LA-3.2

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Kernel & Image

The kernel tells us what a linear map annihilates; the image tells us what it can produce. Together, they completely characterize the map's behavior and reveal its fundamental structure.

3-4 hours Core Level 10 Objectives

Learning Objectives

Define the kernel and image of a linear map
Prove that kernel and image are always subspaces
Characterize injectivity through the kernel
Characterize surjectivity through the image
Compute kernels and images for specific linear maps
Understand the geometric meaning of kernel and image
Calculate the rank and nullity of linear maps
Apply kernel-image concepts to determine map properties
Prove fundamental properties of kernel and image
Relate kernel and image to matrix null space and column space

Prerequisites

Linear map definition (LA-3.1)
Vector spaces and subspaces (LA-2.2)
Span and linear combinations (LA-2.3)
Basis and dimension (LA-2.4)
Linear independence

Historical Context

The concepts of kernel and image emerged as mathematicians sought to understand the structure of linear transformations. The term "kernel" comes from the German "Kern" (core), reflecting the idea that the kernel captures the "core" of what gets collapsed to zero.

These concepts are central to the First Isomorphism Theorem, which states that the image of a linear map is isomorphic to the quotient of the domain by the kernel: $V/\ker(T) \cong \text{im}(T)$ .

1. Definitions

Every linear map $T: V \to W$ has two fundamental subspaces associated with it: the kernel (what gets sent to zero) and the image (what can be reached).

Definition 3.3: Kernel (Null Space)

The kernel (or null space) of a linear map $T: V \to W$ is the set of all vectors in $V$ that $T$ maps to zero:

\ker(T) = \text{null}(T) = \{v \in V : T(v) = 0_W\}

Definition 3.4: Image (Range)

The image (or range) of a linear map $T: V \to W$ is the set of all vectors in $W$ that are outputs of $T$ :

\text{im}(T) = \text{range}(T) = T(V) = \{T(v) : v \in V\}

Remark 3.5: Notation Variations

Different textbooks use different notation:

Kernel: $\ker(T)$ , $\text{null}(T)$ , $N(T)$
Image: $\text{im}(T)$ , $\text{range}(T)$ , $R(T)$ , $T(V)$

Example 3.9: Basic Examples

Zero map $0: V \to W$ :

$\ker(0) = V$ (everything maps to zero)
$\text{im}(0) = \{0_W\}$ (only zero is produced)

Identity map $I_V: V \to V$ :

$\ker(I_V) = \{0_V\}$ (only zero maps to zero)
$\text{im}(I_V) = V$ (everything is reachable)

2. Kernel and Image are Subspaces

Theorem 3.12: Kernel is a Subspace

For any linear map $T: V \to W$ , the kernel $\ker(T)$ is a subspace of $V$ .

Proof:

We verify the three subspace conditions:

Non-empty: $T(0_V) = 0_W$ , so $0_V \in \ker(T)$ .
Closed under addition: If $u, v \in \ker(T)$ , then $T(u + v) = T(u) + T(v) = 0 + 0 = 0$ , so $u + v \in \ker(T)$ .
Closed under scalar multiplication: If $v \in \ker(T)$ and $\alpha \in F$ , then $T(\alpha v) = \alpha T(v) = \alpha \cdot 0 = 0$ , so $\alpha v \in \ker(T)$ .

∎

Theorem 3.13: Image is a Subspace

For any linear map $T: V \to W$ , the image $\text{im}(T)$ is a subspace of $W$ .

Proof:

We verify the three subspace conditions:

Non-empty: $T(0_V) = 0_W$ , so $0_W \in \text{im}(T)$ .
Closed under addition: If $w_1, w_2 \in \text{im}(T)$ , then $w_1 = T(v_1)$ and $w_2 = T(v_2)$ for some $v_1, v_2 \in V$ . Thus $w_1 + w_2 = T(v_1) + T(v_2) = T(v_1 + v_2) \in \text{im}(T)$ .
Closed under scalar multiplication: If $w \in \text{im}(T)$ , then $w = T(v)$ for some $v \in V$ . Thus $\alpha w = \alpha T(v) = T(\alpha v) \in \text{im}(T)$ .

