LINEAR-ALGEBRA · Interactive Practice | Unit 10 · Video 1

One Matrix, Four Worlds: How Rank Reveals the Hidden Architecture of Any Matrix

IKey Formulas

Subspace	Notation	Lives in	Dimension
Column Space	$C(A)$	$\mathbb{R}^m$	$r$
Null Space	$N(A)$	$\mathbb{R}^n$	$n - r$
Row Space	$C(A^T)$	$\mathbb{R}^n$	$r$
Left Null Space	$N(A^T)$	$\mathbb{R}^m$	$m - r$

Rank $r$ = number of pivots after row reduction, with $r \leq \min(m, n)$

Dimension counting: In $\mathbb{R}^n$ : $r + (n - r) = n$ . In $\mathbb{R}^m$ : $r + (m - r) = m$ .

IIVisualization 1 — How Rank Splits Each Space

Rank $r$ carves $\mathbb{R}^n$ into row space $\oplus$ null space, and $\mathbb{R}^m$ into column space $\oplus$ left null space.

Rows m 3 Cols n 3 Rank r 2

💡 Challenge: what must $m$ , $n$ , $r$ satisfy to make both null spaces vanish at once?

IIIVisualization 2 — Reading Rank From Row Reduction

Row reduction reveals the rank as its pivot count — and that single number $r$ fixes all four subspace dimensions.

Matrix

matrix A

echelon form U · green cells are pivots

💡 Challenge: which matrix in the list has a trivial ( $\dim 0$ ) left null space — and why does its shape force that?

IVVisualization 3 — The Null Space as a Column Recipe

The null vector $(1,-2,1)^T$ is a recipe — $c_1 - 2c_2 + c_3 = \mathbf{0}$ — so $A\mathbf{x} = \mathbf{0}$ for every scalar $t$ .

Scale t 1

A · x where x = t·(1, −2, 1)ᵀ

every row cancels to zero

VQuiz Questions

Question 1

A matrix $A$ is $5 \times 7$ with rank $r = 3$ . What is the dimension of the null space $N(A)$ ?

✅ Correct! dim N(A) = n − r = 7 − 3 = 4.

❌ Not quite. Remember: dim N(A) = n − r, where n is the number of columns (7), not rows (5).

Show solution

Solution:

$N(A)$ lives in $\mathbb{R}^n$ where $n = 7$ (the number of columns).

$\dim N(A) = n - r = 7 - 3 = 4$

Row reduction produces $n - r = 4$ free variables, each contributing one basis vector to the null space.

Question 2

For an $m \times n$ matrix $A$ , the column space $C(A)$ and the null space $N(A)$ live in which spaces, respectively?

✅ Correct! Column space lives in R^m (output) and null space lives in R^n (input).

❌ Not quite. Think: columns have m entries → C(A) ⊆ R^m. The vector x in Ax = 0 has n entries → N(A) ⊆ R^n.

Show solution

Solution:

Column space $C(A)$ : Each column has $m$ entries, so $C(A) \subseteq \mathbb{R}^m$ (the output space, where $A\mathbf{x}$ lands).
Null space $N(A)$ : All $\mathbf{x}$ with $A\mathbf{x} = \mathbf{0}$ . Since $\mathbf{x}$ has $n$ components, $N(A) \subseteq \mathbb{R}^n$ (the input space).

They live in different ambient spaces whenever $m \neq n$ .

Question 3

For a $6 \times 4$ matrix $B$ with rank 3, what are the dimensions of all four fundamental subspaces?

(Listed as: dim C(B), dim N(B), dim C(B^T), dim N(B^T))

✅ Correct! The dimensions sum correctly: 3 + 1 = 4 in R⁴ and 3 + 3 = 6 in R⁶.

❌ Not quite. Careful: dim N(B) = n − r uses columns (4), while dim N(Bᵀ) = m − r uses rows (6).

Show solution

Solution:

$B$ is $6 \times 4$ , so $m = 6$ , $n = 4$ , $r = 3$ .

Subspace	Formula	Dimension
$C(B)$	$r$	3
$N(B)$	$n - r = 4 - 3$	1
$C(B^T)$	$r$	3
$N(B^T)$	$m - r = 6 - 3$	3

Check: In $\mathbb{R}^4$ : $3 + 1 = 4$ ✓ In $\mathbb{R}^6$ : $3 + 3 = 6$ ✓

The common trap is option "3, 3, 3, 1" which swaps the two null-space dimensions. Remember: $N(A)$ uses $n$ and $N(A^T)$ uses $m$ .

Question 4

True or False: Row reduction preserves the row space of a matrix (the echelon form has the same row space as the original matrix).

✅ Correct! Row operations keep every new row in the span of the originals, so the row space is unchanged.

❌ Not quite. Each row operation replaces a row with a linear combination of existing rows — the span doesn't change.

Show solution

Solution: True

Each elementary row operation replaces a row with a linear combination of existing rows. The new set of rows spans the same space as the original set, so the row space is unchanged.

This is why the nonzero rows of the echelon form are a valid basis for the row space of the original matrix.

⚠️ Important contrast: Row reduction does not preserve the column space. To find a basis for $C(A)$ , you must identify the pivot columns in the echelon form and then take the corresponding columns from the original matrix $A$ .

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