MIT 18.06SC Lecture 6: Column Space and Null Space

Linear Algebra
MIT 18.06
Vector Spaces
Author

Chao Ma

Published

October 13, 2025

Context

My lecture notes

Column space and null space are two fundamental subspaces associated with any matrix. This lecture shows which vectors \(b\) make \(Ax = b\) solvable and which vectors \(x\) satisfy \(Ax = 0\).

Subspace Properties

  • \(P \cup L\) is not a subspace (P and L are two subspaces)
  • \(P \cap L\) is a subspace

Column Space

Definition

Given a matrix A: \[ A = \begin{bmatrix} 1 & 1 & 2 \\ 2 & 1 & 3 \\ 3 & 1 & 4 \\ 4 & 1 & 5 \end{bmatrix} \]

The column space of A consists of all possible linear combinations of the columns of A.

Key Observation

Because the column space has 4 dimensions (rows) but only 3 columns (subspaces/lines), the entire space cannot be filled. There are many vectors \(b\) outside the column space.

Therefore: We cannot say that for every \(Ax = b\), there is a solution.

Solutions to \(Ax = b\)

Special Case: \(b = \mathbf{0}\)

  • \(b = [0, 0, 0, 0]\) always has a solution
  • This is the origin point, and all subspaces pass through the origin
  • The solution is \(x = [0, 0, 0]\)

Which \(b\) Can Be Solved?

General Rule: \(Ax = b\) has a solution if and only if \(b\) is in the column space of A.

Examples of Solvable \(b\):

  1. \(b = [1, 2, 3, 4]\) has solution \(x = [1, 0, 0]\)
    • This is the first column of A
  2. \(b = [1, 1, 1, 1]\) has solution \(x = [0, 1, 0]\)
    • This is the second column of A
  3. Any linear combination of columns has a solution
    • If \(b = c_1 \cdot \text{col}_1 + c_2 \cdot \text{col}_2 + c_3 \cdot \text{col}_3\)
    • Then \(x = [c_1, c_2, c_3]\) is the solution

Null Space

Definition

The null space of A, denoted \(N(A)\), contains all vectors \(x\) that satisfy: \[ Ax = \mathbf{0} \]

Proof: \(N(A)\) is a Subspace

To prove that the null space is a subspace, we must show it satisfies two properties:

1. Closed Under Addition

If \(v\) and \(w\) are in \(N(A)\), then \(v + w\) is also in \(N(A)\).

Proof: \[ \begin{align} Av &= \mathbf{0} \\ Aw &= \mathbf{0} \\ A(v + w) &= Av + Aw = \mathbf{0} + \mathbf{0} = \mathbf{0} \end{align} \]

Therefore, \(v + w \in N(A)\).

2. Closed Under Scalar Multiplication

If \(x\) is in \(N(A)\) and \(c\) is any scalar, then \(cx\) is also in \(N(A)\).

Proof: \[ \begin{align} Ax &= \mathbf{0} \\ A(cx) &= c(Ax) = c \cdot \mathbf{0} = \mathbf{0} \end{align} \]

Therefore, \(cx \in N(A)\).

Important Contrast: When \(b \neq \mathbf{0}\)

The solution set of \(Ax = b\) (when \(b \neq \mathbf{0}\)) is not a subspace.

Proof: Not Closed Under Scalar Multiplication

Given: \[ Ax = b, \quad b \neq \mathbf{0} \]

For scalar \(c \neq 1\): \[ A(cx) = c(Ax) = cb \neq b \]

Therefore, if \(x\) is a solution, \(cx\) is not a solution (unless \(c = 1\)).

Conclusion: The solution set fails the scalar multiplication property, so it is not a subspace.

Summary

Column Space: - Column space = all possible linear combinations of columns - \(Ax = b\) is solvable \(\Leftrightarrow\) \(b\) is in the column space - Not every \(b \in \mathbb{R}^4\) is in the column space of this particular A

Null Space: - Null space \(N(A) = \{x : Ax = \mathbf{0}\}\) is a subspace - Solution set of \(Ax = b\) (when \(b \neq \mathbf{0}\)) is not a subspace - The null space always contains the zero vector - The null space is closed under addition and scalar multiplication

Source: MIT 18.06SC Linear Algebra, Lecture 6