2.3 Linear maps, null space, range, rank–nullity

A linear map is a function that respects adding and scaling. What it sends to zero (the null space) and what it can reach (the range) trade off exactly. Their dimensions add up to the dimension of the input space.

In JEE, “linear” meant a straight-line graph. Here it means a function that does not care whether you add first or apply it first, which turns out to be far more useful. Those functions are the main subject of linear algebra. Every matrix is one (chapter 2.5), and so are differentiation, integration, rotations, projections and the layers of a neural network. This chapter defines them and proves the most important fact about them: what a map destroys and what it produces always add up to what it started with.

Before you start

2.1: vector spaces and the subspace test. 2.2: span, basis, dimension, and extending an independent set to a basis (Apostol Thm 1.7). Here are three quick checks.

Quick checkWhat is

\dim P_3

, the polynomials of degree at most 3?

Quick checkIn a 5-dimensional space, you have 3 independent vectors. Which statement is true?

Quick checkIs

\{(x, y) : y = 2x + 1\}

a subspace of

\mathbb R^2

The picture

Each map below moves the whole plane. Pick “Squash to a line” and press “Apply the map”. Every point of the plane lands on the green line, the range. The whole red line, the null space, lands on the single point $O$ , the origin. One dimension survived and one was destroyed, and $1 + 1 = 2$ , the dimension of the plane. That count is this chapter’s main theorem.

Two things to notice in every preset. Grid lines stay straight and evenly spaced (or get squashed flat together), and the origin stays put. Those two facts are what “linear” means geometrically. The arrows $\hat\imath$ and $\hat\jmath$ are the standard basis vectors $(1, 0)$ and $(0, 1)$ of $\mathbb R^2$ from chapter 2.2.

The language: functions

Same idea elsewhere· Recap: functions (defined in chapter 0.3)

A function (or map, mapping, transformation) $T: V \to W$ assigns to each $x \in V$ exactly one element $T(x) \in W$ , the value of $T$ at $x$ (also called the image of $x$ ). The set $V$ is the domain and $W$ is the codomain. The set of all outputs, $T(V) = \{T(x) : x \in V\}$ , is the range of $T$ (some books call the whole set the image of $T$ ). The range can be smaller than the codomain.

$T$ is one-to-one if different inputs always give different outputs: $T(x) = T(y)$ only when $x = y$ . $T$ is onto if its range is all of $W$ . We write $x \mapsto T(x)$ , read ” $x$ is sent to $T(x)$ ”. Chapter 0.3 has the full treatment, where one-to-one is called injective and onto is called surjective.

Definitions and theorems

Where this came from· Linear substitutions came first

Before anyone said “linear map”, mathematicians used linear substitutions: replace the variables $x, y$ by $ax + by$ and $cx + dy$ . Gauss already composed such substitutions in his work on quadratic forms (1801), and the problem was always the same. A change of variables should simplify a form or an equation, so what does it keep and what does it lose? In 1858 Arthur Cayley wrote substitutions as arrays and defined their sum and product, which began matrix algebra (the word “matrix” is James Joseph Sylvester’s, from about 1850). The definition used here, a function between vector spaces that respects adding and scaling, comes from Peano’s axioms of 1888 and became standard in the twentieth century.

Definition T

Linear map

Let $V$ and $W$ be vector spaces. A function $T: V \to W$ is a linear map (Apostol: linear transformation) if, for all $x, y \in V$ and every scalar $c$ :

$T(x + y) = T(x) + T(y)$ (it respects addition);
$T(cx) = c\,T(x)$ (it respects scaling).

In one lineA function between vector spaces with $T(x + y) = T(x) + T(y)$ and $T(cx) = c\,T(x)$ .

Why it existsThey are the functions that keep grid lines straight and evenly spaced (or collapse them) and keep the origin fixed, and they can be understood completely using bases.

In words: adding then mapping gives the same result as mapping then adding, and the same holds for scaling. Spagnolie’s picture from Lecture 7: the map sends the parallelogram that builds $x + y$ to the parallelogram that builds $T(x) + T(y)$ .

Quick checkWhich of these maps

\mathbb R^2 \to \mathbb R^2

is linear?

Two consequences come up all the time:

$T(O) = O$ . Here the $O$ on the left is the zero of $V$ and the $O$ on the right is the zero of $W$ . (Likewise, in rule 1 the $+$ on the left adds in $V$ and the $+$ on the right adds in $W$ .) Proof: $O = 0 \cdot O$ in $V$ by ‘Scaling by zero’ from 2.1, then rule 2 gives $T(0 \cdot O) = 0 \cdot T(O)$ , and $0 \cdot T(O) = O$ in $W$ by ‘Scaling by zero’ again. So a map that moves the origin is not linear.
$T$ passes through every linear combination: $T\!\left(\sum c_i x_i\right) = \sum c_i T(x_i)$ . Apply rule 1 repeatedly, then rule 2 to each term.

Why· Why linear maps are so easy to understand

The second consequence says a linear map is completely decided by what it does to a basis. If $e_1, \dots, e_n$ is a basis, every $x$ is $\sum c_i e_i$ , so

T(x) = c_1 T(e_1) + \dots + c_n T(e_n).

Tell me the $n$ vectors $T(e_1), \dots, T(e_n)$ and I can compute $T$ of anything. In the picture above, the whole plane’s motion is fixed by where the two arrows $\hat\imath$ and $\hat\jmath$ land. Chapter 2.5 writes those landing spots as the columns of a matrix.