∎

Theorem 3.14: Image Equals Span of Basis Images

If $\{v_1, \ldots, v_n\}$ is a basis of $V$ , then:

\text{im}(T) = \text{span}\{T(v_1), \ldots, T(v_n)\}

Proof:

Any $w \in \text{im}(T)$ equals $T(v)$ for some $v = \sum \alpha_i v_i \in V$ . By linearity, $w = T(\sum \alpha_i v_i) = \sum \alpha_i T(v_i)$ , which is in the span.

∎

3. Characterizing Injectivity and Surjectivity

The kernel and image provide elegant characterizations of when a linear map is injective (one-to-one) or surjective (onto).

Theorem 3.15: Injectivity Characterization

A linear map $T: V \to W$ is injective if and only if $\ker(T) = \{0\}$ .

Proof:

(⇒) Suppose $T$ is injective. If $v \in \ker(T)$ , then $T(v) = 0 = T(0)$ . By injectivity, $v = 0$ . So $\ker(T) = \{0\}$ .

(⇐) Suppose $\ker(T) = \{0\}$ . If $T(u) = T(v)$ , then $T(u - v) = T(u) - T(v) = 0$ , so $u - v \in \ker(T) = \{0\}$ . Thus $u = v$ , and $T$ is injective.

∎

Theorem 3.16: Surjectivity Characterization

A linear map $T: V \to W$ is surjective if and only if $\text{im}(T) = W$ .

Proof:

This is essentially the definition of surjectivity: $T$ is surjective iff every element of $W$ is hit by some element of $V$ , which is exactly when $\text{im}(T) = W$ .

∎

Corollary 3.2: Bijectivity

$T$ is bijective (an isomorphism) iff $\ker(T) = \{0\}$ and $\text{im}(T) = W$ .

4. Computing Kernel and Image

Here we present systematic methods for computing these fundamental subspaces.

Example 3.10: Computing Kernel

Problem: Find $\ker(T)$ for $T: \mathbb{R}^3 \to \mathbb{R}^2$ defined by $T(x, y, z) = (x + y, 2z - x)$ .

Solution: We need $T(x, y, z) = (0, 0)$ :

\begin{cases} x + y = 0 \\ 2z - x = 0 \end{cases}

From the first equation: $y = -x$ . From the second: $z = x/2$ .

So $\ker(T) = \{(t, -t, t/2) : t \in \mathbb{R}\} = \text{span}\{(1, -1, 1/2)\}$ .

Example 3.11: Computing Image

Problem: Find $\text{im}(T)$ for the same $T$ .

Solution: Apply $T$ to the standard basis:

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

$\text{im}(T) = \text{span}\{(1,-1), (1,0), (0,2)\}$ . Since $(1,-1)$ and $(0,2)$ are linearly independent, $\text{im}(T) = \mathbb{R}^2$ .

Example 3.12: Projection Kernel and Image

For the projection $P: \mathbb{R}^3 \to \mathbb{R}^3$ defined by $P(x, y, z) = (x, y, 0)$ :

$\ker(P) = \{(0, 0, z) : z \in \mathbb{R}\}$ (the z-axis)
$\text{im}(P) = \{(x, y, 0) : x, y \in \mathbb{R}\}$ (the xy-plane)

Note: $\mathbb{R}^3 = \ker(P) \oplus \text{im}(P)$ .

Example 3.13: Differentiation Kernel and Image

For differentiation $D: P_3(\mathbb{R}) \to P_2(\mathbb{R})$ defined by $D(p) = p'$ :

$\ker(D) = \{c : c \in \mathbb{R}\}$ (constant polynomials, dimension 1)
$\text{im}(D) = P_2(\mathbb{R})$ (D is surjective)

5. Rank and Nullity

Definition 3.5: Rank and Nullity

For a linear map $T: V \to W$ :

The rank of $T$ is $\text{rank}(T) = \dim(\text{im}(T))$
The nullity of $T$ is $\text{nullity}(T) = \dim(\ker(T))$

Theorem 3.17: Rank-Nullity Theorem (Preview)

For a linear map $T: V \to W$ with $V$ finite-dimensional:

\dim(V) = \text{rank}(T) + \text{nullity}(T) = \dim(\text{im } T) + \dim(\ker T)

Remark 3.6: Intuition

The domain dimension "splits" into two parts: what gets collapsed to zero (nullity) and what survives as output (rank). This is the central dimension formula of linear algebra.

