LINEAR-ALGEBRA · Unit 10 · Video 2 · Interactive Practice

One Row Reduction, Four Bases

IKey Formulas

Subspace	Basis Source	Dimension	Lives in
Column Space $C(A)$	Pivot columns of A (original!)	$r$	$\mathbb{R}^m$
Null Space $N(A)$	Special solutions from R	$n - r$	$\mathbb{R}^n$
Row Space $C(A^T)$	Nonzero rows of R	$r$	$\mathbb{R}^n$
Left Null Space $N(A^T)$	Zero-row(s) of E (where $EA = R$ )	$m - r$	$\mathbb{R}^m$

Dimension Theorem: $r + (n - r) = n$ and $r + (m - r) = m$

IIVisualization 1 — Row Reduction, Step by Step

Reduce $A$ to its RREF one operation at a time — where do the pivots land, and which columns stay free?

Reduction step 0

pivot free column zero row

💡 Rows 1 and 3 of $A$ are identical, so one row collapses to zero: the rank is $r = 2$ , with pivots in columns 1–2 and columns 3–4 free.

IIIVisualization 2 — Column Space Lives in $A$ , Not $R$

Combinations of $A$ 's pivot columns can reach $(1,1,1)$ — can any combination of $R$ 's columns do the same?

c₁ 1 c₂ 0

pivot columns of A

columns of R

💡 Every column of $R$ has a zero third entry, so no combination can reach $(1,1,1)$ . Row reduction preserves the row space but changes the column space — which is why a basis for $C(A)$ must be read off $A$ .

IVVisualization 3 — Four Bases from One Reduction

Reducing $A$ to $R$ (and recording $E$ ) hands you a basis for each fundamental subspace — which part of $A$ , $R$ , or $E$ supplies which?

Subspace

basis for the selected subspace

💡 The rank $r = 2$ fixes every dimension: $r + (n - r) = n$ and $r + (m - r) = m$ , so the four subspaces balance in two pairs.

VSummary: All Four Bases from One Computation

For $A = \begin{bmatrix} 1 & 2 & 3 & 1 \\ 1 & 1 & 2 & 1 \\ 1 & 2 & 3 & 1 \end{bmatrix}$ , $R = \begin{bmatrix} 1 & 0 & 1 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$ , $E = \begin{bmatrix} -1 & 2 & 0 \\ 1 & -1 & 0 \\ -1 & 0 & 1 \end{bmatrix}$ with rank $r = 2$ :

Subspace	Basis	Dim	Source
Column Space $C(A)$	$\{(1,1,1),\; (2,1,2)\}$	$r = 2$	Pivot columns of A
Null Space $N(A)$	$\{(-1,-1,1,0),\; (-1,0,0,1)\}$	$n - r = 2$	Special solutions from R
Row Space $C(A^T)$	$\{(1,0,1,1),\; (0,1,1,0)\}$	$r = 2$	Nonzero rows of R
Left Null Space $N(A^T)$	$\{(-1,0,1)\}$	$m - r = 1$	Zero-row of E

Dimension check: $2 + 2 = 4 = n$ ✓ and $2 + 1 = 3 = m$ ✓

VIQuiz Questions

Question 1

When extracting the column space basis after row reduction, you identify the pivot column positions from $R$ . From which matrix do you then take the actual basis vectors?

✅ Correct! R tells you which columns are pivot columns, but A provides the actual basis vectors.

❌ Not quite. Row operations change the column space, so the vectors in R's columns differ from A's columns.

Show solution

Solution:

The column space basis must come from the original matrix A.

Row reduction tells us which columns are pivot columns (their positions), but the actual basis vectors must be taken from $A$ , not $R$ .

Why? Row operations change the column space. For example:

Column 1 of $A$ : $(1, 1, 1)$ — clearly in $C(A)$
Column 1 of $R$ : $(1, 0, 0)$ — a completely different vector

Every column of $R$ has $0$ in the third entry, so no combination of $R$ 's columns can produce $(1, 1, 1)$ . Thus $C(A) \neq C(R)$ .

Rule: $R$ identifies the positions; $A$ provides the vectors.

