4(\mathbf{R})

; the quotient

\mathcal{P}_4(\mathbf{R})/U

is the range of

\pi

❌ Not quite.

\pi(p)

is the zero element

U

of the quotient exactly when

p + U = U

, that is, when

p \in U

; and

\pi(x) = x + U \neq U

❌ That is $\dim U$ .

1, x^2, x^4

is a basis of

U

, so

\dim U = 3

; the quotient has dimension

\dim \mathcal{P}_4(\mathbf{R}) - \dim U

❌ Check $\dim \mathcal{P}_4(\mathbf{R})$ . Its basis

1, x, x^2, x^3, x^4

has five vectors, so the dimension is

5

, not

4

❌ Not quite. Use

\dim \mathcal{P}_4(\mathbf{R})/U = \dim \mathcal{P}_4(\mathbf{R}) - \dim U

Show solution

Null space. The zero element of $\mathcal{P}_4(\mathbf{R})/U$ is $0 + U = U$ , and $p + U = U$ exactly when $p \in U$ . So $\pi(p) = 0 + U \iff p \in U$ , that is, $\text{null}\, \pi = U$ .

Dimension. The list $1, x^2, x^4$ is linearly independent (distinct powers of $x$ ), so $\dim U = 3$ , while $\dim \mathcal{P}_4(\mathbf{R}) = 5$ . Hence

\dim \mathcal{P}_4(\mathbf{R})/U = \dim \mathcal{P}_4(\mathbf{R}) - \dim U = 5 - 3 = 2.

A basis of the quotient is $x + U,\ x^3 + U$ : every $a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4$ lies in the translate $a_1 x + a_3 x^3 + U$ , because the difference $a_0 + a_2 x^2 + a_4 x^4$ is in $U$ .

Problem 2 · A Spanning List With a Redundant Vector

Given: $U = \text{span}\big((1, 0, 2, 0),\ (0, 1, 0, -1),\ (2, 1, 4, -1)\big)$ , a subspace of $\mathbf{R}^4$ — find $\dim \mathbf{R}^4/U$ .

✅ Correct! The third vector is

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

, so

\dim U = 2

and

\dim \mathbf{R}^4/U = 4 - 2 = 2

❌ Count a basis, not the spanning list.

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

, so the three vectors span only a

2

-dimensional subspace.

❌ Not quite.

4

\dim \mathbf{R}^4

; the quotient loses the

\dim U

directions that

\pi

sends to the zero element.

❌ Subtract, don't add.

\dim \mathbf{R}^4 = \dim U + \dim \mathbf{R}^4/U

, so the quotient has dimension

\dim \mathbf{R}^4 - \dim U

, which cannot exceed

4

Show solution

Step 1: $\dim U$ . The first two vectors are linearly independent (look at the second coordinates: $0$ and $1$ , and the first coordinates: $1$ and $0$ ). The third is a combination of them:

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

So $(1, 0, 2, 0),\ (0, 1, 0, -1)$ is a basis of $U$ and $\dim U = 2$ .

Step 2: the quotient. Applying the fundamental theorem to $\pi\colon \mathbf{R}^4 \to \mathbf{R}^4/U$ , with $\text{null}\, \pi = U$ and $\text{range}\, \pi = \mathbf{R}^4/U$ :

\dim \mathbf{R}^4/U = \dim \mathbf{R}^4 - \dim U = 4 - 2 = 2.

Problem 3 · The Induced Map of a Matrix

Given: $T \in \mathcal{L}(\mathbf{R}^4, \mathbf{R}^2)$ with $T(x_1, x_2, x_3, x_4) = (x_1 + 2x_2 - x_4,\ -2x_1 - 4x_2 + 2x_4)$ , that is, matrix $\begin{pmatrix} 1 & 2 & 0 & -1 \\ -2 & -4 & 0 & 2 \end{pmatrix}$ — find $\dim \mathbf{R}^4/(\text{null}\, T)$ and $\widetilde{T}\big((3, 0, 5, 1) + \text{null}\, T\big)$ .

What is the dimension of the quotient?

What is the value of the induced map?

✅ Correct!

\text{range}\, T = \text{span}\big((1, -2)\big)

has dimension

1

, so

\dim \mathbf{R}^4/(\text{null}\, T) = 1

; and

\widetilde{T}

simply applies

T

to any representative:

T(3, 0, 5, 1) = (2, -4)

❌ That is $\dim \text{null}\, T$ . The second row is

-2

times the first, so

\text{null}\, T

is cut out by one equation and has dimension

3

; the quotient has dimension

4 - 3

❌ Not quite.

2 = \dim \mathbf{R}^2

, but

T

is not surjective: every output

(t, -2t)

lies on one line, so

\dim \text{range}\, T = 1

❌ Not quite.

\dim \mathbf{R}^4/(\text{null}\, T) = 4 - \dim \text{null}\, T = \dim \text{range}\, T

❌ That is $\pi$ , not $\widetilde{T}$ .

\pi

sends a vector to its translate;

\widetilde{T}

sends a translate to a vector of

W = \mathbf{R}^2

, namely

Tv

for any representative

v

❌ Not quite.

\widetilde{T}

sends only the zero element

\text{null}\, T

0

, and

(3, 0, 5, 1) \notin \text{null}\, T

because the first component of

T(3, 0, 5, 1)

3 + 0 - 1 = 2 \neq 0

❌ Check where $\widetilde{T}$ lands. Its codomain is

W = \mathbf{R}^2

, so the output is a vector, not a translate; besides,

\text{null}\, T

lives in

\mathbf{R}^4

, not

\mathbf{R}^2

❌ Not quite. By definition

\widetilde{T}(v + \text{null}\, T) = Tv

; compute

T(3, 0, 5, 1)

Show solution

Step 1: dimensions. The second component of $T$ is $-2$ times the first, so $Tx = 0$ exactly when $x_1 + 2x_2 - x_4 = 0$ . That one equation leaves three free coordinates; a basis of $\text{null}\, T$ is $(-2, 1, 0, 0),\ (0, 0, 1, 0),\ (1, 0, 0, 1)$ , so $\dim \text{null}\, T = 3$ and

\dim \mathbf{R}^4/(\text{null}\, T) = 4 - 3 = 1.

