Lesson 3: Kernel (Null Space) and Range (Column Space)
Estimated time: 40-50 minutes
Learning Objectives
- Define the kernel (null space) and range (image/column space) of a linear transformation
- Compute the kernel by solving Ax = 0
- Compute the range as the column space of the standard matrix
- Connect kernel and range to the existence and uniqueness of solutions
The Kernel
Kernel (Null Space): The kernel of a linear transformation T : R^n → R^m is the set of all vectors that T maps to the zero vector:
ker(T) = { x in R^n : T(x) = 0 } = { x : Ax = 0 }
The kernel is always a subspace of the domain R^n.
The kernel tells you about "information loss" -- it contains everything that gets crushed to zero. If the kernel is just {0}, no information is lost and T is one-to-one.
Worked Example 1: Computing the Kernel
Let T(x, y, z) = (x + y, y + z) with matrix A = [1 1 0; 0 1 1].
Solve Ax = 0:
[0 1 1 | 0] → [0 1 1 | 0]
x - z = 0 and y + z = 0, so x = z, y = -z, z free.
ker(T) = { z(1, -1, 1) : z in R } = span{(1, -1, 1)}.
The kernel is a line through the origin in R^3.
Worked Example 2: Trivial Kernel
Let T(x, y) = (x + y, x - y) with matrix A = [1 1; 1 -1].
Solve Ax = 0: Row reduce [1 1 | 0; 1 -1 | 0] → [1 0 | 0; 0 1 | 0].
Only solution: x = 0, y = 0. ker(T) = {0}.
T is one-to-one: different inputs always produce different outputs.
The Range
Range (Image/Column Space): The range of T is the set of all possible outputs:
range(T) = { T(x) : x in R^n } = { Ax : x in R^n } = Col(A)
The range is always a subspace of the codomain R^m, and it equals the column space of A.
Worked Example 3: Computing the Range
Using A = [1 1 0; 0 1 1] from Example 1:
The range is Col(A) = span of the columns of A.
Columns: (1, 0), (1, 1), (0, 1). Row reduce A:
[0 1 1] → [0 1 1]
Pivots in columns 1 and 2, so range(T) = span{(1, 0), (1, 1)} = R^2.
T maps onto all of R^2 -- it is onto (surjective).
Worked Example 4: Range That Is Not All of R^m
Let T(x, y) = (x + y, 2x + 2y, x - y) with A = [1 1; 2 2; 1 -1].
Row reduce: [1 1; 2 2; 1 -1] → [1 1; 0 0; 0 -2] → [1 0; 0 1; 0 0].
Pivots in both columns. range(T) = span{(1, 2, 1), (1, 2, -1)} -- a 2-dimensional plane in R^3.
T is NOT onto because the range is a plane, not all of R^3.
Connection to Solutions of Ax = b
Three connected ideas:
(1) Ax = b is consistent if and only if b is in range(T) = Col(A).
(2) The solution is unique if and only if ker(T) = {0}.
(3) If x_p is one solution to Ax = b, then ALL solutions are: x = x_p + x_h, where x_h is any vector in ker(T).
Interpretation
The kernel gives the "ambiguity" in solutions. A bigger kernel means more solutions to any consistent system. ker(T) = {0} means every consistent system has exactly one solution.
One-to-One and Onto
One-to-one (injective): T is one-to-one if and only if ker(T) = {0}.
Onto (surjective): T is onto if and only if range(T) = R^m (the entire codomain).
For T : R^n → R^m with matrix A:
- One-to-one: A has a pivot in every column (n pivots)
- Onto: A has a pivot in every row (m pivots)
- Both (bijective): possible only when m = n and A is invertible
Check Your Understanding
1. Find ker(T) where T(x,y) = (x+2y, 3x+6y).
2. What is the range of T from the previous question?
3. T : R^3 → R^3 has ker(T) = {0}. Must T be onto?
4. If Ax = b has a solution x_p and ker(T) = span{v1, v2}, describe all solutions.
Key Takeaways
- Kernel = null space of A = all x with Ax = 0; measures "information loss"
- Range = column space of A = all possible outputs; determines which equations b = Ax are solvable
- ker(T) = {0} if and only if T is one-to-one
- range(T) = R^m if and only if T is onto
- General solution = particular solution + any vector in the kernel