Worked example

A linear map

T: \mathbb R^2 \to \mathbb R^2

has

T(1, 0) = (2, 1)

and

T(0, 1) = (-1, 3)

. Find

T(4, -2)

Solution.

Write the input in the standard basis. $(4, -2) = 4(1, 0) + (-2)(0, 1)$ . The values of $T$ are known only on the basis vectors, so express the input with them.
Pass $T$ through the combination. $T(4, -2) = 4\,T(1, 0) - 2\,T(0, 1)$ . Linearity: $T$ respects addition and scaling, so it passes through any linear combination.
Substitute the known values. $4(2, 1) - 2(-1, 3) = (8, 4) + (2, -6) = (10, -2)$ . Ordinary arithmetic in $\mathbb R^2$ .

Answer. $T(4, -2) = (10, -2)$ . The same method gives $T$ of any input, so two vectors of information determine the whole map. This is why a matrix with two columns can describe $T$ .

Your answerSame map:

T(1, 0) = (2, 1)

and

T(0, 1) = (-1, 3)

. What is the SECOND entry of

T(3, 5)

Examples, and how to check linearity

Matrix times vector. For a 2 by 2 array $A$ , define $T(x_1, x_2) = (a_{11}x_1 + a_{12}x_2,\ a_{21}x_1 + a_{22}x_2)$ . Each output entry is a fixed combination of the inputs, so sums and multiples pass straight through. For the first entry: $a_{11}(x_1 + y_1) + a_{12}(x_2 + y_2) = (a_{11}x_1 + a_{12}x_2) + (a_{11}y_1 + a_{12}y_2)$ , and the second entry works the same way. Spagnolie’s examples:

$\begin{bmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{bmatrix}$ rotates the plane by $\theta$ ;
$\begin{bmatrix}1 & 0\\ 0 & 0\end{bmatrix}$ projects onto the $x$ -axis: $(x_1, x_2) \mapsto (x_1, 0)$ .

Differentiation. $D(f) = f'$ is linear on differentiable functions, because $(af + g)' = af' + g'$ . Taking $a = 1$ gives the sum rule from school, and taking $g = 0$ gives the constant-multiple rule. Those are exactly the two conditions.

Integration. $T(f) = \int_0^1 f(t)\,dt + f(1)$ is linear from continuous functions on $[0, 1]$ to $\mathbb R$ (Spagnolie’s Lecture 7). Both parts respect sums and multiples: $\int_0^1 (af + g) = a\int_0^1 f + \int_0^1 g$ , and $(af + g)(1) = af(1) + g(1)$ .

Not linear: $T(f) = f + 1$ . $T(0) = 1 \ne 0$ . One failure is enough.

Not linear: $T(x_1, x_2) = (x_1, 1)$ (Apostol 2.4 #7). $T(O) = (0, 1) \ne O$ .

Not linear: doubling the angle (Apostol 2.4 #15). Describe a point by its distance $r$ from the origin and its angle $\theta$ measured from the positive $x$ -axis. $T$ keeps $r$ and doubles $\theta$ . Then $T(1, 0) = (1, 0)$ (angle $0$ stays $0$ ) and $T(0, 1) = (-1, 0)$ (angle $90°$ becomes $180°$ ). Since $(1, 1) = (1, 0) + (0, 1)$ , additivity would force $T(1, 1) = T(1,0) + T(0,1) = (0, 0)$ . But $(1, 1)$ has angle $45°$ and radius $\sqrt 2$ , so $T(1, 1)$ has angle $90°$ and radius $\sqrt 2$ , which is $(0, \sqrt 2)$ . The two answers differ, so $T$ is not linear.

Worked example

Show that

T(x_1, x_2, x_3) = (2x_1 - x_3,\ x_2 + 3x_3)

is a linear map from

\mathbb R^3

\mathbb R^2

Solution. The plan: check both conditions for arbitrary inputs, with letters and no numbers.

Name the inputs. Let $x = (x_1, x_2, x_3)$ and $y = (y_1, y_2, y_3)$ , and let $c$ be a scalar. Linearity must hold for ALL inputs, so we assume nothing about them.
Check addition. $T(x + y) = (2(x_1 + y_1) - (x_3 + y_3),\ (x_2 + y_2) + 3(x_3 + y_3))$ . Regroup each entry: $(2x_1 - x_3) + (2y_1 - y_3)$ and $(x_2 + 3x_3) + (y_2 + 3y_3)$ . That is $T(x) + T(y)$ . Each output entry is a fixed combination of the inputs, so the entry for $x + y$ splits into the entry for $x$ plus the entry for $y$ .
Check scaling. $T(cx) = (2cx_1 - cx_3,\ cx_2 + 3cx_3) = c\,(2x_1 - x_3,\ x_2 + 3x_3) = c\,T(x)$ . Factor $c$ out of each entry.

Answer. Both conditions hold, so $T$ is linear. As a quick sanity check, $T(0, 0, 0) = (0, 0)$ .

Worked example

T(x_1, x_2) = (x_1 x_2,\ x_1)

linear?

Solution.

Test the zero vector first. $T(0, 0) = (0, 0)$ . A linear map must send $O$ to $O$ . This one passes, and passing proves nothing yet.
Look for a failure of scaling. Take $x = (1, 1)$ and $c = 2$ . Then $T(x) = (1, 1)$ , so $c\,T(x) = (2, 2)$ . The product $x_1 x_2$ is suspicious, because scaling both inputs by 2 scales that product by 4.
Compute the other side. $T(cx) = T(2, 2) = (4, 2)$ . Linearity needs $T(cx) = c\,T(x)$ .
Compare. $(4, 2) \ne (2, 2)$ . One specific failure is enough to prove “not linear”.

