Recipe

Find the null space and verify its dimension

Use when: you're given a matrix $A$ , need $\operatorname{Nul}(A)$ itself, and want the pivot/free column count as a built-in check on the answer

Goal: a spanning set for $\operatorname{Nul}(A)$ together with confirmation that pivot columns plus free columns equals $A$ 's total column count

Taught in Null space, revisited · extends Find the null space of a matrix

Quick recipe

Row-reduce $A$ to RREF
Identify the pivot columns and the free columns
Solve the pivot variables in terms of the free ones, and write the spanning set
Confirm pivot columns plus free columns equal $A$ 's total columns

1
The right-hand side of $A\mathbf x=\mathbf 0$ stays zero through every row operation, so row-reduce $A$ itself, no augmented column needed.

Example

For $A=\begin{pmatrix}1&-1&2\\3&-2&5\end{pmatrix}$ : $R_2\to R_2-3R_1$ gives $(0,1,-1)$ ; clearing column 2 out of row 1 with $R_1\to R_1+R_2$ finishes it: $\operatorname{rref}(A) = \begin{pmatrix}1&0&1\\0&1&-1\end{pmatrix}$

Why: this is the same row-reduction step the null-space recipe already established — nothing about finding $\operatorname{Nul}(A)$ itself changes here.
2
Every column with a pivot is a bound variable; every column without one is free.

Example

Columns 1 and 2 of the RREF each have a pivot; column 3 does not, so $x_3$ is the one free variable.

Why: this split is what both halves of this recipe are built on — the free columns build the spanning set, and counting both kinds together is this recipe's new final check.
3
Isolate each pivot variable using its RREF row, substitute into $\mathbf x$ , then factor out the free variable.

Example

Row 1: $x_1+x_3=0 \Rightarrow x_1=-x_3$ . Row 2: $x_2-x_3=0 \Rightarrow x_2=x_3$ . So $\mathbf x = x_3\begin{pmatrix}-1\\1\\1\end{pmatrix} \quad\Longrightarrow\quad \boxed{\operatorname{Nul}(A) = \operatorname{span}\left\{\begin{pmatrix}-1\\1\\1\end{pmatrix}\right\}}$ Check: $A(-1,1,1)^{\mathsf T}$ has first entry $1(-1)-1(1)+2(1)=0$ and second entry $3(-1)-2(1)+5(1)=0$ . ✓

Why: this is exactly the null-space recipe's back-substitution and spanning-vector construction, unchanged.
4
Count the pivots found in Step 2, count the free columns found in the same step, and add them — the sum must equal $A$ 's number of columns, $n$ , every single time.

Example

$A$ has 2 pivot columns and 1 free column: $2+1=3$ , matching $A$ 's 3 columns exactly.

Why: every column is either a pivot column or a free column, never both and never neither, so the two counts partitioning $n$ isn't a coincidence to double-check case by case — it's this section's seed of the rank-nullity relationship, and a wrong pivot count almost always shows up as a mismatch here first.

Common mistakes

Stopping after Step 3 and never running the Step 4 count — the whole point of revisiting this recipe here is catching a miscounted pivot before it corrupts $\dim\operatorname{Nul}(A)$ .
Confusing $\operatorname{Nul}(A)$ 's home space ( $\mathbb R^n$ , one entry per column of $A$ ) with $\operatorname{Col}(A)$ 's ( $\mathbb R^m$ , one entry per row) when writing the spanning vector's size.
Recounting pivots from a partially row-reduced matrix instead of the final RREF.
Treating the pivot-plus-free count as evidence about $\operatorname{Nul}(A)$ 's spanning vectors themselves, rather than just a count — the check confirms the NUMBER of free variables, not that any individual vector was computed correctly.