Image & Kernel: Injectivity and Surjectivity

The Image

The image (or range) of a linear transformation

T: V \to W

is the set of all outputs:

Image Definition

\text{Im}(T) = \{T(\mathbf{v}) : \mathbf{v} \in V\} \subseteq W

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THEOREM

The image is a subspace of

W

.

It contains

T(\mathbf{0}) = \mathbf{0}

, and if

T(\mathbf{u})

and

T(\mathbf{v})

are in the image, then so is

cT(\mathbf{u}) + dT(\mathbf{v}) = T(c\mathbf{u} + d\mathbf{v})

— closure under both operations follows from linearity.

When

T(\mathbf{x}) = A\mathbf{x}

for a matrix

A

, the image is the column space of

A

: the set of all vectors expressible as linear combinations of the columns. The dimension of the image equals the rank of

A

.

The image answers the reachability question: a vector

\mathbf{w} \in W

is in the image if and only if the equation

T(\mathbf{v}) = \mathbf{w}

— equivalently,

A\mathbf{x} = \mathbf{w}

— has a solution.

A map reaching the whole plane

The output fills the plane, so the image is everything and the map is onto. The image is spanned by where the basis vectors land, which is why it is the column space of the matrix by another name. Shrink the image by making the columns dependent on the kernel and image explorer.

Whether the image fills the codomain is precisely the question of surjectivity dealt with below.

The Kernel

The kernel (or null space) of

T: V \to W

is the set of all inputs that map to zero:

Kernel Definition

\ker(T) = \{\mathbf{v} \in V : T(\mathbf{v}) = \mathbf{0}\} \subseteq V

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THEOREM

The kernel is a subspace of

V

.

It contains

\mathbf{0}

(since

T(\mathbf{0}) = \mathbf{0}

), and if

T(\mathbf{u}) = \mathbf{0}

and

T(\mathbf{v}) = \mathbf{0}

, then

T(c\mathbf{u} + d\mathbf{v}) = cT(\mathbf{u}) + dT(\mathbf{v}) = \mathbf{0}

, so

c\mathbf{u} + d\mathbf{v} \in \ker(T)

.

When

T(\mathbf{x}) = A\mathbf{x}

, the kernel is the null space of

A

: all solutions to the homogeneous system

A\mathbf{x} = \mathbf{0}

. Its dimension is the nullity, equal to

n - \text{rank}(A)

.

Pitfall: Placing the kernel in the codomain

Reading the kernel as the zero output gives

\ker(T) = \{(0, 0)\}

for

T(x, y, z) = (x, y)

, so

T

looks injective. The kernel is a set of inputs in

\mathbb{R}^3

: solving

T(\mathbf{v}) = \mathbf{0}

gives the line

\text{Span}\{(0, 0, 1)\}

.

The kernel measures the information lost by

T

. Vectors in the kernel are collapsed to

\mathbf{0}

— they represent directions that the transformation annihilates. A larger kernel means more information is destroyed.

A line crushed to the origin

The image has collapsed to a line, and a whole line of inputs is now being sent to the origin — that line is the kernel. The two losses are the same loss counted twice: one dimension gone from the image is one dimension gained by the kernel, which is rank-nullity in its shortest form. Watch both change together on the kernel and image explorer.

A kernel containing only the origin is what injectivity means, which is why the two properties are so often checked together.

Kernel and Image Notation

Notation

Kernel and Image Notation

A lowercase German survivor next to a capitalized rival with a double life, the colon that declares a map, and the dictionary between map-speak and matrix-speak.

\to

and

\mapsto

— the four-subspace notation; bold vectors — vector notation.

\ker(T)

·

\text{Im}(T)

The kernel and image of T

Both defined above — the notational quirk is the mismatched capitalization, faithfully mirroring the literature: lowercase

\ker

, from the German Kern of Emmy Noether's school, against a usually-capitalized

\text{Im}

.