Answer. Not linear, because $T(2x) \ne 2T(x)$ for $x = (1, 1)$ . The map sends $O$ to $O$ and is still not linear.

Quick checkA function

T

satisfies

T(O) = O

. Must

T

be linear?

Watch out· The school 'linear function' y = mx + c is not linear here

$f(x) = 3x + 2$ has a straight-line graph, but $f(0) = 2 \ne 0$ , so it is not a linear map. Its proper name is an affine map: a linear map plus a shift. Only $f(x) = mx$ is linear in this book’s sense.

Your answerFor the school function

f(x) = 3x + 2

, compute

f(1 + 1) - \big(f(1) + f(1)\big)

. A linear map would give

0

What a map kills, and what it reaches

Definition T

Null space

The null space of $T: V \to W$ is

N(T) = \{x \in V : T(x) = O\}.

Many books call it the kernel, written $\ker T$ . It is part of the domain $V$ .

In one lineAll inputs that $T$ sends to zero, $N(T) = \{x \in V : T(x) = O\}$ . Also called the kernel.

Why it existsIt measures what $T$ destroys. $T$ is one-to-one exactly when only $O$ is destroyed.

Definition T

Range

The range of $T$ is $T(V) = \{T(x) : x \in V\}$ , also written $R(T)$ or $\operatorname{im} T$ . It is part of the codomain $W$ .

In one lineAll outputs of $T$ , $T(V) = \{T(x) : x \in V\}$ . Also called the image.

Why it existsIt measures what $T$ can produce. $T(x) = b$ has a solution exactly when $b$ is in the range.

Theorem T Apostol Thms 2.1 and 2.2

Null space and range are subspaces

For a linear map $T: V \to W$ , the null space $N(T)$ is a subspace of $V$ and the range $T(V)$ is a subspace of $W$ .

In one line $N(T)$ is a subspace of the domain and $T(V)$ is a subspace of the codomain.

Why it is trueLinearity carries sums and multiples through $T$ , so both sets pass the subspace test.

ProofSpagnolie Lecture 7; Apostol 2.2

The idea: The subspace test, twice. Each closure check is one use of linearity.

T(O) = O

, so

O \in N(T)

and

O \in T(V)

Shown above: every linear map sends

O

O

x, y \in N(T)

T(x + y) = T(x) + T(y) = O + O = O

, so

x + y \in N(T)

Null space, addition. Linearity, then the assumption that both are sent to

O

x \in N(T)

T(cx) = c\,T(x) = cO = O

, so

cx \in N(T)

Null space, scaling. Then

cO = O

, proved as a problem in 2.1.

w_1 = T(x_1)

and

w_2 = T(x_2)

, then

w_1 + w_2 = T(x_1) + T(x_2) = T(x_1 + x_2)

, which is in

T(V)

Range, addition. An element of the range is

T

of something. Name the somethings.

c\,w_1 = c\,T(x_1) = T(cx_1) \in T(V)

. Both sets pass the subspace test.

Range, scaling.

Proof workshopGive the reason for each line: the null space and the range are subspaces

Each line of the proof is shown in order. Choose the reason that justifies it, then press Check.

$T(O) = O$ , so $O \in N(T)$ and $O \in T(V)$ .
If $x, y \in N(T)$ , then $T(x + y) = T(x) + T(y) = O + O = O$ .
If $x \in N(T)$ , then $T(cx) = c\,T(x) = cO = O$ .
If $w_1 = T(x_1)$ and $w_2 = T(x_2)$ , then $w_1 + w_2 = T(x_1 + x_2) \in T(V)$ .
$c\,w_1 = c\,T(x_1) = T(cx_1) \in T(V)$ .
So $N(T)$ and $T(V)$ are subspaces.

Quick check

T: \mathbb R^3 \to \mathbb R^2

is linear. The null space

N(T)

is a subspace of which space?

Definition T

Nullity

The nullity of $T$ is $\dim N(T)$ .

In one lineThe dimension of the null space: how many independent directions $T$ destroys.

Why it existsIt is the number that rank–nullity trades against the rank.

Definition T

Rank

The rank of $T$ is $\dim T(V)$ .

In one lineThe dimension of the range: how many independent directions survive in the output.

Why it existsIt is the true size of what $T$ produces. For a matrix it is the number of independent columns (chapter 2.6).

Where this came from· Who named rank and nullity

The idea of rank grew out of a very old question. When does a system of linear equations have a solution, and how many free choices remain? Georg Frobenius gave the word “rank” its modern meaning in the late 1870s, in work on systems of linear equations and forms. Sylvester later coined the word “nullity” for the count of free choices. Rank–nullity says the two always add up to the number of unknowns, which is the budget idea in the box below.

The main theorem

Theorem T Apostol Thm 2.3

Rank–nullity theorem

If $V$ is finite-dimensional and $T: V \to W$ is linear, then $T(V)$ is finite-dimensional and

\dim N(T) + \dim T(V) = \dim V.

In one lineFor a linear map out of a finite-dimensional space, $\dim V = \operatorname{rank} T + \operatorname{nullity} T$ .