6. Common Mistakes

Mistake 1: Confusing Kernel and Image Spaces

$\ker(T) \subseteq V$ (in the domain), while $\text{im}(T) \subseteq W$ (in the codomain). They live in different spaces!

Mistake 2: Trivial Kernel Means Trivial Map

$\ker(T) = \{0\}$ means $T$ is injective, not trivial. The trivial (zero) map has $\ker(T) = V$ !

Mistake 3: Image Equals Codomain

$\text{im}(T)$ is often a proper subspace of $W$ . Only when $T$ is surjective do we have $\text{im}(T) = W$ .

Mistake 4: Forgetting 0 is Always in Both

$0 \in \ker(T)$ and $0 \in \text{im}(T)$ always. Neither kernel nor image is ever empty.

7. Key Takeaways

Kernel

$\ker(T) = \{v : T(v) = 0\}$ . Subspace of $V$ . $\ker(T) = \{0\}$ iff $T$ is injective.

Image

$\text{im}(T) = T(V)$ . Subspace of $W$ . $\text{im}(T) = W$ iff $T$ is surjective.

Dimension Formula

$\dim V = \text{rank}(T) + \text{nullity}(T)$ . The fundamental relationship.

Matrix Connection

For matrix $A$ : $\ker(A)$ = null space, $\text{im}(A)$ = column space.

8. Worked Examples

Example 1: Matrix Transformation

Problem: Let $T: \mathbb{R}^3 \to \mathbb{R}^2$ be defined by $T(x, y, z) = (x + 2y - z, 2x + 4y - 2z)$ . Find $\ker(T)$ and $\text{im}(T)$ .

Solution:

For the kernel, solve $T(x, y, z) = (0, 0)$ :

\begin{cases} x + 2y - z = 0 \\ 2x + 4y - 2z = 0 \end{cases}

The second equation is 2× the first, so we have one independent equation: $x + 2y - z = 0$ , i.e., $z = x + 2y$ .

\ker(T) = \{(x, y, x + 2y) : x, y \in \mathbb{R}\} = \text{span}\{(1, 0, 1), (0, 1, 2)\}

For the image, note $T(x,y,z) = (x + 2y - z)(1, 2)$ , so $\text{im}(T) = \text{span}\{(1, 2)\}$ .

Verification: $\dim(\ker T) + \dim(\text{im } T) = 2 + 1 = 3 = \dim \mathbb{R}^3$ ✓

Example 2: Polynomial Space

Problem: Let $T: P_2 \to P_2$ be defined by $T(p)(x) = p(x) - p(0)$ . Find $\ker(T)$ and $\text{im}(T)$ .

Solution:

For $p(x) = a + bx + cx^2$ : $T(p)(x) = (a + bx + cx^2) - a = bx + cx^2$

$\ker(T) = \{p : T(p) = 0\} = \{p : p(x) = p(0)\}$ = constant polynomials = $\text{span}\{1\}$

$\text{im}(T) = \{bx + cx^2 : b, c \in \mathbb{R}\} = \text{span}\{x, x^2\}$

Verification: $1 + 2 = 3 = \dim P_2$ ✓

Example 3: Trace Map

Problem: Let $\text{tr}: M_n(\mathbb{R}) \to \mathbb{R}$ be the trace map. Find its kernel and image.

Solution:

$\ker(\text{tr}) = \{A \in M_n : \text{tr}(A) = 0\}$ = traceless matrices

Dimension: $n^2 - 1$ (one constraint on $n^2$ entries)

$\text{im}(\text{tr}) = \mathbb{R}$ since $\text{tr}(cI) = cn$ for any $c$ .

Verification: $(n^2 - 1) + 1 = n^2 = \dim M_n$ ✓

Example 4: Determining Injectivity

Problem: Is $T: \mathbb{R}^2 \to \mathbb{R}^3$ defined by $T(x, y) = (x + y, x - y, 2x)$ injective?

