LINEAR-ALGEBRA · Interactive Practice | Unit 9 · Video 3

Every Matrix Splits Its Columns in Two

IKey Formulas

Formula	Name	Description
$\text{rank}(A) + \text{nullity}(A) = n$	Rank-Nullity Theorem	Pivot columns + free columns = total columns
$\text{rank}(A) = r$	Rank	Number of pivot positions after row reduction
$\text{nullity}(A) = n - r$	Nullity	Number of free variables = dimension of null space
$\dim(\text{Col}(A)) = r$	Column space dimension	Always equals the rank

IIVisualization 1 — Pivot Columns vs Free Columns

Row reduction sorts the columns of $A$ into two kinds — which anchor $C(A)$ , and which are already combinations of them?

Inspect column

matrix A · click a column

pivot → basis for C(A) free → combination of pivots

reduced form R · read the coefficients

💡 A basis for $C(A)$ is the pivot columns of the original $A$ , not of $R$ : row reduction keeps which columns are dependent, but changes the column vectors themselves.

IIIVisualization 2 — A Column Changes Sides

As $t$ drives the two rows of $A=\begin{bmatrix}1&2&3\\2&4&t\end{bmatrix}$ into dependence, which column loses its pivot — and where does it go?

Parameter t 6

matrix A → echelon U

after R₂ ← R₂ − 2R₁

the three columns

IVVisualization 3 — The Four Fundamental Subspaces

The rank $r$ fixes all four fundamental subspaces at once — so where do the leftover dimensions go?

Rows m 3 Cols n 4 Rank r 2

Row space

Null space

A
→

Column space

Left null space

💡 The column space and row space always share the same dimension $r$ — this equality of column rank and row rank is why "rank" is a single well-defined number.

VQuiz Questions

Question 1 · Direct Application

A matrix $A$ is $3 \times 5$ (3 rows, 5 columns) and has rank 3. What is the nullity of $A$ ?

✅ Correct! nullity = n − rank = 5 − 3 = 2.

❌ Not quite. Remember: nullity = n − rank, where n is the number of columns (not rows).

Show solution

Solution:

By the Rank-Nullity Theorem: $\text{rank}(A) + \text{nullity}(A) = n$ $3 + \text{nullity}(A) = 5$ $\text{nullity}(A) = 5 - 3 = 2$

The nullity is 2, meaning the null space is 2-dimensional (there are 2 free variables after row reduction).

Question 2 · Common Pitfall

After row reducing a matrix $A$ , the echelon form has pivots in columns 1 and 2. To find a basis for the column space of $A$ , which columns should you use?

✅ Correct! Always take the basis vectors from the original matrix. Row reduction only tells you which columns to pick.

❌ Not quite. Row operations change the column vectors themselves. The echelon form tells you which columns to choose, but the actual vectors must come from the original matrix.

Show solution

Solution:

Row reduction identifies which columns are pivot columns, but you must take those columns from the original matrix $A$ , not from the echelon form.

Row operations preserve column dependencies (which columns are linear combinations of others) but change the actual column vectors. The echelon form's columns span a different subspace than the original columns.

Answer: Columns 1 and 2 of the original matrix $A$ .

Question 3 · Multi-Step Reasoning

Consider the matrix from the video: $A = \begin{bmatrix} 1 & 2 & 3 & 1 \\ 1 & 1 & 2 & 1 \\ 1 & 2 & 3 & 1 \end{bmatrix}$

Its echelon form is $\begin{bmatrix} 1 & 2 & 3 & 1 \\ 0 & -1 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$ . How many special solutions (basis vectors for the null space) does this matrix have?

✅ Correct! There are 2 free variables (columns 3 and 4), producing exactly 2 special solutions that form a basis for the null space.

❌ Not quite. Count the non-pivot columns — each one corresponds to a free variable and produces exactly one special solution.

Show solution

Solution:

From the echelon form, identify the pivots:

Pivot in column 1 (the 1 in row 1)
Pivot in column 2 (the −1 in row 2)

So $\text{rank} = 2$ and there are $n - r = 4 - 2 = 2$ free columns (columns 3 and 4).

Each free variable produces exactly one special solution:

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

Answer: 2 special solutions, one per free variable.

Question 4 · Conceptual Understanding

True or False: A $4 \times 6$ matrix can have nullity equal to 0.

✅ Correct! A 4×6 matrix has at most rank 4, so nullity ≥ 6 − 4 = 2. The null space must be at least 2-dimensional.

❌ Not quite. Think about the maximum possible rank: can a matrix with 4 rows really have 6 pivots? Remember rank ≤ min(m, n).

Show solution

Solution:

If nullity = 0, the Rank-Nullity Theorem gives: $\text{rank}(A) + 0 = 6 \implies \text{rank}(A) = 6$

But $\text{rank}(A) \leq \min(m, n) = \min(4, 6) = 4$ . A matrix with only 4 rows can have at most 4 pivots.

So $\text{rank}(A) = 6$ is impossible, meaning nullity = 0 is also impossible. In fact: $\text{nullity}(A) = 6 - \text{rank}(A) \geq 6 - 4 = 2$

A $4 \times 6$ matrix must always have nullity $\geq 2$ .

Answer: False.

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