CasesThe capital on

\text{Im}

is partly self-defense: lowercase

\operatorname{im}

exists, but

\text{Im}

has a double life —

\text{Im}(z)

is the imaginary part of a complex number, and in complex vector spaces both readings can occur in one paragraph.

Also written

\operatorname{im}(T)

,

\operatorname{ran}(T)

, or the word “range” — with range itself ambiguous, as the

R(A)

wars at the four subspaces attest.

Do not confuseEach other's habitat.

\ker(T) \subseteq V

lives in the domain,

\text{Im}(T) \subseteq W

in the codomain — swapping the homes is the commonest conceptual slip the notation guards against.

T: V \to W

T, from V to W

The signature: colon binds the name to its spaces, the arrow points from domain to codomain. It declares where the map operates before any formula says what it does.

CasesLinear maps inherit a matrix privilege:

T\mathbf{v}

without parentheses, written like the product

A\mathbf{x}

it secretly is. General functions keep their parentheses; linearity is what licenses the juxtaposition.

Also written

V \xrightarrow{\;T\;} W

— the name riding on the arrow, standard in diagram-heavy texts.

Do not confuseThe such-that colon. In

\{\mathbf{v} : T(\mathbf{v}) = \mathbf{0}\}

the colon means “such that” — the kernel's own definition above uses it — while in

T: V \to W

it binds a name. Same mark, adjacent lines, different grammar.

\text{Im}(T) = \text{Col}(A)

,

\; \ker(T) = \text{Null}(A)

Image is column space; kernel is null space

The dictionary between the two dialects: map-speak (

\ker

,

\text{Im}

) and matrix-speak (

\text{Null}

,

\text{Col}

) name identical subspaces whenever

T(\mathbf{x}) = A\mathbf{x}

— the translation both definitions above perform in their closing paragraphs.

CasesThe dimensions translate too: rank–nullity reads

\dim \text{Im}(T) + \dim \ker(T) = \dim V

in map dialect and

\text{rank}(A) + \text{nullity}(A) = n

in matrix dialect — as in The Rank-Nullity Theorem for Maps below. One theorem, two vocabularies.

Also written“Nullity” — the named dimension of the kernel; the rare case where a dimension got its own word instead of a

\dim

expression.

Do not confuseTwo different theories. Nothing differs but vocabulary — a reader fluent in one dialect and not the other misses that every fact transfers verbatim.

All linear algebra symbols →↑ Transformations

Injectivity

A linear transformation

T

is injective (one-to-one) if different inputs always produce different outputs:

T(\mathbf{u}) = T(\mathbf{v})

implies

\mathbf{u} = \mathbf{v}

.

For linear maps, injectivity has an elegant equivalent:

Injectivity Kernel Criterion

T \text{ injective} \iff \ker(T) = \{\mathbf{0}\}

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The proof is short. If

T(\mathbf{u}) = T(\mathbf{v})

, then

T(\mathbf{u} - \mathbf{v}) = T(\mathbf{u}) - T(\mathbf{v}) = \mathbf{0}

, so

\mathbf{u} - \mathbf{v} \in \ker(T)

. If the kernel is trivial,

\mathbf{u} - \mathbf{v} = \mathbf{0}

and

\mathbf{u} = \mathbf{v}

.

THEOREM

For matrix transformations, injectivity is equivalent to full column rank:

\text{rank}(A) = n

.

This means every column is a pivot column, no free variables exist in

A\mathbf{x} = \mathbf{0}

, the columns are linearly independent, and the determinant is nonzero (in the square case).

Injectivity means the transformation preserves distinctness — no two different inputs are confused with each other.

Surjectivity

A linear transformation

T: V \to W

is surjective (onto) if

\text{Im}(T) = W

— every vector in the codomain is the image of some vector in the domain.

THEOREM

For matrix transformations, surjectivity is equivalent to full row rank:

\text{rank}(A) = m

.

This means every row contains a pivot, the column space is all of

\mathbb{R}^m

, and the system

A\mathbf{x} = \mathbf{b}

has a solution for every right-hand side

\mathbf{b}

.