Solution: Check if $\ker(T) = \{0\}$ .

Solve $T(x, y) = (0, 0, 0)$ :

\begin{cases} x + y = 0 \\ x - y = 0 \\ 2x = 0 \end{cases}

From equations 1 and 2: $x = 0$ and $y = 0$ .

So $\ker(T) = \{(0, 0)\}$ , and $T$ is injective.

Example 5: Determining Surjectivity

Problem: Is $T: \mathbb{R}^3 \to \mathbb{R}^2$ defined by $T(x, y, z) = (x - y, y + z)$ surjective?

Solution: Check if $\text{im}(T) = \mathbb{R}^2$ .

Compute $T$ on the standard basis:

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

$\text{im}(T) = \text{span}\{(1,0), (-1,1), (0,1)\}$

Since $(1,0)$ and $(0,1)$ are in the span, $\text{im}(T) = \mathbb{R}^2$ .

$T$ is surjective.

9. Advanced Topics

Theorem 3.18: Preimage of a Subspace

Let $T: V \to W$ be linear and $U \leq W$ be a subspace. Then the preimage $T^{-1}(U) = \{v \in V : T(v) \in U\}$ is a subspace of $V$ .

Proof:

$T(0) = 0 \in U$ , so $0 \in T^{-1}(U)$
If $v_1, v_2 \in T^{-1}(U)$ , then $T(v_1), T(v_2) \in U$ , so $T(v_1 + v_2) = T(v_1) + T(v_2) \in U$
If $v \in T^{-1}(U)$ , then $T(\alpha v) = \alpha T(v) \in U$

∎

Theorem 3.19: Image of a Subspace

Let $T: V \to W$ be linear and $U \leq V$ be a subspace. Then $T(U) = \{T(u) : u \in U\}$ is a subspace of $W$ .

Theorem 3.20: Composition and Kernel/Image

For linear maps $S: W \to U$ and $T: V \to W$ :

$\ker(T) \subseteq \ker(S \circ T)$
$\text{im}(S \circ T) \subseteq \text{im}(S)$
If $T$ is surjective, $\text{im}(S \circ T) = \text{im}(S)$
If $S$ is injective, $\ker(S \circ T) = \ker(T)$

Remark 3.7: First Isomorphism Theorem

The First Isomorphism Theorem states that for any linear map $T: V \to W$ :

V / \ker(T) \cong \text{im}(T)

This says the quotient of $V$ by its kernel is isomorphic to the image. The map $v + \ker(T) \mapsto T(v)$ is a well-defined isomorphism.

10. Connection to Matrices

When a linear map is represented by a matrix, the kernel and image have concrete interpretations.

Definition 3.6: Null Space and Column Space

For a matrix $A \in M_{m \times n}(F)$ :

Null space: $\text{null}(A) = \{x \in F^n : Ax = 0\}$
Column space: $\text{col}(A) = \{Ax : x \in F^n\} = \text{span of columns of } A$

Remark 3.8: Computing via Row Reduction

To find the null space of $A$ :

Row reduce $A$ to echelon form
Identify free variables (columns without pivots)
Express the general solution parametrically

To find the column space of $A$ :

Row reduce $A$ to echelon form
The pivot columns of the original matrix form a basis

Example 3.14: Matrix Example

Problem: Find the null space and column space of

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

Solution:

Row reduce: $A \to \begin{pmatrix} 1 & 2 & 1 \\ 0 & 0 & 0 \end{pmatrix}$

Pivot column: 1. Free variables: $x_2, x_3$ .

$\text{null}(A) = \text{span}\{(-2, 1, 0), (-1, 0, 1)\}$

$\text{col}(A) = \text{span}\{(1, 2)\}$ (first column of original $A$ )

11. Quick Reference Summary

Concept	Definition	Key Property
Kernel	$\ker(T) = \{v : T(v) = 0\}$	Subspace of $V$
Image	$\text{im}(T) = \{T(v) : v \in V\}$	Subspace of $W$
Rank	$\dim(\text{im } T)$	= column rank of matrix
Nullity	$\dim(\ker T)$	= # free variables
Injective	$\ker(T) = \{0\}$	nullity = 0
Surjective	$\text{im}(T) = W$	rank = dim W

12. Additional Practice Problems

Problem 1

Let $T: \mathbb{R}^4 \to \mathbb{R}^3$ be defined by $T(x, y, z, w) = (x + y, y + z, z + w)$ . Find bases for $\ker(T)$ and $\text{im}(T)$ .