Why it is trueTake a basis of the null space and extend it to a basis of $V$ . The extra vectors are exactly the ones whose images form a basis of the range.

Spagnolie stated this in Lecture 7 and left the proof to the book (“Read the proof in 2.3”). Here it is in full. It uses only two results from chapter 2.2: a subspace of a finite-dimensional space has a basis, and an independent set can be extended to a basis.

Why· The picture behind the proof

Use the “Squash to a line” map again. In it, $T(x, y) = (x + 2y,\ 0.5x + y)$ . The red line is the null space, the vectors along $(-2, 1)$ , all sent to $O$ . The green line is the range, the multiples of $(1, 0.5)$ . Now follow the proof in the picture.

Start with a basis of the null space. Here it is one vector, $e_1 = (-2, 1)$ , on the red line, so $k = 1$ .
Extend it to a basis of the whole plane by adding one vector off the red line, say $e_2 = (1, 0)$ , which is $\hat\imath$ . So $n = 2$ .
Every point of the plane is $c_1 e_1 + c_2 e_2$ . The map kills the $e_1$ part, so all that survives is $c_2\,T(e_2)$ .
$T(e_2) = (1, 0.5)$ lies on the green line and spans it. So the range has one basis vector, and $k + (n - k) = 1 + 1 = 2$ .

ProofApostol 2.3

The idea: Start with a basis of what T destroys, e₁…e_k. Extend it to a basis of all of V by adding e_{k+1}…e_n. Claim: the images of the ADDED vectors form a basis of the range. Then rank = n − k.

Let

e_1, \dots, e_k

be a basis of

N(T)

N(T)

is a subspace of the finite-dimensional space

V

, so it has a basis (a problem in 2.2). Call its size

k

, the nullity. If

N(T) = \{O\}

, then

k = 0

and the basis is empty; the proof still works.In the picture:

e_1

is a vector along the red line.

Extend it to a basis

e_1, \dots, e_k, e_{k+1}, \dots, e_n

V

Apostol Thm 1.7(a) from 2.2: any independent set extends to a basis.

n = \dim V

.In the picture:

e_2

is a vector off the red line, such as

\hat\imath

Claim:

T(e_{k+1}), \dots, T(e_n)

is a basis of

T(V)

This is what we will show. Then

\dim T(V) = n - k

, which rearranges to the theorem.

Let

w = T(x) \in T(V)

, and write

x = \sum_{i=1}^n c_i e_i

Spanning. Take any output

T(x)

and expand

x

in the basis of

V

w = \sum_{i=1}^n c_i T(e_i) = \sum_{i=k+1}^n c_i T(e_i)

, since

T(e_i) = O

for

i \le k

. So the claimed vectors span

T(V)

Linearity passes

T

through the combination. The first

k

basis vectors are in the null space, so their images are

O

and drop out.In the picture: the red part of every point is sent to

O

, and only the part along

e_2

lands on the green line.

Suppose

\sum_{i=k+1}^n c_i T(e_i) = O

Independence. Suppose some combination of the claimed vectors is

O

, and show every coefficient is zero.

Then

T\!\left(\sum_{i=k+1}^n c_i e_i\right) = O

, so

z = \sum_{i=k+1}^n c_i e_i

is in

N(T)

Linearity backwards: a combination of images is the image of the combination.

z = \sum_{i=1}^k d_i e_i

for some scalars

d_i

Everything in

N(T)

is a combination of its basis

e_1, \dots, e_k

\sum_{i=1}^k d_i e_i - \sum_{i=k+1}^n c_i e_i = O

Subtract the two expressions for

z

. This is a combination of the whole basis

e_1, \dots, e_n

equal to

O

All

c_i = 0

. So

T(e_{k+1}), \dots, T(e_n)

is independent.

e_1, \dots, e_n

is a basis, hence independent, so every coefficient is zero. In particular every

c_i

is zero.

\dim T(V) = n - k

, so

\dim N(T) + \dim T(V) = k + (n - k) = n = \dim V

It spans and is independent, so it is a basis. It has

n - k

vectors.

Quick checkIn the proof, why are

T(e_1), \dots, T(e_k)

left out of the basis of the range?

Proof workshopRebuild the rank–nullity proof

Tap the lines in the order a proof would write them. Each line can only use what is already above it.

Why· What the theorem means

Think of $\dim V$ as a budget of independent directions. A linear map spends each one in one of two ways: it destroys it (sends it to $O$ ) or keeps it alive in the output. Nothing is created and nothing goes missing. A map from $\mathbb R^3$ to $\mathbb R^2$ must destroy at least one direction, because its range has room for at most 2. A map from $\mathbb R^2$ to $\mathbb R^3$ can reach at most a plane in $\mathbb R^3$ .

Your answerA linear map

T: P_5 \to \mathbb R^3

is onto. What is

\dim N(T)

Worked example

Let

T: \mathbb R^7 \to \mathbb R^4

be linear and onto. What is

\dim N(T)

? Then decide whether a linear map

S: \mathbb R^6 \to \mathbb R^3

can have a null space of dimension 2.

Solution, part 1.

Translate “onto”. The range is all of $\mathbb R^4$ , so the rank is $\dim \mathbb R^4 = 4$ . Rank is the dimension of the range, and onto means the range is the whole codomain.
Apply rank–nullity to the domain. $\dim N(T) + 4 = \dim \mathbb R^7 = 7$ . The theorem compares the total with the dimension of the DOMAIN, here 7.
Solve. $\dim N(T) = 3$ . No computation with vectors was needed. Only the dimensions entered.