Surjectivity means the transformation has no blind spots — every output is reachable from some input. Failure of surjectivity means the image is a proper subspace of the codomain: certain vectors in

W

are inherently unreachable, no matter what input is chosen.

Equivalent condition	Injectivity (one-to-one)	Surjectivity (onto)
Map-level definition	T(u) = T(v) ⇒ u = v	every w ∈ W is the image of some v
Kernel / image	ker(T) = {0}	Im(T) = W
Rank condition	rank(A) = n (full column rank)	rank(A) = m (full row rank)
Columns / rows of A	columns are linearly independent	columns span ℝᵐ (column space = ℝᵐ)
System Ax = b behavior	at most one solution for every b	at least one solution for every b
Echelon-form trait	no free variables in Ax = 0	every row contains a pivot
Determinant (square A only)	det(A) ≠ 0	det(A) ≠ 0 (collapses to the same test)

Bijectivity and Isomorphisms

A linear transformation that is both injective and surjective is bijective. A bijective linear transformation is called an isomorphism — it establishes that the domain and codomain are structurally identical as vector spaces.

For a map

T: V \to W

between spaces of equal dimension, the three conditions collapse:

Bijectivity Equal Dim Case

\dim V = \dim W \Rightarrow \bigl(T \text{ injective} \iff T \text{ surjective} \iff T \text{ bijective}\bigr)

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Pitfall: Taking equal dimensions as bijective

Calling

A = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}

an isomorphism because it maps

\mathbb{R}^2

to

\mathbb{R}^2

fails:

A(-2, 1) = \mathbf{0}

, so it is neither injective nor surjective. Its kernel is the line

\text{Span}\{(-2, 1)\}

and its image only the line

\text{Span}\{(1, 2)\}

.

Checking any one of the three establishes the other two. This is because the rank-nullity theorem forces

\dim(\text{Im}(T)) + \dim(\ker(T)) = n

(where

n = \dim(V) = \dim(W)

), and

\dim(\text{Im}(T)) \leq n

. If the kernel is trivial (injective), the image has dimension

n

and must equal all of

W

(surjective). If the image is all of

W

(surjective), the kernel must have dimension

0

(injective).

For matrix transformations between spaces of the same dimension, bijectivity is equivalent to the matrix being square and invertible.

Between finite-dimensional spaces of the same dimension, one-to-one and onto stand or fall together, so checking either one settles both.

The Rank-Nullity Theorem for Maps

For a linear transformation

T: V \to W

with

V

finite-dimensional:

Rank-Nullity for Maps

\dim\text{Im}(T) + \dim\ker(T) = \dim V

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The domain dimensions split between what the map preserves and what it destroys. The image captures the dimensions that survive; the kernel captures the dimensions that are annihilated.

For matrix transformations

T(\mathbf{x}) = A\mathbf{x}

, this becomes

\text{rank}(A) + \text{nullity}(A) = n

— the familiar rank-nullity theorem in concrete language.

Pitfall: Counting the codomain in rank-nullity

Subtracting the rank from the number of rows gives, for a

3 \times 5

matrix of rank

3

, nullity

3 - 3 = 0

, as if

A\mathbf{x} = \mathbf{0}

had only the zero solution. The count runs over the

n = 5

columns:

\text{nullity}(A) = 5 - 3 = 2

.

The theorem constrains the interplay between injectivity and surjectivity. If

\dim(V) > \dim(W)

, the image can have at most

\dim(W)

dimensions, forcing the kernel to have at least

\dim(V) - \dim(W)

dimensions — the map cannot be injective. If

\dim(V) < \dim(W)

, the image cannot fill all of

W

— the map cannot be surjective.

Dimension Constraints

The rank-nullity theorem imposes hard limits on what a linear transformation can achieve.

T: V \to W

can be injective only if

\dim(V) \leq \dim(W)

. A map from a larger space to a smaller one must collapse some directions — the kernel is forced to be nontrivial.

T: V \to W

can be surjective only if

\dim(V) \geq \dim(W)

. A map from a smaller space to a larger one cannot cover all directions — the image is a proper subspace.