Problem 2

Let $T: P_3 \to P_3$ be defined by $T(p)(x) = xp'(x)$ . Determine if $T$ is injective, surjective, or neither.

Problem 3

Let $T: M_{2 \times 2} \to M_{2 \times 2}$ be defined by $T(A) = A - A^T$ . Find $\ker(T)$ and $\text{im}(T)$ .

Problem 4

If $T: V \to V$ satisfies $T^2 = 0$ (nilpotent of order 2), prove that $\text{im}(T) \subseteq \ker(T)$ .

Problem 5

Let $S, T: V \to V$ be linear operators with $ST = 0$ . Prove that $\text{im}(T) \subseteq \ker(S)$ .

Problem 6

Let $T: V \to W$ be linear with $\dim V = n$ . Prove that if $\dim(\ker T) = k$ , then $T$ maps any set of $n - k + 1$ linearly independent vectors to a linearly dependent set.

13. Theoretical Insights

Theorem 3.21: Kernel Chain

For a linear operator $T: V \to V$ :

\{0\} = \ker(T^0) \subseteq \ker(T) \subseteq \ker(T^2) \subseteq \ker(T^3) \subseteq \cdots

This chain eventually stabilizes: there exists $k$ such that $\ker(T^k) = \ker(T^{k+1}) = \cdots$

Theorem 3.22: Image Chain

For a linear operator $T: V \to V$ :

V = \text{im}(T^0) \supseteq \text{im}(T) \supseteq \text{im}(T^2) \supseteq \text{im}(T^3) \supseteq \cdots

This chain also stabilizes eventually.

Theorem 3.23: Idempotent Decomposition

If $P: V \to V$ is idempotent ( $P^2 = P$ ), then:

V = \ker(P) \oplus \text{im}(P)

Moreover, $\ker(P) = \text{im}(I - P)$ and $\text{im}(P) = \ker(I - P)$ .

Proof:

Direct sum: We show $\ker(P) \cap \text{im}(P) = \{0\}$ .

If $v \in \ker(P) \cap \text{im}(P)$ , then $P(v) = 0$ and $v = P(u)$ for some $u$ .

Thus $v = P(u) = P^2(u) = P(P(u)) = P(v) = 0$ .

Spanning: Any $v \in V$ can be written as:

v = (v - P(v)) + P(v)

where $P(v - P(v)) = P(v) - P^2(v) = P(v) - P(v) = 0$ , so $v - P(v) \in \ker(P)$ .

∎

Remark 3.9: Projections and Complements

Idempotent operators are called projections. The decomposition $V = \ker(P) \oplus \text{im}(P)$ shows that projections split the space into two complementary subspaces. This is fundamental in many areas, including functional analysis and quantum mechanics.

Theorem 3.24: Dimension Bounds for Composition

For $T: V \to W$ and $S: W \to U$ linear:

$\text{rank}(S \circ T) \leq \min\{\text{rank}(S), \text{rank}(T)\}$
$\text{nullity}(S \circ T) \geq \text{nullity}(T)$
$\text{rank}(S \circ T) \geq \text{rank}(S) + \text{rank}(T) - \dim W$ (Sylvester)

14. More Worked Examples

Example: Integration Operator

Problem: Let $T: P_2 \to P_3$ be defined by $T(p)(x) = \int_0^x p(t)\,dt$ . Find $\ker(T)$ and $\text{im}(T)$ .

Solution:

For $p(x) = a + bx + cx^2$ :

T(p)(x) = ax + \frac{b}{2}x^2 + \frac{c}{3}x^3

$\ker(T) = \{0\}$ since $T(p) = 0$ implies $a = b = c = 0$ .