Solution, part 2.

Assume such an $S$ exists and find the rank it would force. The nullity would be 2, so the rank is $6 - 2 = 4$ . Rank–nullity with $\dim \mathbb R^6 = 6$ .
Compare with the codomain. The range of $S$ is a subspace of $\mathbb R^3$ , so its dimension is at most 3. A subspace of a 3-dimensional space has dimension at most 3 (a problem in 2.2).
Conclude. $4 \le 3$ is false. The assumption led to a contradiction.

Answer. $\dim N(T) = 3$ , and no such $S$ exists. In fact every linear map $\mathbb R^6 \to \mathbb R^3$ has nullity at least $6 - 3 = 3$ .

Examples and non-examples

Projection onto the $x$ -axis (Apostol 2.4 #3). $T(x_1, x_2) = (x_1, 0)$ . The null space is $\{(0, x_2)\}$ , the $y$ -axis, with nullity 1. The range is $\{(w_1, 0)\}$ , the $x$ -axis, with rank 1. Check: $1 + 1 = 2 = \dim \mathbb R^2$ .

A map from $\mathbb R^3$ to $\mathbb R^2$ (Lecture 7). $T(x_1, x_2, x_3) = (x_1, 0)$ . Its null space is $\{(0, a, b)\}$ , the $yz$ -plane, with nullity 2. Its range is the $x$ -axis of $\mathbb R^2$ , with rank 1. Check: $2 + 1 = 3$ .

Worked example

Find the null space, range, rank and nullity of

T: \mathbb R^3 \to \mathbb R^3

T(x_1, x_2, x_3) = (x_1 + x_2,\ x_2 + x_3,\ x_1 - x_3)

Solution.

Find the null space by solving $T(x) = O$ . The equations are $x_1 + x_2 = 0$ , $x_2 + x_3 = 0$ and $x_1 - x_3 = 0$ . The null space is the set of inputs sent to $O$ , so we set the output to $O$ .
Solve. The first gives $x_1 = -x_2$ and the second gives $x_3 = -x_2$ . The third then reads $-x_2 + x_2 = 0$ , which is always true. So $x_2 = s$ is free and $x = s(-1, 1, -1)$ . The third equation adds no new condition. This is why one unknown stays free.
Read off the null space. $N(T) = L(\{(-1, 1, -1)\})$ , so the nullity is 1. One basis vector means dimension 1.
Find the range from the images of the standard basis. $T(1, 0, 0) = (1, 0, 1)$ , $T(0, 1, 0) = (1, 1, 0)$ , $T(0, 0, 1) = (0, 1, -1)$ . Every input is a combination of the standard basis, and $T$ passes through combinations, so the range is the span of these three images.
Remove the waste. $(1, 1, 0) = (1, 0, 1) + (0, 1, -1)$ , so the second image is a combination of the others. The other two are independent, since neither is a multiple of the other. A basis of the span needs independence, so we drop the dependent vector (chapter 2.2).
Read off the range. $T(V) = L(\{(1, 0, 1), (0, 1, -1)\})$ is the plane of outputs $(a, b, a - b)$ , and the rank is 2. Two basis vectors mean dimension 2.
Check with rank–nullity. $1 + 2 = 3 = \dim \mathbb R^3$ . The theorem must hold. If it did not, we would have made an arithmetic mistake.

Answer. $N(T) = \{s(-1, 1, -1)\}$ with nullity 1, and $T(V)$ is the plane $\{(a, b, a - b)\}$ with rank 2.

Your answerLet

T: \mathbb R^3 \to \mathbb R^3

T(x_1, x_2, x_3) = (x_1 + x_2 + x_3,\ 2x_1 + 2x_2 + 2x_3,\ 3x_1 + 3x_2 + 3x_3)

. What is its rank?

Worked example Apostol 2.4 #10

Show that

T(x_1, x_2) = (2x_1 - x_2,\ x_1 + x_2)

has nullity 0, and that it is onto

\mathbb R^2

Solution.

Set $T(x) = O$ . $2x_1 - x_2 = 0$ and $x_1 + x_2 = 0$ . This finds the null space.
Solve. The first gives $x_2 = 2x_1$ . Put it into the second: $x_1 + 2x_1 = 3x_1 = 0$ , so $x_1 = 0$ and $x_2 = 0$ . Substitution leaves one equation in one unknown.
Read off the nullity. $N(T) = \{O\}$ , so the nullity is 0. The only input sent to $O$ is $O$ itself.
Use rank–nullity. The rank is $2 - 0 = 2$ . $\dim N(T) + \dim T(V) = \dim \mathbb R^2$ .
Conclude onto. The range is a 2-dimensional subspace of $\mathbb R^2$ , so it is all of $\mathbb R^2$ . A subspace with the full dimension is the whole space (a problem in 2.2).
Check by hitting a target. To reach $(w_1, w_2)$ , take $x_1 = (w_1 + w_2)/3$ and $x_2 = (2w_2 - w_1)/3$ . Then $2x_1 - x_2 = w_1$ and $x_1 + x_2 = w_2$ . An explicit formula confirms that every target is reached.

Answer. Nullity 0, rank 2, and $T$ is onto. Compare this with the “Squash to a line” map, which has nullity 1 and is not onto. When the domain and codomain have the same dimension, the two properties go together.