T

can be bijective only if

\dim(V) = \dim(W)

. This is necessary but not sufficient — even with equal dimensions, the map must still have full rank.

These constraints apply to all linear maps, not just matrix transformations. They are consequences of the rank-nullity theorem and the dimension theory of vector spaces.

Dimension relation	Injective possible?	Surjective possible?	Bijective possible?	Forced consequence
dim(V) < dim(W)	✓	✗	✗	image is a proper subspace of W
dim(V) = dim(W)	✓	✓	✓	injective ⇔ surjective ⇔ bijective (one check establishes all three)
dim(V) > dim(W)	✗	✓	✗	kernel forced to have dim ≥ dim(V) − dim(W)

Comparing dimensions can rule out injectivity or surjectivity, but it can never guarantee either.

Computing the Image and Kernel

For a matrix transformation

T(\mathbf{x}) = A\mathbf{x}

, the image and kernel are computed by row reduction.

The kernel is the null space of

A

: solve

A\mathbf{x} = \mathbf{0}

, reduce to echelon form, and express the solution in parametric vector form. Each free variable contributes one basis vector for

\ker(T)

.

The image is the column space of

A

: row reduce

A

, identify the pivot columns, and take the corresponding columns of the original matrix

A

as a basis for

\text{Im}(T)

.

Worked Example

For

A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 1 & 3 & 4 \end{pmatrix}

, row reduction gives

\begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix}

. Pivots in columns

1

and

2

. The image has basis

\{(1, 0, 1), (2, 1, 3)\}

— two-dimensional. The kernel has one free variable (

x_3 = t

), giving

\ker(T) = \text{Span}\{(-1, -1, 1)\}

— one-dimensional. Check:

2 + 1 = 3 = n

.

Pitfall: Using pivot columns of the echelon form

Taking the pivot columns of the reduced matrix gives

\{(1, 0, 0), (2, 1, 0)\}

, the plane

x_3 = 0

, which contains

(1, 0, 0)

, a vector

A

never outputs. The image is the plane

x_3 = x_1 + x_2

spanned by

(1, 0, 1)

and

(2, 1, 3)

.

Both answers come from one reduction, but they are read off in different places. The kernel is a set of inputs, so its basis is built from the solution of the reduced system. The image is a set of outputs, so the echelon form only says which columns to keep, and the vectors themselves are the columns that

A

produces. Here the third column is redundant, since

(3, 1, 4) = (1, 0, 1) + (2, 1, 3)

.

The Fundamental Decomposition

The rank-nullity theorem has a structural interpretation that goes beyond dimension counting. The domain

V

decomposes as a direct sum:

The Fundamental Decomposition

V = \ker(T) \oplus (\text{a complement of } \ker(T))

THEOREM

The transformation

T

kills everything in the kernel and maps the complement bijectively onto the image.

Every vector

\mathbf{v} \in V

splits as

\mathbf{v} = \mathbf{v}_k + \mathbf{v}_c

where

\mathbf{v}_k \in \ker(T)

and

\mathbf{v}_c

is in the complement. Then

T(\mathbf{v}) = T(\mathbf{v}_c)

, and the restriction of

T

to the complement is a bijection onto

\text{Im}(T)

.

Fact: The

+ C

of calculus is a kernel

On functions defined on an interval, differentiation is linear, and its kernel is the constant functions, since only constants have derivative

0

. So two antiderivatives of the same

f

differ by a kernel element, and the

+ C

in

\int f\,dx = F + C

is that kernel added to one answer.

For matrix transformations, the four fundamental subspaces provide the natural complement: the row space of

A

is the orthogonal complement of the null space in

\mathbb{R}^n

, and

A

maps the row space bijectively onto the column space. The null-space component is destroyed; the row-space component survives intact.