$\text{im}(T) = \{ax + bx^2 + cx^3 : a, b, c \in \mathbb{R}\}$ = polynomials with zero constant term.

Conclusion: $T$ is injective but not surjective.

Example: Transpose Operator

Problem: Let $T: M_n \to M_n$ be defined by $T(A) = A^T$ . Find $\ker(T)$ .

Solution:

$\ker(T) = \{A : A^T = 0\} = \{0\}$

since $A^T = 0$ implies $A = (A^T)^T = 0^T = 0$ .

$\text{im}(T) = M_n$ since $T(A^T) = A$ for any $A$ .

Conclusion: $T$ is an isomorphism (bijective linear map).

Example: Skew-Symmetric Part

Problem: Let $T: M_n \to M_n$ be defined by $T(A) = \frac{1}{2}(A - A^T)$ . Analyze $T$ .

Solution:

$\ker(T) = \{A : A = A^T\}$ = symmetric matrices

$\text{im}(T) = \{B : B = -B^T\}$ = skew-symmetric matrices

For $n = 2$ : $\dim(\ker T) = 3$ , $\dim(\text{im } T) = 1$

Check: $3 + 1 = 4 = \dim M_2$ ✓

Example: Evaluation Map

Problem: Let $T: P_n \to \mathbb{R}^{n+1}$ be defined by $T(p) = (p(0), p(1), \ldots, p(n))$ . Analyze $T$ .

Solution:

A polynomial $p$ of degree $\leq n$ with $n + 1$ roots must be zero.

So $\ker(T) = \{0\}$ , and $T$ is injective.

Since $\dim P_n = n + 1 = \dim \mathbb{R}^{n+1}$ , $T$ is also surjective.

Conclusion: $T$ is an isomorphism.

15. Applications

Solving Linear Systems

For a linear system $Ax = b$ :

Existence: Solutions exist iff $b \in \text{col}(A)$
Uniqueness: Solution is unique iff $\text{null}(A) = \{0\}$
General solution: $x = x_p + x_h$ where $x_h \in \text{null}(A)$

Differential Equations

For a linear differential operator $L: C^\infty \to C^\infty$ :

$\ker(L)$ = solution space of the homogeneous equation $L(y) = 0$
For $L(y) = f$ : general solution = particular solution + $\ker(L)$

Computer Graphics

Transformations in computer graphics use kernel/image concepts:

Projection: 3D to 2D has 1-dimensional kernel (the viewing direction)
Shadows: Projection onto a plane, kernel = light ray direction
Data compression: Projecting to lower-dimensional subspace

Coding Theory

Linear codes use kernel and image:

Code: $C = \ker(H)$ for parity check matrix $H$
Encoding: Maps messages to codewords (image of generator matrix)
Error detection: $Hy = 0$ iff $y$ is a valid codeword

17. Geometric Interpretation

Understanding kernel and image geometrically provides powerful intuition for linear maps.

Projection onto a Plane

Consider projection from $\mathbb{R}^3$ onto the xy-plane:

Kernel: The z-axis (what gets "flattened" to the origin)
Image: The xy-plane (all possible outputs)
Geometry: Points on a vertical line all map to the same point

Rotation

Consider rotation by angle $\theta$ in $\mathbb{R}^2$ :

Kernel: $\{0\}$ only (rotation preserves distances)
Image: All of $\mathbb{R}^2$ (every point is reachable)
Conclusion: Rotation is an isomorphism

Reflection

Consider reflection about the x-axis in $\mathbb{R}^2$ :

Kernel: $\{0\}$ (only the origin is fixed with opposite sign)
Image: All of $\mathbb{R}^2$
Fixed points: The x-axis (where $T(v) = v$ )

Shear

Consider the shear $T(x, y) = (x + ky, y)$ :

Kernel: $\{0\}$ (shear is injective)
Image: All of $\mathbb{R}^2$ (shear is surjective)
Geometry: Horizontal lines slide along themselves

18. Study Tips

Tip 1: Always Verify with Rank-Nullity

After computing kernel and image, check that $\dim(\ker T) + \dim(\text{im } T) = \dim V$ . This catches computational errors.

