LINEAR-ALGEBRA · Unit 11 · Video 4 · Interactive Practice

One Equation, Four Subspaces

IKey Formulas

Formula	Name	Description
$\dim C(A^T) = r$	Row Space Dimension	Equals the rank
$\dim N(A) = n - r$	Rank-Nullity Theorem	Columns minus rank
$\dim C(A) = r$	Column Space Dimension	Equals the rank
$\dim N(A^T) = m - r$	Left Null Space Dim	Rows minus rank

For an $m \times n$ matrix of rank $r$ : row space and null space are orthogonal complements in $\mathbb{R}^n$ ; column space and left null space are orthogonal complements in $\mathbb{R}^m$ .

IIVisualization 1 — Null-Space Membership

For $A = \begin{pmatrix} 1 & 1 & 1 & 1 \end{pmatrix}$ , a vector lies in $N(A)$ exactly when its four components sum to zero.

Because $A$ has rank $1$ and $n = 4$ columns, this null space has dimension $n - r = 3$ — the three basis vectors above span it.

IIIVisualization 2 — Orthogonal Complements

For $A = \begin{pmatrix} a & b \end{pmatrix}$ , the null space is the line through the origin perpendicular to the row space.

a 1 b 1

The complement of a line is a line in $\mathbb{R}^2$ ( $1+1=2$ ), a plane in $\mathbb{R}^3$ ( $1+2=3$ ), and a $3$ -space in the video's $\mathbb{R}^4$ ( $1+3=4$ ).

IVVisualization 3 — The Dimension Balance

For any $m \times n$ matrix of rank $r$ , the four subspace dimensions always satisfy $r+(n-r)=n$ and $r+(m-r)=m$ .

m (rows) 1 n (cols) 4 r (rank) 1

Rank can't exceed $\min(m,n)$ ; push $r$ past it and it stays capped. At $r=0$ the whole space is null space, and the dimensions still sum correctly.

VQuiz Questions

Question 1

The matrix $A = \begin{pmatrix} 1 & 1 & 1 & 1 \end{pmatrix}$ is a $1 \times 4$ matrix with rank 1.

What is the dimension of the null space $N(A)$ ?

✅ Correct! By rank-nullity: dim N(A) = n − r = 4 − 1 = 3.

❌ Not quite. Use the rank-nullity theorem: dim N(A) = n − rank(A). Here n = 4 and rank = 1.

Show solution

Solution:

By the rank-nullity theorem: $\dim N(A) = n - r = 4 - 1 = 3$

The matrix has $n = 4$ columns and rank $r = 1$ (one pivot), so the null space has dimension $4 - 1 = 3$ .

The three basis vectors are: $(-1, 1, 0, 0)$ , $(-1, 0, 1, 0)$ , $(-1, 0, 0, 1)$ .

Question 2

Which of the following vectors is in the null space of $A = \begin{pmatrix} 1 & 1 & 1 & 1 \end{pmatrix}$ ?

✅ Correct! The components sum to zero: 1 + (−1) + 0 + 0 = 0.

❌ Not quite. A vector is in N(A) when its components sum to zero. Try adding up the components of each option.

Show solution

Solution:

A vector $\mathbf{v}$ is in $N(A)$ if and only if $v_1 + v_2 + v_3 + v_4 = 0$ . Check each:

$(1, 1, 1, 1)$ : $1 + 1 + 1 + 1 = 4 \neq 0$ ✗
$(1, -1, 0, 0)$ : $1 + (-1) + 0 + 0 = 0$ ✓
$(-1, 1, 1, 1)$ : $-1 + 1 + 1 + 1 = 2 \neq 0$ ✗
$(0, 0, 0, 1)$ : $0 + 0 + 0 + 1 = 1 \neq 0$ ✗

Only $(1, -1, 0, 0)$ has components summing to zero.

Question 3

A matrix $B$ has size $3 \times 7$ with rank 2.

What is the dimension of the null space $N(B)$ ?

✅ Correct! dim N(B) = n − r = 7 − 2 = 5.

❌ Not quite. Remember: dim N(B) = (number of columns) − rank. Be careful not to use the number of rows!

Show solution

Solution:

By the rank-nullity theorem: $\dim N(B) = n - r = 7 - 2 = 5$

Here $n = 7$ (columns) and $r = 2$ (rank), so the null space has dimension $7 - 2 = 5$ .

Common mistakes:

Using $m$ instead of $n$ : $3 - 2 = 1$ gives the left null space dimension, not the null space.
Answering with the rank itself ( $r = 2$ ) confuses rank with nullity.

Question 4

True or False: For a $1 \times n$ matrix of rank 1, the left null space $N(A^T)$ always equals $\{\mathbf{0}\}$ (contains only the zero vector).

✅ Correct! dim N(Aᵀ) = m − r = 1 − 1 = 0, so the left null space is just {0}.

❌ Not quite. Compute dim N(Aᵀ) = m − r. What do you get when m = 1 and r = 1?

Show solution

Solution: True

The left null space has dimension $m - r$ .

For a $1 \times n$ matrix of rank 1: $m = 1$ and $r = 1$ , so: $\dim N(A^T) = m - r = 1 - 1 = 0$

A subspace of dimension 0 contains only the zero vector: $N(A^T) = \{\mathbf{0}\}$ .

This is exactly what we saw in the video: the equation $A^T y = 0$ with $A^T$ being a column of ones forces $y = 0$ .

Solved: 0 / 4