The same split governs equations. If

T(\mathbf{v}_0) = \mathbf{w}

, then every

\mathbf{v}_0 + \mathbf{k}

with

\mathbf{k} \in \ker(T)

is also a solution of

T(\mathbf{v}) = \mathbf{w}

, and any two solutions differ by a vector of the kernel. The complete solution set is therefore one particular solution shifted by the whole kernel: a single point when the kernel is trivial, a line, a plane or a larger flat piece when it is not.

Summary: Image and Kernel Side by Side

The whole page revolves around a single duality: the image lives in the codomain and records what T reaches, while the kernel lives in the domain and records what T destroys. Every concept introduced — injectivity, surjectivity, bijectivity, rank-nullity, dimension constraints, the fundamental decomposition — is a relation between these two subspaces. The table below sets them side by side across every attribute the page has touched, so the symmetry and the link supplied by rank-nullity are visible at a glance.

Setting the two side by side invites reading them as mirror images, and for the first few attributes they very nearly are. They are not, and the places where the symmetry fails carry more information than the places where it holds — so the entries below are worth reading in pairs, with attention to which lines match and which reverse.

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Transformations · image and kernel

Image and kernel, side by side

Two subspaces that live in different spaces and answer opposite questions. Every line below states the same attribute for both, and reading them in pairs is what makes the symmetry — and where it breaks — visible.

7attributes

What each one is3

1

Definition§ 1

outputs produced vs inputs destroyed

\operatorname{im}(T) = \{T(\mathbf{v})\}

vs

\ker(T) = \{\mathbf{v} : T(\mathbf{v}) = \mathbf{0}\}

The image collects what comes out; the kernel collects what goes in and vanishes. Both are subspaces rather than arbitrary sets — which is what makes dimension the right way to measure them.

2

Where it lives§ 1

codomain vs domain

\operatorname{im}(T) \subseteq W

,

\ker(T) \subseteq V

They sit in different spaces, so they cannot be compared directly, intersected, or added. This is the asymmetry the side-by-side layout most easily hides — the two columns are not two views of one object.

3

Matrix counterpart§ 8

for

T(\mathbf{x}) = A\mathbf{x}

\operatorname{Col}(A)

vs

\operatorname{Null}(A)

The column space and null space under different names. Everything on this page is therefore computable by row reduction, and the abstract statement and the matrix statement never diverge.

How each is computed and measured2

4

Dimension§ 6

summing to

n = \dim V

\operatorname{rank}(A)

vs

n - \operatorname{rank}(A)

Rank-nullity. The two dimensions are not independently chosen — fixing one fixes the other, because every dimension of the domain either survives or collapses. Note the sum is

\dim V

, not

\dim W

: the theorem is about the domain.

5

How to compute it§ 8

both from one reduction

pivot columns of the original

A

vs RREF parametric solution

The image basis is taken from

A

itself — row operations change the column space — while the kernel basis comes from the reduced form. Opposite rules from the same computation, and the pair most often confused.

What the extreme cases mean2

6

Trivial§ 3

T

is the zero map vs

T

is injective

\operatorname{im}(T) = \{\mathbf{0}\}

vs

\ker(T) = \{\mathbf{0}\}

Here the symmetry breaks hardest. A trivial image means the map does nothing at all; a trivial kernel means it loses nothing — the strongest possible statement in the other direction. Same word, opposite significance.

7

Full§ 4

T

is surjective vs

T

is the zero map

\operatorname{im}(T) = W

vs

\ker(T) = V

And again reversed. A full image is surjectivity, the ideal end; a full kernel is the zero map, the degenerate end. Read with the row above, each column runs from degenerate to ideal, but in opposite directions.

The pairing is not decorative: rank-nullity ties the two dimensions together, so the subspaces are not independent objects that happen to be worth comparing. Every direction of the domain is accounted for exactly once — either it survives into the image or it collapses into the kernel.

Image and kernel, side by side·/linear-algebra/transformations/image-kernelLearn Math Class

The clearest break is at the extremes. A trivial kernel is the best case — the map destroys nothing and is injective — while a trivial image is the worst, since the map sends everything to zero. Reverse them and the same reversal happens: a full image means surjectivity, a full kernel means the zero map again. The kernel column runs from ideal to degenerate; the image column runs the other way.