Tip 2: Start with Standard Basis

When computing the image, apply $T$ to the standard basis vectors first. The image is the span of these outputs.

Tip 3: Kernel = Solve Homogeneous System

Finding the kernel always reduces to solving $T(v) = 0$ . For matrix maps, this is $Ax = 0$ —use row reduction!

Tip 4: Think Geometrically

Visualize the kernel as what gets "crushed" to zero, and the image as the "shadow" or "footprint" of the domain in the codomain.

What's Next?

Now that you understand kernel and image, the next steps explore:

Rank-Nullity Theorem: The fundamental dimension formula connecting kernel and image
Isomorphisms: Bijective linear maps and when spaces are "the same"
Matrix Representation: Representing linear maps as matrices
Change of Basis: How matrix representations change with basis choice
Dual Spaces: Linear functionals and the dual perspective

The Rank-Nullity Theorem is the crown jewel of this material—it precisely quantifies the trade-off between kernel and image. Understanding this trade-off is essential for all of linear algebra and its applications.

Kernel & Image Practice

Questions

Correct

Accuracy

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

is linear and

\ker(T) = \{0\}

, is

T

injective?

Hard

Not attempted

What is

\ker(T)

where

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

Medium

Not attempted

T

is linear and injective, what is

\dim(\ker T)

Easy

Not attempted

Is the image of a linear map always a subspace?

Easy

Not attempted

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

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

. What is

\text{im}(T)

Medium

Not attempted

\text{im}(T) = W

, then

T

is:

Easy

Not attempted

What is

\ker(I_V)

where

I_V

is the identity on

V

Easy

Not attempted

\ker(T) \neq \{0\}

, then

T

is:

Medium

Not attempted

What is

\ker(0)

where

0: V \to W

is the zero map?

Easy

Not attempted

\dim V = 5

and

\dim(\ker T) = 2

, what is

\dim(\text{im } T)

Medium

Not attempted

For the differentiation map

D: P_3 \to P_2

, what is

\ker(D)

Medium

Not attempted

T: V \to V

satisfies

T^2 = T

, then

V = ?

Hard

Not attempted

Frequently Asked Questions

Why is ker(T) also called the null space?

Because it's the set of vectors T sends to zero (null). The terminology 'null space' is common for matrices, while 'kernel' is used for abstract linear maps. Both refer to the same concept.

How do kernel and image relate to matrices?

For T represented by matrix A: ker(T) = null space of A (solution space of Ax = 0), im(T) = column space of A. This connects abstract linear algebra to matrix computations and row reduction.

What's the geometric meaning of the kernel?

The kernel is what T 'collapses' to zero. For a projection onto a plane, the kernel is the line perpendicular to that plane. For differentiation, the kernel is constant functions. It measures the 'dimension loss' of the map.

Can the kernel and image intersect non-trivially?

For T: V → V, yes! If v is in both ker(T) and im(T), then T(v) = 0 and v = T(u) for some u. This means T²(u) = 0. Nilpotent operators have this property.

What's the rank of a linear map?

rank(T) = dim(im(T)). This equals the column rank of any matrix representing T. The Rank-Nullity theorem says dim(V) = rank(T) + nullity(T), where nullity(T) = dim(ker T).

How do I compute the kernel of a linear map?

Set T(v) = 0 and solve for v. For maps given by matrices, this reduces to solving the homogeneous system Ax = 0 using row reduction.

How do I compute the image of a linear map?

Method 1: Apply T to a basis of V and take the span. Method 2: For matrix A, the image is the column space—find a basis by identifying pivot columns after row reduction.

What happens to kernel and image under composition?

For S ∘ T: ker(T) ⊆ ker(S ∘ T) and im(S ∘ T) ⊆ im(S). The composition can only make the kernel larger and the image smaller.

When is a linear map determined by its kernel?

Not by kernel alone! Knowing ker(T) tells you what T sends to 0, but not where other vectors go. However, if you also know im(T), the Rank-Nullity theorem constrains the possibilities.

What's the relationship between kernel, image, and invertibility?

T is invertible iff ker(T) = {0} AND im(T) = W. For maps between spaces of the same finite dimension, either condition implies both (and invertibility).