Linear transformations — definition, kernel, image: Question bank

True or false — justify

Every linear map satisfies

T (0_{V}) = 0_{W}

True. Write

0_{V} = 0 \cdot v

; homogeneity gives

T (0_{V}) = 0 \cdot T (v) = 0_{W}

. It is a free sanity check for linearity.

T (0) = 0

, then

T

is linear.

False.

T (0) = 0

is necessary but not sufficient. E.g.

T (x, y) = (x^{2}, y)

sends

0

0

yet fails additivity, so passing this one test proves nothing on its own.

The kernel of

T : V \to W

is a subset of

W

False. The kernel is a set of inputs that map to zero, so

ker T \subseteq V

. Only the output value

0_{W}

lives in

W

The image of

T : V \to W

is a subspace of

V

False. Outputs live in

W

, so

im T \subseteq W

. It is a subspace of the codomain, not the domain.

Both the kernel and image always contain the zero vector (of their respective spaces).

True.

T (0_{V}) = 0_{W}

shows

0_{V} \in ker T

and

0_{W} \in im T

; every subspace must contain its zero.

A linear map can send two different nonzero vectors to the same output.

True, precisely when the kernel is nontrivial: if

T (u) = T (v)

then

u - v \in ker T

, so a nonzero kernel produces genuine collisions.

ker T = {0}

then

T

is surjective.

False. Trivial kernel gives injectivity, not surjectivity. Concretely

T (x, y) = (x, y, 0)

from

R^{2} \to R^{3}

has

ker T = {0}

(injective) but its image is only the flat

x y

-plane, so it misses all of

W

with

z \neq = 0

For

T (x) = A x

, the image is the row space of

A

False. Since

A x = \sum x_{i} (column_{i})

, the image is the column space of

A

, not the row space.

Rank–nullity says

dim V = dim W

minus the nullity.

False. It says

dim V = nullity + rank

, both measured relative to the domain

V

(and requires

V

finite-dimensional);

dim W

never appears in the formula.

An affine map

T (x) = A x + b

with

b \neq = 0

is linear.

False.

T (0) = b \neq = 0

violates

T (0) = 0

, so it is affine (a linear map followed by a shift), not linear.

Spot the error

ker T

is a subspace because it contains

0

Containing

0

is necessary but not the reason. A subspace must be closed under addition and scaling:

T (u) = T (v) = 0 \Rightarrow T (c u + d v) = 0

. Containing

0

alone (e.g. a single point) is not enough.

T

is injective, so its kernel could contain a nonzero vector as long as it's unique."

Wrong. Any nonzero

v \in ker T

gives

T (v) = T (0) = 0

with

v \neq = 0

— two inputs, one output — so

T

is not injective. Injective forces

ker T = {0}

exactly.

dim (im T)

can exceed

dim V

Impossible. Rank

= dim V - nullity \leq dim V

, since nullity

\geq 0

. A map can never produce more independent outputs than it had input dimensions.

T (x, y) = (x + 1, y)

is linear because its graph is straight."

A straight graph is not linearity. Plug in the origin:

T (0, 0) = (1, 0) \neq = (0, 0)

, so the whole picture is shifted off the origin by the

+ 1

; it is affine, not linear.

"Because the columns of

A

span the image, the number of columns equals the rank."

No. The rank is the number of independent columns. Extra dependent columns inflate the count without enlarging the span, so rank

\leq

number of columns.

T

maps a

3

-dim space onto a

4

-dim space, so it can be surjective."

Impossible. Rank

\leq dim V = 3 < 4

, so the image can have at most dimension

3

and cannot fill a

4

-dimensional codomain.

Why questions

Why is

T

completely determined by its values on a basis?

Any

x = \sum x_{i} e_{i}

, and linearity gives

T (x) = \sum x_{i} T (e_{i})

. Fixing the

T (e_{i})

fixes every output, which is why stacking them as columns yields the matrix.

Why does a nonzero kernel mean the input cannot be recovered?

Two inputs differing by a kernel element share an output (

T (u) = T (u + k)

when

k \in ker T

), so from a given output you cannot tell which of them you started with — the map is not invertible on those directions.

Why must additivity and homogeneity be checked, not just one?

They control two independent operations (adding, scaling), so one can hold while the other fails. Concrete example:

T (x, y) = (∣ x ∣, y)

obeys homogeneity only for

c \geq 0

and breaks additivity since

∣ - 1 + 1∣ = 0 \neq = ∣ - 1∣ + ∣1∣ = 2

— checking one property would have wrongly "passed" it.

Why is the image called the column space for matrix maps?

Because

A x

is a weighted sum of

A

's columns, so every reachable output is a linear combination of columns — exactly the span of the columns.

Why does rank–nullity hold no matter which basis you pick?

The proof extends a basis of

ker T

to a basis of

V

; the counts

p

(nullity) and

q

(rank) are dimensions, which are basis-independent, so the identity

dim V = p + q

is intrinsic.

Why can't the image ever bend away from the origin?

The image is a subspace: it is closed under scaling and addition and contains

0_{W}

, so it must be a flat set passing through the origin — a line, plane, or higher flat, never a shifted or curved region.

Edge cases

What is the kernel of the zero map

T (v) = 0

for all

v

The entire domain:

ker T = V

, nullity

= dim V

, and rank

= 0

since the image is just

{0}

. Rank–nullity still holds:

dim V = dim V + 0

What is the kernel of the identity map

T (v) = v

Only

{0}

, since

v = 0

is the sole solution of

T (v) = 0

. Nullity

= 0

, rank

= dim V

, and the map is bijective.

dim V < dim W

, can

T

ever be surjective?

No. Rank

\leq dim V < dim W

, so the image cannot fill

W

; such a map can be injective but never onto.

dim V > dim W

, must

T

have a nontrivial kernel?

Yes. Nullity

= dim V - rank \geq dim V - dim W > 0

, so at least one nonzero vector is crushed to

0

;

T

cannot be injective.

For a linear

T : V \to V

on a finite-dimensional space, is injective equivalent to surjective?

Yes. Rank–nullity forces nullity

= 0 ⟺

rank

= dim V

, so trivial kernel and full image occur together — injective, surjective and bijective coincide.

Can a linear map from

R^{n}

R^{m}

have an empty image?

No. The image always contains

0_{W} = T (0_{V})

, so it is at minimum

{0}

and never empty; "smallest possible" image is the zero subspace.

Is a subspace of dimension

0

still a valid kernel or image?

Yes. The zero subspace

{0}

has dimension

0

and is a legitimate kernel (injective maps) or image (the zero map). Dimension zero is allowed, not degenerate-invalid.

Connections

Injective surjective bijective — the kernel/image tests here decide exactly these three properties.
Rank–Nullity Theorem — the counting law behind every "edge case" above.
Column space and null space — the matrix names for image and kernel.
Change of basis — why rank and nullity are basis-independent.