This matches $\text{range}\, T = \{(t, -2t) : t \in \mathbf{R}\} = \text{span}\big((1, -2)\big)$ , of dimension $1$ , as $\mathbf{R}^4/(\text{null}\, T) \cong \text{range}\, T$ requires.

Step 2: the value. By definition $\widetilde{T}(v + \text{null}\, T) = Tv$ :

T(3, 0, 5, 1) = (3 + 0 - 1,\ -6 - 0 + 2) = (2, -4).

Check with another representative. $(3, 0, 5, 1) - (2, 0, 0, 0) = (1, 0, 5, 1)$ , and $T(1, 0, 5, 1) = (1 - 1, -2 + 2) = (0, 0)$ , so $(2, 0, 0, 0)$ lies in the same translate. Indeed $T(2, 0, 0, 0) = (2, -4)$ : the same output.

Problem 4 · Differentiation Through the Quotient

Given: $D \in \mathcal{L}\big(\mathcal{P}_3(\mathbf{R})\big)$ , $Dp = p'$ , and its induced map $\widetilde{D}\colon \mathcal{P}_3(\mathbf{R})/(\text{null}\, D) \to \mathcal{P}_3(\mathbf{R})$ — find the dimension of the quotient and decide which statement about $\widetilde{D}$ is correct.

What is the dimension of the quotient?

Which statement is correct?

✅ Correct!

\text{null}\, D = \mathcal{P}_0(\mathbf{R})

, so the quotient has dimension

4 - 1 = 3

, and

\widetilde{D}

is an isomorphism onto

\text{range}\, D = \mathcal{P}_2(\mathbf{R})

, also of dimension

3

❌ Not quite.

4 = \dim \mathcal{P}_3(\mathbf{R})

; the constants are sent to

0

D

, and they collapse to the zero element of the quotient.

❌ That is $\dim \text{null}\, D$ .

\text{null}\, D

is the constants, of dimension

1

; the quotient has dimension

4 - 1

❌ Check $\text{null}\, D$ .

D(x) = 1 \neq 0

, so

x \notin \text{null}\, D

; only the constants satisfy

p' = 0

, and

\dim \text{null}\, D = 1

❌ Not quite. Find

\text{null}\, D

(the polynomials with

p' = 0

) and subtract its dimension from

\dim \mathcal{P}_3(\mathbf{R}) = 4

❌ Not surjective.

\text{range}\, \widetilde{D} = \text{range}\, D = \mathcal{P}_2(\mathbf{R})

: no derivative of a cubic has an

x^3

term, so

x^3

is never an output.

❌ Same translate.

(x^3 + 7) - (x^3 - 2) = 9 \in \text{null}\, D

, so

x^3 + 7 + \text{null}\, D = x^3 - 2 + \text{null}\, D

: equal outputs from one element of the quotient do not break injectivity.

❌ Not quite. Different representatives of one translate are exactly what the well-definedness check handles: their difference

9

is in

\text{null}\, D

, so

D

gives both the output

3x^2

❌ Not quite.

\widetilde{D}

is always injective, and its range equals

\text{range}\, D

Show solution

Step 1: $\text{null}\, D$ and the quotient. $p' = 0$ exactly when $p$ is constant, so $\text{null}\, D = \mathcal{P}_0(\mathbf{R})$ , of dimension $1$ . Then

\dim \mathcal{P}_3(\mathbf{R})/(\text{null}\, D) = 4 - 1 = 3.

Each element of the quotient is a polynomial "up to its constant term": $x^3 + 7 + \text{null}\, D = x^3 - 2 + \text{null}\, D = x^3 + \text{null}\, D$ .

Step 2: $\widetilde{D}$ . $\widetilde{D}(p + \text{null}\, D) = p'$ , so $\widetilde{D}(x^3 + \text{null}\, D) = 3x^2$ , whichever representative is used.

Injective: $\widetilde{D}(p + \text{null}\, D) = 0$ means $p' = 0$ , so $p \in \text{null}\, D$ and $p + \text{null}\, D$ is the zero element.
Range: $\text{range}\, \widetilde{D} = \text{range}\, D = \mathcal{P}_2(\mathbf{R})$ , since $D(a_1 x + \tfrac{a_2}{2} x^2 + \tfrac{a_3}{3} x^3) = a_1 + a_2 x + a_3 x^2$ , while no output has degree $3$ .

So $\widetilde{D}$ is an isomorphism from $\mathcal{P}_3(\mathbf{R})/(\text{null}\, D)$ onto $\mathcal{P}_2(\mathbf{R})$ , both of dimension $3$ , but not onto the codomain $\mathcal{P}_3(\mathbf{R})$ .

Solved: 0 / 4

The Quotient Map and V/null T ≅ range T

IKey Formulas

IIT~\widetilde{T}T Is Well Defined

IIIThe Quotient Map and dim⁡V/U\dim V/UdimV/U

IVThe Triangle T=T~∘πT = \widetilde{T} \circ \piT=T∘π

VQuiz Questions

II $\widetilde{T}$ Is Well Defined

IIIThe Quotient Map and $\dim V/U$

IVThe Triangle $T = \widetilde{T} \circ \pi$