Second derivative on polynomials (Lecture 7). Let $V = P_{n-1}$ and $T(f) = f''$ . (We use $P_{n-1}$ so that $\dim V = n$ , and take $n \ge 2$ .) The null space is $\{a + bt\}$ , because $f'' = 0$ exactly for these, so the nullity is 2. Differentiating twice lowers the degree by 2, and every polynomial of degree at most $n - 3$ is reached by integrating it twice. So the range is all polynomials of degree at most $n - 3$ , with rank $n - 2$ . Check: $2 + (n - 2) = n$ .

Your answerLet

T(f) = f''

P_6

. What is the rank of

T

Differentiation on all differentiable functions. $N(D) = \{\text{constant functions}\}$ , because $f' = 0$ exactly when $f$ is constant. See the application below. The domain is infinite-dimensional, so rank–nullity does not apply.

Worked example

Let

D: P_3 \to P_3

D(f) = f'

. Show that

D

is linear, and find its null space, range, rank and nullity.

Solution.

Check linearity. $D(af + g) = (af + g)' = af' + g' = a\,D(f) + D(g)$ . The derivative rules from school are exactly the two conditions. The output of $D$ has degree at most 2, so it lands in $P_3$ .
Write a general input and its image. $f = c_0 + c_1 t + c_2 t^2 + c_3 t^3$ gives $D(f) = c_1 + 2c_2 t + 3c_3 t^2$ . Differentiate term by term.
Find the null space. $D(f) = 0$ forces $c_1 = 0$ , $2c_2 = 0$ and $3c_3 = 0$ . So $f = c_0$ , a constant. A polynomial is the zero polynomial exactly when all its coefficients are 0.
Read off the nullity. $N(D) = L(\{1\})$ , so the nullity is 1. The constants have basis $\{1\}$ .
Find the range. The target $p + qt + rt^2$ is reached by $c_1 = p$ , $c_2 = q/2$ , $c_3 = r/3$ (and any $c_0$ ). So $D(P_3) = P_2$ , with basis $1, t, t^2$ and rank 3. We solved for an input that hits an arbitrary target in $P_2$ . Nothing of degree 3 can be hit, because differentiating lowers the degree.
Check. $1 + 3 = 4 = \dim P_3$ . Rank–nullity.

Answer. $N(D)$ is the constants, nullity 1. The range is $P_2$ , rank 3. Since $P_2$ is smaller than the codomain $P_3$ , $D$ is not onto. (On all differentiable functions the null space of $D$ is again the constants, which is why every antiderivative carries a " $+\,C$ ".)

Worked example

Prove: for a linear map

T

, we have

T(x) = T(y)

exactly when

x - y \in N(T)

Solution. The plan: both directions use the identity $T(x - y) = T(x) - T(y)$ , which comes from linearity.

Prove the identity. $T(x - y) = T(x + (-1)y) = T(x) + (-1)\,T(y) = T(x) - T(y)$ . Condition 1 splits the sum, and condition 2 takes the scalar $-1$ out.
( $\Rightarrow$ ) Assume $T(x) = T(y)$ . Then $T(x - y) = T(x) - T(y) = O$ . Use the identity, then the assumption that the two outputs are equal.
Conclude this direction. So $x - y \in N(T)$ . That is the definition of the null space.
( $\Leftarrow$ ) Assume $x - y \in N(T)$ . Then $O = T(x - y) = T(x) - T(y)$ . The definition of the null space, then the identity.
Conclude this direction. Add $T(y)$ to both sides to get $T(x) = T(y)$ . Axioms 3 to 6 from 2.1 let us move a term across.

Answer. Two inputs have the same output exactly when they differ by a vector of the null space. This lemma is the heart of “one-to-one iff $N(T) = \{O\}$ ”, and of the fact that all solutions of $T(x) = b$ are one solution plus the null space.

Worked example

For

T(x_1, x_2, x_3) = (x_1 + x_2,\ x_2 + x_3,\ x_1 - x_3)

, find ALL solutions of

T(x) = (3, 5, -2)

We found $N(T) = \{s(-1, 1, -1)\}$ and the range $\{(a, b, a - b)\}$ in an example above.

Solution.

Check that a solution exists. The target $(3, 5, -2)$ has the form $(a, b, a - b)$ with $a = 3$ , $b = 5$ , since $3 - 5 = -2$ . $T(x) = b$ is solvable exactly when $b$ is in the range.
Find one solution by guessing a free value. Try $x_2 = 0$ . Then $x_1 + x_2 = 3$ gives $x_1 = 3$ , and $x_2 + x_3 = 5$ gives $x_3 = 5$ . The third entry is $3 - 5 = -2$ , as required. Any one solution will do. We call it the particular solution $x_0 = (3, 0, 5)$ .
Add the whole null space. Every solution is $x_0 + s(-1, 1, -1)$ . If $T(x) = b$ and $T(x_0) = b$ , then $T(x - x_0) = O$ , so $x - x_0$ is in the null space (the lemma above).
Check one member. For $s = 1$ : $x = (2, 1, 4)$ , and $T(x) = (3, 5, -2)$ . Confirms that the family really solves the equation.

Answer. $x = (3, 0, 5) + s(-1, 1, -1)$ for any real $s$ : a line of solutions, parallel to the null space.

Quick checkCan a linear map

T: \mathbb R^3 \to \mathbb R^2

be one-to-one?

Where it is used

Used inside maths· Inside maths: the + C in every integral

Why does every antiderivative come with " $+\,C$ "? Because $D(f) = f'$ is linear, and its null space is the constants. Suppose $D(F) = g$ and $D(G) = g$ . By linearity, $D(G - F) = D(G) - D(F) = g - g$ , the zero function. So $G - F$ is in the null space, which means $G = F + C$ for a constant $C$ . In general, if $T(x) = b$ has at least one solution $x_0$ , then all its solutions are $x_0$ plus the null space. The reason is the same two lines: $T(x) = b$ exactly when $T(x - x_0) = T(x) - T(x_0) = b - b = O$ , that is, when $x - x_0 \in N(T)$ . The same reasoning gives “particular solution plus complementary function” for differential equations in MATH 376, and “one solution plus free variables” for $Ax = b$ in chapter 2.6.

Used in the real world· Cameras: the null space is the line of sight

A simple (orthographic) camera looking along the $z$ -axis records a 3D point $(x, y, z)$ as the 2D point $(x, y)$ . That is a linear map from $\mathbb R^3$ to $\mathbb R^2$ . Its null space is the $z$ -axis: every point straight in front of the lens lands on the same pixel. Rank–nullity says $3 = 2 + 1$ , so exactly one direction of information is lost in every photo, and that direction is depth. So in this idealised camera, depth cannot be recovered from the image alone. Real cameras use perspective, which is not linear, and phones add a second lens or a depth sensor to measure depth directly.

Used in the real world· AI: what a neural-network layer cannot see

A dense layer computes $y = Wx$ , a linear map from $\mathbb R^{2048}$ to $\mathbb R^{512}$ when $W$ is a 512 by 2048 weight matrix. Its rank is at most 512, so by rank–nullity its null space has dimension at least $2048 - 512 = 1536$ . Any change to the input along those 1536 directions produces exactly the same output. The layer cannot see it. Designing bottleneck layers (autoencoders, the “down-projection” in transformers) means choosing which directions to throw away.

Traps

Watch out· Check T(0) = 0 first

It is the fastest way to catch a non-linear map. $T(x) = x + 1$ , $T(x_1, x_2) = (x_1, 1)$ and $T(f) = f + 1$ all fail it at once. But passing it proves nothing: $T(x) = x^2$ has $T(0) = 0$ and is not linear.

Watch out· Null space lives in the domain, range lives in the codomain

For $T: \mathbb R^3 \to \mathbb R^2$ , the null space is a subspace of $\mathbb R^3$ and the range is a subspace of $\mathbb R^2$ . Rank–nullity adds their dimensions and compares the total with the dimension of the DOMAIN. The codomain’s dimension does not appear in it.

Watch out· Examples cannot prove linearity

To show $T$ is linear you must check both conditions for ARBITRARY $x$ , $y$ and $c$ . One counterexample shows it is not linear, as with the angle-doubling map, where one pair of vectors breaks additivity.

Problems

checkApostol 2.4 #7

Is $T(x_1, x_2) = (x_1, 1)$ linear?

Solution

No. $T(0, 0) = (0, 1) \ne (0, 0)$ .

check

A linear map $T: \mathbb R^5 \to \mathbb R^3$ has rank 3. What is its nullity? Can a linear map $\mathbb R^5 \to \mathbb R^3$ have nullity 1?

Solution

Nullity $= 5 - 3 = 2$ . Nullity 1 would force rank 4, but the range sits inside $\mathbb R^3$ , so the rank is at most 3. So no.

check

Is $T(f) = f(0)$ , from functions to $\mathbb R$ , linear? What is its null space?

Solution

Yes: $(af + g)(0) = af(0) + g(0)$ . The null space is all functions with $f(0) = 0$ .

useApostol 2.4 #13

Let $T$ reflect the plane in a line through the origin at angle $\alpha$ . Show $T$ is linear and find its null space and range.

Solution

Let $v = (-\sin\alpha, \cos\alpha)$ , a unit vector perpendicular to the mirror. Then $T(x) = x - 2(x \cdot v)v$ : subtract twice the part of $x$ that sticks out from the mirror, which flips that part to the other side. The dot product $x \cdot v$ is linear in $x$ , so $T$ is linear. Reflecting twice puts every point back where it started, so $T(T(x)) = x$ . If $T(x) = O$ , then $x = T(T(x)) = T(O) = O$ , so $N(T) = \{O\}$ . Rank–nullity then gives rank 2, and a 2-dimensional subspace of $\mathbb R^2$ is all of $\mathbb R^2$ (a problem in 2.2). So the range is $\mathbb R^2$ .

useApostol 2.4 #28

Let $V$ be the convergent real sequences. For $x = (x_n)$ with limit $a$ , define $T(x) = (a - x_n)$ . Show $T$ is linear and describe its null space and range.

Solution

If $x_n \to a$ and $y_n \to b$ , then $x_n + y_n \to a + b$ , so $T(x + y) = (a + b - x_n - y_n) = T(x) + T(y)$ . Similarly $T(cx) = cT(x)$ . The null space is the sequences with $x_n = a$ for all $n$ : the constant sequences. Every $T(x)$ converges to $a - a = 0$ . Every sequence $y_n \to 0$ is reached: take $x = -y$ , whose limit is $0$ , so $T(x) = (0 + y_n) = y$ . So the range is all sequences converging to $0$ .

use

Find the null space, range, rank and nullity of $T: P_3 \to P_3$ , $T(f) = f' - f(0)$ . (Here $f(0)$ means the constant polynomial with that value.)