What ties them together is dimension rather than structure. Rank-nullity says the two dimensions sum to

\dim V

, so neither can be chosen independently of the other — every direction in the domain is accounted for exactly once, surviving into the image or collapsing into the kernel. Note the sum is over the domain: the codomain plays no part, which is why a map can have a small image without having a large kernel.

Common Pitfalls

Four mistakes account for many wrong answers about image and kernel. Each is marked with a box where it happens in the sections above; here each is worked through in full, with the wrong and the right working side by side.

Placing the kernel in the codomain

The kernel is treated as a set of outputs, usually as the zero output itself, and written as a subspace of

W

. It is a set of inputs: it lives in

V

, and its vectors have as many entries as

A

has columns.

Wrong working: for

T(x, y, z) = (x, y)

, with matrix

A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}

, the kernel is described as the zero output,

\ker(T) = \{(0, 0)\} \subseteq \mathbb{R}^2

. The kernel looks trivial, so

T

is declared injective.

Right working: solve

A\mathbf{x} = \mathbf{0}

. The equations are

x = 0

and

y = 0

, and

z

is free, so

\ker(T) = \text{Span}\{(0, 0, 1)\} \subseteq \mathbb{R}^3

a line, one-dimensional.

T

is not injective:

T(0, 0, 5) = T(0, 0, 0) = (0, 0)

.

Why it happens: the definition contains

\mathbf{0}

, the zero vector of

W

, and it is easy to read the kernel as where the zero is rather than as what is sent to it. The equation

A\mathbf{x} = \mathbf{0}

also has the zero with

m

entries on its right-hand side.

How to check: count entries. Kernel vectors have

n

entries, one per column of

A

; image vectors have

m

entries, one per row. Rank-nullity agrees: the rank is

2

and

n = 3

, so the nullity must be

1

, not

0

.

Taking equal dimensions as bijective

The equal-dimension rule is read as: if

\dim V = \dim W

, then

T

is bijective. The rule says only that the three properties then stand or fall together.

Wrong working:

A = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}

maps

\mathbb{R}^2

to

\mathbb{R}^2

. The dimensions are equal, so

T

is declared an isomorphism and its inverse is looked for.

Right working: test one of the three properties.

A\begin{pmatrix} -2 \\ 1 \end{pmatrix} = \begin{pmatrix} -2 + 2 \\ -4 + 4 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}

So

\ker(T) = \text{Span}\{(-2, 1)\}

and

T

is not injective. By the equal-dimension rule it is not surjective either: the image is only the line

\text{Span}\{(1, 2)\}

, and

(1, 0)

is never reached. Also

\det A = 4 - 4 = 0

, so there is no inverse.

Why it happens: the formula puts the word bijective right next to

\dim V = \dim W

, and the dimension test of the Dimension Constraints section is easy to remember as a sufficient condition when it is only a necessary one.

How to check: a dimension count can only rule a property out. To confirm bijectivity, one real test is still needed: a trivial kernel, full rank, or, for a square matrix,

\det A \neq 0

.

Counting the codomain in rank-nullity

The nullity is computed as

\dim W - \text{rank}

, that is

m - \text{rank}(A)

, using the number of rows instead of the number of columns.

Wrong working: take the

3 \times 5

matrix

A = \begin{pmatrix} 1 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 & 1 \end{pmatrix}

of rank

3

, a map

T: \mathbb{R}^5 \to \mathbb{R}^3

. Then nullity

= 3 - 3 = 0

, so

A\mathbf{x} = \mathbf{0}

would have only the zero solution and

T

would be injective.

Right working:

\text{nullity}(A) = n - \text{rank}(A) = 5 - 3 = 2

. The matrix is already in reduced form, with

x_4

and

x_5

free, and the kernel has the basis

(-1, 0, -1, 1, 0), \qquad (0, -1, -1, 0, 1)

Why it happens:

m

and

n

are both sizes of

A

, and the rank is bounded by both. The image also sits in the codomain, so it is tempting to think the codomain is the space being split. The theorem splits the domain.