Question 2

For the $3 \times 4$ matrix $A$ from the video with rank $r = 2$ , what is the dimension of the null space $N(A)$ ?

✅ Correct! dim N(A) = n - r = 4 - 2 = 2. One special solution per free variable.

❌ Not quite. Remember: null space dimension = n - r, where n is the number of columns (not rows).

Show solution

Solution:

The null space dimension formula is:

$\dim N(A) = n - r$

where $n$ is the number of columns and $r$ is the rank:

$\dim N(A) = 4 - 2 = 2$

The two basis vectors (special solutions) come from setting each free variable to 1 in turn:

$x_3 = 1,\, x_4 = 0$ : back-substitute to get $\mathbf{s}_1 = (-1, -1, 1, 0)$
$x_3 = 0,\, x_4 = 1$ : back-substitute to get $\mathbf{s}_2 = (-1, 0, 0, 1)$

Each free column contributes exactly one special solution.

Question 3

True or False: Row operations preserve the row space of a matrix. That is, $A$ and its reduced row echelon form $R$ have the same row space.

✅ Correct! Row operations are reversible linear combinations of rows, so the row space stays the same.

❌ Not quite. Think about what each row operation does: it replaces a row with a linear combination of existing rows. And E is invertible, so the process is reversible.

Show solution

Solution: True

Row operations preserve the row space because:

Each row operation replaces a row with a linear combination of existing rows, so new rows stay in the span of old rows.
The elimination matrix $E$ is invertible ( $EA = R$ , $A = E^{-1}R$ ), so the process is reversible — old rows can be recovered from new ones.

Since each set of rows lies in the span of the other, the row spaces are identical: $C(A^T) = C(R^T)$ .

Important contrast: Row operations do not preserve the column space! This asymmetry is why we read the row space basis from $R$ (nonzero rows) but the column space basis from $A$ (pivot columns of the original).

Question 4

The elimination matrix for our example is $E = \begin{bmatrix} -1 & 2 & 0 \\ 1 & -1 & 0 \\ -1 & 0 & 1 \end{bmatrix}$ , where $EA = R$ . Which vector forms a basis for the left null space $N(A^T)$ , and why?

✅ Correct! The zero row of R points to the corresponding row of E, which gives the left null space basis vector.

❌ Not quite. Which row of R is the zero row? The matching row of E is the left null space basis vector, because (that row of E) times A equals the zero row.

Show solution

Solution:

The left null space basis comes from the rows of $E$ that correspond to zero rows in $R$ .

Since $EA = R$ , each row of $E$ times $A$ gives the corresponding row of $R$ :

Row 1 of $E$ times $A$ = row 1 of $R$ = $(1, 0, 1, 1) \neq \mathbf{0}$
Row 2 of $E$ times $A$ = row 2 of $R$ = $(0, 1, 1, 0) \neq \mathbf{0}$
Row 3 of $E$ times $A$ = row 3 of $R$ = $(0, 0, 0, 0) = \mathbf{0}$ ✓

So row 3 of $E$ , which is $(-1, 0, 1)$ , satisfies $(-1, 0, 1) \cdot A = \mathbf{0}^T$ .

Verification: $(-1)\begin{bmatrix}1 & 2 & 3 & 1\end{bmatrix} + (0)\begin{bmatrix}1 & 1 & 2 & 1\end{bmatrix} + (1)\begin{bmatrix}1 & 2 & 3 & 1\end{bmatrix} = \begin{bmatrix}0 & 0 & 0 & 0\end{bmatrix}$

Rows 1 and 3 of $A$ are identical, so their difference vanishes. The left null space has dimension $m - r = 3 - 2 = 1$ .

Solved: 0 / 4

One Row Reduction, Four Bases

IKey Formulas

IIVisualization 1 — Row Reduction, Step by Step

IIIVisualization 2 — Column Space Lives in AAA, Not RRR

IVVisualization 3 — Four Bases from One Reduction

VSummary: All Four Bases from One Computation

VIQuiz Questions

IIIVisualization 2 — Column Space Lives in $A$ , Not $R$