Kernel & Image of Linear Transformations

🎯 Learning Objectives

• Define the kernel (null space) of a linear transformation
• Define the image (range) of a linear transformation
• Find bases for the kernel and image of a linear transformation
• State and apply the rank-nullity theorem for linear transformations
• Determine whether a linear transformation is injective or surjective

📋 Prerequisites

• Linear Transformations (Week 5) — definition and properties
• Null Space & Column Space (Week 5) — the matrix version of kernel and image
• Basis & Dimension (Week 4) — finding bases for subspaces

1. Intuition: What Gets Crushed vs. What Gets Hit

1. Kernel (null space): Which vectors in $V$ get mapped to $\mathbf{0}$? These are the "crushed" vectors — they lose all information.
2. Image (range): Which vectors in $W$ can be reached? These are the "hit" vectors — the possible outputs. (Diagram) Key insight: The kernel measures non-injectivity (how many vectors collapse to the same output). The image measures non-surjectivity (how much of $W$ is covered).

2. Kernel

2.1 Definition

Definition (Kernel). Let $T: V \to W$ be a linear transformation. The kernel (or null space) of $T$ is:

• $\mathbf{0}_V \in \ker(T)$ because $T(\mathbf{0}_V) = \mathbf{0}_W$ ✓
• If $\mathbf{u}, \mathbf{v} \in \ker(T)$, then $T(\mathbf{u}+\mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v}) = \mathbf{0} + \mathbf{0} = \mathbf{0}$, so $\mathbf{u}+\mathbf{v} \in \ker(T)$ ✓
• If $\mathbf{v} \in \ker(T)$ and $c \in \mathbb{R}$, then $T(c\mathbf{v}) = cT(\mathbf{v}) = c \cdot \mathbf{0} = \mathbf{0}$, so $c\mathbf{v} \in \ker(T)$ ✓

2.2 Finding a Basis for the Kernel

Example 1: Kernel of a matrix transformation

$T: \mathbb{R}^3 \to \mathbb{R}^2$, $T(\mathbf{x}) = A\mathbf{x}$ where

.

Solve $A\mathbf{x} = \mathbf{0}$: Row 2 = 2(Row 1), so RREF is

.

$x_1 + 2x_2 - x_3 = 0$, so $x_1 = -2x_2 + x_3$. Free: $x_2, x_3$.

Basis for $\ker(T)$: $\{(-2,1,0), (1,0,1)\}$. $\dim(\ker(T)) = 2$. Example 2: Kernel of a polynomial transformation

$T: P_2 \to P_1$, $T(p) = p'$ (derivative).

$\ker(T) = \{p \in P_2 \mid p'(x) = 0\} = \{\text{constant polynomials}\}$.

Basis: $\{1\}$. $\dim(\ker(T)) = 1$. (Constant polynomials are degree 0.)

3. Image

3.1 Definition

Definition (Image). Let $T: V \to W$ be a linear transformation. The image (or range) of $T$ is:

• $\mathbf{0}_W \in \text{im}(T)$ because $T(\mathbf{0}_V) = \mathbf{0}_W$ ✓
• If $\mathbf{w}_1, \mathbf{w}_2 \in \text{im}(T)$, then $\mathbf{w}_1 = T(\mathbf{v}_1)$, $\mathbf{w}_2 = T(\mathbf{v}_2)$. Then $\mathbf{w}_1 + \mathbf{w}_2 = T(\mathbf{v}_1) + T(\mathbf{v}_2) = T(\mathbf{v}_1 + \mathbf{v}_2) \in \text{im}(T)$ ✓
• If $\mathbf{w} \in \text{im}(T)$ and $c \in \mathbb{R}$, $c\mathbf{w} = cT(\mathbf{v}) = T(c\mathbf{v}) \in \text{im}(T)$ ✓

3.2 Finding a Basis for the Image

Example 3: Image of a matrix transformation

$T: \mathbb{R}^3 \to \mathbb{R}^2$, $T(\mathbf{x}) = A\mathbf{x}$ where

.

Columns: $(1,2)$, $(2,4)$, $(-1,-2)$. Columns 2 and 3 are multiples of column 1.

$\text{im}(T) = \text{span}\{(1,2)\}$, a line through origin.

Basis: $\{(1,2)\}$. $\dim(\text{im}(T)) = 1$. Example 4: Image of a derivative transformation

$T: P_2 \to P_1$, $T(p) = p'$.

$T(1) = 0$, $T(x) = 1$, $T(x^2) = 2x$.

$\text{im}(T) = \text{span}\{0, 1, 2x\} = \text{span}\{1, 2x\} = P_1$ (all polynomials of degree $\leq 1$).

Basis: $\{1, x\}$. $\dim(\text{im}(T)) = 2$.

4. Rank-Nullity Theorem for Linear Transformations

Theorem (Rank-Nullity). For a linear transformation $T: V \to W$ where $V$ is finite-dimensional:

• Nullity = $\dim(\ker(T))$
• Rank = $\dim(\text{im}(T))$

Example 5: Checking rank-nullity

For $T: P_2 \to P_1$, $T(p) = p'$: $\dim(\ker(T)) = 1$ (constants) $\dim(\text{im}(T)) = 2$ ($P_1$) $\dim(P_2) = 3$ $1 + 2 = 3$ ✓

5. Injectivity, Surjectivity, and Isomorphism

5.1 Injectivity (One-to-One)

Theorem. $T$ is injective iff $\ker(T) = \{\mathbf{0}\}$. Proof: ($\Rightarrow$) If $T(\mathbf{v}) = \mathbf{0}$, and $T(\mathbf{0}) = \mathbf{0}$, then by injectivity $\mathbf{v} = \mathbf{0}$. ($\Leftarrow$) If $T(\mathbf{u}) = T(\mathbf{v})$, then $T(\mathbf{u}-\mathbf{v}) = \mathbf{0}$, so $\mathbf{u}-\mathbf{v} \in \ker(T) = \{\mathbf{0}\}$, hence $\mathbf{u} = \mathbf{v}$.

5.2 Surjectivity (Onto)

Theorem. $T$ is surjective iff $\text{im}(T) = W$ (i.e., $\dim(\text{im}(T)) = \dim(W)$).

5.3 Isomorphism

Definition (Isomorphism). A linear transformation $T: V \to W$ is an isomorphism if it is both injective and surjective (bijective). In that case, $V$ and $W$ are called isomorphic. Key fact: $V \cong W$ (isomorphic) iff $\dim(V) = \dim(W)$. Example 6: Isomorphism

$T: \mathbb{R}^3 \to P_2$, $T(a,b,c) = a + bx + cx^2$ is an isomorphism.

$\ker(T)$: $a + bx + cx^2 = 0$ ⇒ $a=b=c=0$ ⇒ $\ker(T) = \{\mathbf{0}\}$ (injective) ✓

$\text{im}(T) = P_2$ (surjective) ✓

$\dim(\mathbb{R}^3) = \dim(P_2) = 3$ ✓

6. Edge Cases & Gotchas

7. Common Pitfalls

❌ Pitfall 1: Confusing kernel of $T$ with kernel of its matrix

❌ Pitfall 2: Thinking image = codomain

❌ Pitfall 3: Forgetting rank-nullity counts domain dimension

8. Formula Summary Table

9. 📝 Practice Questions

Q1: Kernel of a matrix transformation

Find a basis for $\ker(T)$ where $T: \mathbb{R}^3 \to \mathbb{R}^2$, $T(x,y,z) = (x+y, 2x+2y)$.

Solution: Solve $x+y=0$, $2x+2y=0$ ⇒ $y=-x$, $z$ free. $\ker(T) = \{(x,-x,z) \mid x,z \in \mathbb{R}\} = \text{span}\{(1,-1,0), (0,0,1)\}$. Basis: $\{(1,-1,0), (0,0,1)\}$. Nullity = 2. Q2: Image of a matrix transformation

Find a basis for $\text{im}(T)$ where $T(x,y) = (x+2y, 2x+4y, -x-2y)$.

Solution: Standard matrix: $T(1,0) = (1,2,-1)$, $T(0,1) = (2,4,-2)$.

. Column 2 = 2(column 1). Basis for $C(A)$: $\{(1,2,-1)\}$. $\text{im}(T) = \text{span}\{(1,2,-1)\}$, a line. Rank = 1. Q3: Rank-Nullity on a polynomial map

For $T: P_2 \to P_2$, $T(p)(x) = p(0) + p(1)x + p(2)x^2$, find nullity and rank.

Solution: Let $p(x) = a + bx + cx^2$. $p(0) = a$, $p(1) = a + b + c$, $p(2) = a + 2b + 4c$. $T(p) = a + (a+b+c)x + (a+2b+4c)x^2$.

Kernel: Solve $a = 0$, $a+b+c = 0$, $a+2b+4c = 0$. From first: $a=0$. Then $b+c = 0$, $2b+4c = 0$ ⇒ $b = -c$, $2(-c)+4c = 2c = 0$ ⇒ $c=0$, $b=0$. $\ker(T) = \{0\}$. Nullity = 0. Rank = $\dim(P_2) - 0 = 3$. $\text{im}(T) = P_2$. Q4: Injective but not surjective

Show $T: \mathbb{R}^2 \to \mathbb{R}^3$, $T(x,y) = (x, y, 0)$ is injective but not surjective.

Solution: $\ker(T)$: $(x,y,0) = (0,0,0)$ ⇒ $x=y=0$. $\ker(T) = \{\mathbf{0}\}$, injective ✓.

$\text{im}(T) = \{(x,y,0)\}$ = $xy$-plane in $\mathbb{R}^3$. Not all of $\mathbb{R}^3$ (e.g., $(0,0,1)$ not reachable). Not surjective.

Check rank-nullity: $0 + 2 = 2 = \dim(\mathbb{R}^2)$ ✓ Q5: Surjective but not injective

Show $T: \mathbb{R}^3 \to \mathbb{R}^2$, $T(x,y,z) = (x, 0)$ is surjective? Actually no. Let me fix.

$T: \mathbb{R}^3 \to \mathbb{R}^2$, $T(x,y,z) = (x, y)$ is surjective (every $(a,b)$ is reached by $(a,b,0)$), but not injective ($(0,0,1)$ also maps to $(0,0)$).

$\ker(T) = \{(0,0,z)\}$, dimension 1. $\text{im}(T) = \mathbb{R}^2$, dimension 2. Rank-nullity: $1 + 2 = 3 = \dim(\mathbb{R}^3)$ ✓ Q6: Isomorphism

Is $T: \mathbb{R}^3 \to \mathbb{R}^3$, $T(x,y,z) = (x, x+y, x+y+z)$ an isomorphism?

Solution: Matrix of $T$:

.

$\det(A) = 1 \neq 0$, so $A$ is invertible. $T$ is an isomorphism.

$\ker(T) = \{\mathbf{0}\}$, $\text{im}(T) = \mathbb{R}^3$. Q7: Kernel of an integral operator

$T: C[0,1] \to \mathbb{R}$, $T(f) = \int_0^1 f(x) dx$. What is $\ker(T)$?

Solution: $\ker(T) = \{f \in C[0,1] \mid \int_0^1 f(x) dx = 0\}$. These are functions with zero average (zero net area). Examples: $f(x) = x - \frac{1}{2}$, $\sin(2\pi x)$. Note: $\ker(T)$ is infinite-dimensional. Q8: Image of a differentiation map

$T: P_3 \to P_2$, $T(p) = p'$. Find $\text{im}(T)$.

Solution: $T(1) = 0$, $T(x) = 1$, $T(x^2) = 2x$, $T(x^3) = 3x^2$. $\text{im}(T) = \text{span}\{0, 1, 2x, 3x^2\} = \text{span}\{1, x, x^2\} = P_2$. So $\text{im}(T) = P_2$ (surjective). $\dim(\text{im}(T)) = 3$. Nullity = $\dim(P_3) - 3 = 4 - 3 = 1$. Q9: Determining if a vector is in the image

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

. Is $(3,1)$ in $\text{im}(T)$?

Solution: Solve $T(\mathbf{x}) = (3,1)$: $x + 2z = 3$, $y + z = 1$. Choose $z = 0$: $x = 3$, $y = 1$. So $\mathbf{x} = (3,1,0)$ works. Yes, $(3,1) \in \text{im}(T)$. Q10: Kernel and image of projection

$T: \mathbb{R}^3 \to \mathbb{R}^3$, $T(x,y,z) = (x, y, 0)$ (projection onto $xy$-plane). Find bases for $\ker(T)$ and $\text{im}(T)$.

Solution: $\ker(T)$: $(x,y,0) = (0,0,0)$ ⇒ $x=0, y=0$. So $\ker(T) = \{(0,0,z)\} = \text{span}\{(0,0,1)\}$. Basis: $\{(0,0,1)\}$.

$\text{im}(T)$: $T(1,0,0) = (1,0,0)$, $T(0,1,0) = (0,1,0)$, $T(0,0,1) = (0,0,0)$. $\text{im}(T) = \text{span}\{(1,0,0), (0,1,0)\}$ = $xy$-plane. Basis: $\{(1,0,0), (0,1,0)\}$.

Rank-nullity: $1 + 2 = 3$ ✓

🔗 Cross-References

• Next topic: Matrix Representation of Linear Transformations
• Week 5 (Rank-Nullity Theorem): The matrix version of the same idea
• Week 7 (Similarity): Two matrix representations of the same $T$
• BSCS2004 (ML Foundations): Feature maps, kernel methods in SVM Join Discord Previous5.2 Linear TransformationsNext6.2 Matrix Representation