LINEAR-ALGEBRA · Interactive Practice | Unit 7 · Video 1

When Elimination Hits a Zero

IKey Formulas

Formula	Name	Description
$\text{rank}(A) = r$	Rank	Number of pivots in echelon form
Free variables $= n - r$	Free variable count	Total columns minus rank
$\dim(\text{Null}(A)) = n - r$	Null space dimension	Equals number of free variables
Pivot columns $\to$ independent	Column classification	Columns containing pivots are linearly independent

IIVisualization 1 — When a Zero Pivot Appears

Elimination drives this $3\times4$ matrix toward echelon form — what happens when a pivot position turns up zero?

Step 0

matrix

pivot free column zero pivot

Step 0 · Original A

Hidden structure:

c_2 = 2c_1

and

R_3 = R_1 + R_2

— elimination will expose both.

Step 1 · Clear column 1

R_2 \leftarrow R_2 - 2R_1

. Pivot

a_{11}=1

; row 2 becomes

[\,0\;0\;2\;4\,]

Step 2 · Zero in the pivot

R_3 \leftarrow R_3 - 3R_1

leaves

(2,2)=0

with

(3,2)=0

below it — no row to swap, so skip column 2.

Step 3 · Echelon form U

R_3 \leftarrow R_3 - R_2

zeroes row 3. Pivots in columns 1 and 3 → rank

=2

; columns 2 and 4 are free.

IIIVisualization 2 — Pivot Columns vs. Free Columns

What decides whether a column keeps a pivot or turns free once elimination finishes?

Matrix

original A

echelon form U

IVVisualization 3 — How Rank Splits the Variables

Rank $r$ splits $n$ columns into $r$ pivot and $n-r$ free variables — when is the null space trivial?

Rows m 3 Cols n 4 Rank r 2

💡 The row count $m$ only caps how large $r$ can be ( $r \le \min(m,n)$ ); the free-variable count $n-r$ — and with it the null-space dimension — depends on $n$ and $r$ alone.

VQuiz Questions

Question 1

Given the echelon form: $U = \begin{bmatrix} 1 & 2 & 2 & 2 \\ 0 & 0 & 2 & 4 \\ 0 & 0 & 0 & 0 \end{bmatrix}$

What is the rank of the original matrix $A$ ?

✅ Correct! There are 2 pivots (at positions (1,1) and (2,3)), so the rank is 2.

❌ Not quite. Count the leading nonzero entries — one per nonzero row of U.

Show solution

Solution:

The rank equals the number of pivots (leading nonzero entries in each nonzero row) in echelon form.

Looking at $U$ :

Row 1: Leading entry is $1$ in column 1 → Pivot #1
Row 2: Leading entry is $2$ in column 3 → Pivot #2
Row 3: All zeros → no pivot

$\text{rank}(A) = \text{number of pivots} = 2$

Question 2

In the same echelon form $U$ above, which variables are the free variables?

✅ Correct! Columns 2 and 4 have no pivots, so $x_2$ and $x_4$ are free variables.

❌ Not quite. Free variables correspond to columns without pivots. Check which columns contain the leading nonzero entry of each row.

Show solution

Solution:

Columns with pivots → pivot variables. Columns without pivots → free variables.

From $U = \begin{bmatrix} 1 & 2 & 2 & 2 \\ 0 & 0 & 2 & 4 \\ 0 & 0 & 0 & 0 \end{bmatrix}$ :

Column	Pivot?	Variable type
Column 1	Yes (the 1 in row 1)	$x_1$ is pivot
Column 2	No	$x_2$ is free ✓
Column 3	Yes (the 2 in row 2)	$x_3$ is pivot
Column 4	No	$x_4$ is free ✓

The free variables are $x_2$ and $x_4$ .

Question 3

True or False: Row reduction (elimination) preserves the column space of a matrix.

That is, if $U$ is obtained from $A$ by row operations, then the column space of $U$ equals the column space of $A$ .

✅ Correct! Elimination preserves the null space (and row space), but not the column space. For column space analysis, always use the original matrix $A$ .

❌ Not quite. Row operations modify the entries in each column. The columns of $U$ are different vectors from the columns of $A$ , so the spaces they span are generally different.

Show solution

Solution: False.

Row operations change the actual entries in each column, so the column space of $U$ is generally different from the column space of $A$ .

What elimination does preserve:

Property	Preserved?	Why?
Null space	✅ Yes	$Ax = 0$ and $Ux = 0$ have identical solutions
Row space	✅ Yes	Row operations keep rows in the same span
Column space	❌ No	The columns of $U$ are different vectors
Rank	✅ Yes	Same number of pivots

Rule of thumb: For null space questions, use $U$ . For column space questions, go back to $A$ .

Question 4

A $5 \times 7$ matrix $A$ has rank 3. What is the dimension of the null space of $A$ ?

✅ Correct! With $n = 7$ columns and rank $r = 3$ , the null space has dimension $7 - 3 = 4$ .

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

Show solution

Solution:

The null space dimension equals the number of free variables:

$\dim(\text{Null}(A)) = n - r$

where $n$ = number of columns and $r$ = rank.

$\dim(\text{Null}(A)) = 7 - 3 = 4$

The null space is a 4-dimensional subspace of $\mathbb{R}^7$ .

Important: The number of rows ( $m = 5$ ) does not appear in this formula. Only $n$ (columns) and $r$ (rank) determine the null space dimension.

$r = 3$ pivot variables (determined by back-substitution)
$n - r = 4$ free variables (can be set to any value)

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