Let's work through a concrete example and extract the bases for its fundamental subspaces. Here's a simple $3\times 4$ matrix:

Step 1: Row Reduce to Find Rank and Row Space

We perform row operations to reduce ( A ) to row echelon form:

From this, we see:

• Rank = 2 (two non-zero rows)
• Row space basis: the non-zero rows of RREF

Step 2: Null Space Basis

We solve ( A\vec{x} = \vec{0} ). From RREF, we express leading variables in terms of free variables:

Let ( $x_2=s$ ), ($x_3=t$ )

So the general solution is:

Thus, the null space basis is:

Step 3: Column Space (Vector Space of A)

We look at the original matrix and identify pivot columns from RREF: columns 1 and 4.

So the column space basis is:

Summary Table

Rank of A

• The rank is the number of leading 1s (pivot positions) in the RREF.
• There are 2 pivot positions → Rank(A) = 2

Nullity of A

• Nullity is the number of free variables in the homogeneous system $A\overrightarrow{x}=\overrightarrow{0}$
• Total number of columns = 4
• Nullity = Number of columns − Rank = ( 4 - 2 = 2 )

So:

This satisfies the Rank–Nullity Theorem: