7: Kernel and Image

Every linear transformation has two fundamental subspaces attached to it: the kernel (what gets destroyed) and the image (what gets produced). Understanding these reveals what a transformation does geometrically,where it collapses, where it stretches, and what it can reach.

The Image

(Image of a Linear Transformation)

Let $T: V \to W$ be a linear transformation. The image (or range) of $T$ is the set of all outputs:

Key fact: The image is always a subspace of $W$.

Proof:

1. $T(\mathbf{0}) = \mathbf{0}$, so $\mathbf{0} \in \text{im}(T)$
2. If $w_1 = T(v_1)$ and $w_2 = T(v_2)$, then $w_1 + w_2 = T(v_1) + T(v_2) = T(v_1 + v_2) \in \text{im}(T)$
3. If $w = T(v)$, then $cw = cT(v) = T(cv) \in \text{im}(T)$

(Image for Matrices: Column Space)

For a matrix transformation $T(\mathbf{x}) = A\mathbf{x}$ where $A$ is $m \times n$:

Why? Write $A = [\mathbf{a}_1 \mid \mathbf{a}_2 \mid \cdots \mid \mathbf{a}_n]$ in column form. Then:

Every output is a linear combination of the columns,so the image is exactly the span of the columns.

Interpretation: The column space tells you what vectors can be reached by the transformation. If $\mathbf{b} \in \text{col}(A)$, then $A\mathbf{x} = \mathbf{b}$ has a solution. If $\mathbf{b} \notin \text{col}(A)$, the system is inconsistent.

The Kernel

(Kernel of a Linear Transformation)

Let $T: V \to W$ be a linear transformation. The kernel (or null space) of $T$ is the set of all inputs that map to zero:

Key fact: The kernel is always a subspace of $V$.

Proof:

1. $T(\mathbf{0}) = \mathbf{0}$, so $\mathbf{0} \in \ker(T)$
2. If $T(v_1) = \mathbf{0}$ and $T(v_2) = \mathbf{0}$, then $T(v_1 + v_2) = T(v_1) + T(v_2) = \mathbf{0}$
3. If $T(v) = \mathbf{0}$, then $T(cv) = cT(v) = c\mathbf{0} = \mathbf{0}$

(Kernel for Matrices: Null Space)

For a matrix transformation $T(\mathbf{x}) = A\mathbf{x}$:

The null space is the solution set to the homogeneous system $A\mathbf{x} = \mathbf{0}$.

Interpretation: The kernel captures all the redundancy in the transformation,the directions that get completely flattened to nothing.

(Computing the Kernel)

To find a basis for $\ker(A)$:

1. Row reduce $A$ to $\text{rref}(A)$
2. Identify the free variables (columns without pivots)
3. For each free variable, set it to $1$ and all others to $0$, then solve
4. These solutions form a basis for the kernel

Example: Find $\ker(A)$ for

Row reduce:

Pivots in columns 1 and 3. Free variables: $x_2$ and $x_4$.

Setting $x_2 = 1, x_4 = 0$:

• From row 2: $x_3 = 0$
• From row 1: $x_1 = -2$
• Solution: $(-2, 1, 0, 0)$

Setting $x_2 = 0, x_4 = 1$:

• From row 2: $x_3 = 1$
• From row 1: $x_1 = -1$
• Solution: $(-1, 0, 1, 1)$

Basis for $\ker(A)$:

The kernel is a 2-dimensional subspace of $\mathbb{R}^4$.

The Rank-Nullity Theorem

(Rank and Nullity)

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

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

For a matrix $A$, rank$(A) = \dim(\text{col}(A))$ equals the number of pivot columns.

(The Rank-Nullity Theorem)

For any linear transformation $T: V \to W$ where $V$ is finite-dimensional:

For an $m \times n$ matrix $A$:

Interpretation: The dimension of the input space is split between:

• Dimensions that survive and get mapped somewhere meaningful (the rank)
• Dimensions that get crushed to zero (the nullity)

This is conservation of dimension,you can’t create or destroy dimensions, only redistribute them.

(Proof Sketch)

Let $\{v_1, \ldots, v_k\}$ be a basis for $\ker(T)$. Extend this to a basis $\{v_1, \ldots, v_k, u_1, \ldots, u_r\}$ for $V$.

Claim: $\{T(u_1), \ldots, T(u_r)\}$ is a basis for $\text{im}(T)$.

Spanning: Any $w = T(v) \in \text{im}(T)$ can be written as:

so $T(v) = d_1T(u_1) + \cdots + d_rT(u_r)$ (since $T(v_i) = \mathbf{0}$).

Linear independence: If $d_1T(u_1) + \cdots + d_rT(u_r) = \mathbf{0}$, then $T(d_1u_1 + \cdots + d_ru_r) = \mathbf{0}$, so $d_1u_1 + \cdots + d_ru_r \in \ker(T)$. This means it’s a combination of $v_1, \ldots, v_k$, forcing all $d_i = 0$ by independence.

Thus $\dim(V) = k + r = \text{nullity}(T) + \text{rank}(T)$.

Injectivity and Surjectivity Revisited

The kernel and image directly characterize injectivity and surjectivity:

(Kernel and Injectivity)

Why? If $\ker(T) = \{\mathbf{0}\}$ and $T(v_1) = T(v_2)$, then $T(v_1 - v_2) = \mathbf{0}$, so $v_1 - v_2 \in \ker(T)$, meaning $v_1 = v_2$.

For matrices: $A$ is injective iff $\text{null}(A) = \{\mathbf{0}\}$ iff there’s a pivot in every column iff rank$(A) = n$.

(Image and Surjectivity)

For matrices: $A$ ($m \times n$) is surjective iff $\text{col}(A) = \mathbb{R}^m$ iff there’s a pivot in every row iff rank$(A) = m$.

(The Counting Argument)

For an $m \times n$ matrix $A$:

• Injective requires rank$(A) = n$ (all columns are pivots)
• Surjective requires rank$(A) = m$ (all rows have pivots)
• Bijective requires rank$(A) = m = n$ (square and full rank)

Since rank$(A) \leq \min(m, n)$:

• If $n > m$: Cannot be injective (not enough room for pivots in every column)
• If $m > n$: Cannot be surjective (not enough columns to pivot in every row)

The Row Space

There’s a fourth fundamental subspace hiding in every matrix.

(Row Space)

The row space of $A$, denoted $\text{row}(A)$, is the span of the rows of $A$:

This is a subspace of $\mathbb{R}^n$ (same as the domain).

(Key Property)

Row operations preserve the row space. So:

This means the nonzero rows of $\text{rref}(A)$ form a basis for $\text{row}(A)$.

Note: This is different from column space! Row operations change column relationships but preserve row span.

(Row Space and Null Space are Orthogonal)

For any matrix $A$:

Every vector in the row space is orthogonal to every vector in the null space.

Why? If $\mathbf{x} \in \ker(A)$, then $A\mathbf{x} = \mathbf{0}$. This means each row $\mathbf{r}_i$ satisfies $\mathbf{r}_i \cdot \mathbf{x} = 0$. So $\mathbf{x}$ is orthogonal to every row, hence to every linear combination of rows.

The Four Fundamental Subspaces

Every $m \times n$ matrix $A$ defines four fundamental subspaces:

where $r = \text{rank}(A)$.

(Orthogonal Complements)

These subspaces pair up as orthogonal complements:

• In $\mathbb{R}^n$: $\text{row}(A) \perp \ker(A)$, and together they span $\mathbb{R}^n$
• In $\mathbb{R}^m$: $\text{col}(A) \perp \ker(A^T)$, and together they span $\mathbb{R}^m$

Dimensions check:

• $\dim(\text{row}(A)) + \dim(\ker(A)) = r + (n - r) = n$ ✓
• $\dim(\text{col}(A)) + \dim(\ker(A^T)) = r + (m - r) = m$ ✓

Geometric Summary

The kernel and image tell you exactly how a transformation reshapes space:

The image is the “shadow” of the domain in the codomain,all the places you can reach. Its dimension (the rank) is how many independent directions survive the transformation.

The kernel is the “blind spot” in the domain,all the inputs that become invisible. Its dimension (the nullity) is how many directions get flattened.

The rank-nullity theorem says the domain splits cleanly: some directions die, some survive, and nothing is lost in the accounting.