How to check: count the columns without a pivot; that is the number of free variables,

n - \text{rank}(A)

. And when

\dim V > \dim W

, the kernel can never be trivial, so a nullity of

0

is impossible.

Using pivot columns of the echelon form

For the image, the pivot columns are taken from the echelon form instead of from

A

. Row operations keep the null space and the relations between the columns, but they change the column space.

Wrong working: in the worked example the echelon form is

\begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix}

, with pivots in columns

1

and

2

. Its pivot columns give the basis

\{(1, 0, 0), (2, 1, 0)\}

, which spans the plane

x_3 = 0

.

Right working: take columns

1

and

2

of the original

A

,

\{(1, 0, 1), (2, 1, 3)\}

. They span the plane

x_3 = x_1 + x_2

, and all three columns of

A

lie in it. The vector

(1, 0, 0)

does not: the system

A\mathbf{x} = (1, 0, 0)

has no solution, since row

3

minus rows

1

and

2

gives

0 = -1

.

Why it happens: the kernel basis really does come from the reduced form, so the same habit is carried over to the image, and the echelon form is the matrix already on the page.

How to check: test each proposed basis vector

\mathbf{w}

by solving

A\mathbf{x} = \mathbf{w}

. A quicker sign: when the echelon form has a row of zeros, all its columns have a

0

in that position, while the columns of

A

usually do not.

Image and Kernel FAQ

Does the kernel live in the domain or the codomain?

+

The kernel sits in the domain, since it collects inputs that get sent to zero, while the image sits in the codomain, collecting outputs actually reached. Swapping their homes is the commonest conceptual slip in the topic. The notation guards against it:

\ker(T) \subseteq V

and

\text{Im}(T) \subseteq W

.Read more →

Why is $\text{Im}$ capitalised?

+

Partly self-defence. Lowercase

\operatorname{im}

does exist, but

\text{Im}

leads a double life as the imaginary part of a complex number, and in a complex vector space both meanings can appear within a paragraph. The capital keeps them visually apart where confusing them would be easy and costly.Read more →

Why can you write $T\mathbf{v}$ without parentheses?

+

Linearity licenses it. Because

T

behaves exactly like the matrix product it secretly is, the juxtaposition of

A\mathbf{x}

carries over. General functions keep their parentheses. Watch the colon too, which does two jobs on adjacent lines: in

T : V \to W

it binds a name, while in

\{\mathbf{v} : T(\mathbf{v}) = \mathbf{0}\}

it means "such that".Read more →

Image and Kernel of Linear Transformation

What a Transformation Hits and What It Kills

The Image

The Kernel

Kernel and Image Notation

Kernel and Image Notation

Injectivity

Surjectivity

Bijectivity and Isomorphisms

The Rank-Nullity Theorem for Maps

Dimension Constraints

Computing the Image and Kernel

The Fundamental Decomposition

Summary: Image and Kernel Side by Side

Image and kernel, side by side

Common Pitfalls

Image and Kernel FAQ

Does the kernel live in the domain or the codomain?

Why is $\text{Im}$ capitalised?

Why can you write $T\mathbf{v}$ without parentheses?

Image and Kernel of Linear Transformation

What a Transformation Hits and What It Kills

The Image

The Kernel

Kernel and Image Notation

Kernel and Image Notation

Injectivity

Surjectivity

Bijectivity and Isomorphisms

The Rank-Nullity Theorem for Maps

Dimension Constraints

Computing the Image and Kernel

The Fundamental Decomposition

Summary: Image and Kernel Side by Side

Image and kernel, side by side

Common Pitfalls

Image and Kernel FAQ

Does the kernel live in the domain or the codomain?

Why is Im\text{Im}Im capitalised?

Why can you write TvT\mathbf{v}Tv without parentheses?

Why is $\text{Im}$ capitalised?

Why can you write $T\mathbf{v}$ without parentheses?