Solution

Write $f = c_0 + c_1 t + c_2 t^2 + c_3 t^3$ . Then $T(f) = (c_1 - c_0) + 2c_2 t + 3c_3 t^2$ . $T(f) = 0$ forces $c_2 = c_3 = 0$ and $c_1 = c_0$ , so $N(T) = \{c(1 + t)\}$ and the nullity is 1. The range is all of $P_2$ , since any $p + qt + rt^2$ is hit with $c_2 = q/2$ , $c_3 = r/3$ , $c_0 = 0$ , $c_1 = p$ . So the rank is 3. Check: $1 + 3 = 4 = \dim P_3$ .

prove

Prove that a linear map $T$ is one-to-one (that is, $T(x) = T(y)$ implies $x = y$ ) if and only if $N(T) = \{O\}$ .

Solution

If $T$ is one-to-one and $T(x) = O = T(O)$ , then $x = O$ . So $N(T) = \{O\}$ .

Conversely, suppose $N(T) = \{O\}$ and $T(x) = T(y)$ . Then $T(x - y) = T(x) - T(y) = O$ , so $x - y \in N(T)$ . Hence $x - y = O$ and $x = y$ .

prove

Let $\dim V = \dim W = n$ and let $T: V \to W$ be linear. Prove that $T$ is one-to-one if and only if it is onto (its range is all of $W$ ).

Solution

By the previous problem, one-to-one means nullity 0. By rank–nullity, nullity 0 is the same as rank $n$ . A subspace of $W$ with dimension $n = \dim W$ is all of $W$ (a problem in 2.2), so rank $n$ means the range is $W$ , which is onto. Each step reverses.

prove

Let $T: V \to W$ be linear and $x_1, \dots, x_k \in V$ . Prove: if $T(x_1), \dots, T(x_k)$ are independent, then so are $x_1, \dots, x_k$ . Is the converse true?

Solution

Suppose $\sum c_i x_i = O$ . Apply $T$ : $\sum c_i T(x_i) = T(O) = O$ . Independence of the images forces all $c_i = 0$ .

The converse is false. $x_1 = (0, 1)$ is independent by itself, but the projection $T(x_1, x_2) = (x_1, 0)$ sends it to $O$ . The converse holds when $N(T) = \{O\}$ .

proveApostol 2.4, adapted

Prove that a linear map is determined by its values on a basis: if $S$ and $T$ are linear, $e_1, \dots, e_n$ is a basis of $V$ , and $S(e_i) = T(e_i)$ for every $i$ , then $S = T$ .

Solution

Any $x \in V$ is $\sum c_i e_i$ . Then $S(x) = \sum c_i S(e_i) = \sum c_i T(e_i) = T(x)$ , using linearity of each map.

stretchAxler 3B

Prove that there is no linear map $T: \mathbb R^5 \to \mathbb R^2$ whose null space is $\{(x_1, \dots, x_5) : x_1 = 3x_2 \text{ and } x_3 = x_4 = x_5\}$ .

Solution

That set is $\{(3a, a, b, b, b)\} = L(\{(3,1,0,0,0), (0,0,1,1,1)\})$ , which has dimension 2. Rank–nullity would then give rank $5 - 2 = 3$ . But the range sits inside $\mathbb R^2$ , so the rank is at most 2. So no such map exists.

stretch

Let $T: V \to V$ be linear with $V$ finite-dimensional and $T \circ T = 0$ (applying $T$ twice gives $O$ for every input). Prove $\operatorname{rank} T \le \tfrac12 \dim V$ .

Solution

If $w = T(x)$ is in the range, then $T(w) = T(T(x)) = O$ , so $w \in N(T)$ . Thus $T(V) \subseteq N(T)$ , so $\operatorname{rank} T \le \operatorname{nullity} T$ . Adding $\operatorname{rank} T$ to both sides and using rank–nullity: $2 \operatorname{rank} T \le \operatorname{rank} T + \operatorname{nullity} T = \dim V$ .

Summary

A linear map respects addition and scaling. Consequences: $T(O) = O$ , and $T$ passes through linear combinations.
A linear map is determined by what it does to a basis. This is why matrices can describe them (2.5).
Null space $N(T)$ : inputs sent to $O$ , a subspace of the domain. Range $T(V)$ : all outputs, a subspace of the codomain.
Rank $= \dim T(V)$ , nullity $= \dim N(T)$ .
Rank–nullity: $\operatorname{nullity} + \operatorname{rank} = \dim V$ . Proof: extend a basis of the null space to a basis of $V$ ; the images of the added vectors are a basis of the range.
$T$ is one-to-one exactly when $N(T) = \{O\}$ . Between spaces of equal finite dimension, one-to-one and onto are the same thing.
If $T(x) = b$ has a solution $x_0$ , all its solutions are $x_0$ plus the null space. That one fact explains the $+\,C$ in integrals, the complementary function in ODEs, and the free variables of linear systems.

Watch and read

The same ideas explained by someone else. Use these when an explanation here does not land.

2.3 Linear maps, null space, range, rank–nullity

The picture

The language: functions

Definitions and theorems

Linear map

Examples, and how to check linearity

What a map kills, and what it reaches

Null space

Range

Null space and range are subspaces

Nullity

Rank

The main theorem

Rank–nullity theorem

Examples and non-examples

Where it is used

Traps

Problems

Summary

Watch and read

Where this sits

Comes from

Leads to

Same idea elsewhere