Recipe

Find the left null space of a matrix

Use when: you're given a matrix $A$ and asked for every row vector $\mathbf y$ with $\mathbf y^{\mathsf T}A=\mathbf 0^{\mathsf T}$ , rather than an ordinary (right) null space

Goal: a spanning set for the left null space, found by running the ordinary null-space recipe on $A^{\mathsf T}$ instead of $A$

Taught in The left null space · extends Find the null space of a matrix

Quick recipe

Transpose $A$ to turn the problem into an ordinary null space
Row-reduce $A^{\mathsf T}$ and identify its free columns
Solve the pivot variables in terms of the free one, and write the spanning set
Verify by multiplying the row vector against $A$ directly, not $A^{\mathsf T}$

1
Take $A^{\mathsf T}$ — swap rows and columns. The defining equation $\mathbf y^{\mathsf T}A=\mathbf 0^{\mathsf T}$ becomes $A^{\mathsf T}\mathbf y=\mathbf 0$ once transposed, an ordinary null-space problem in $\mathbf y$ .

Example

For $C = \begin{pmatrix}1&3\\2&1\\1&-2\end{pmatrix}$ : $C^{\mathsf T} = \begin{pmatrix}1&2&1\\3&1&-2\end{pmatrix}$

Why: this single transpose is what lets every already-proven null-space fact — subspace, found by RREF and free variables, dimension from pivot/free counting — transfer over unchanged, just applied to $A^{\mathsf T}$ instead of $A$ .
2
Run ordinary row reduction on $A^{\mathsf T}$ and read off which columns end up without a pivot, exactly as in the null-space recipe.

Example

$R_2 \to R_2-3R_1$ turns row 2 into $(0,-5,-5)$ ; dividing by $-5$ gives $(0,1,1)$ ; clearing column 2 out of row 1 with $R_1\to R_1-2R_2$ finishes it: $\operatorname{rref}(C^{\mathsf T}) = \begin{pmatrix}1&0&-1\\0&1&1\end{pmatrix}$ Columns 1 and 2 have pivots; column 3 is free.

Why: the null-space recipe's row-reduce-and-free-variable machinery doesn't care whether the matrix in front of it is called $A$ or $A^{\mathsf T}$ — it's still just a matrix.
3
Isolate each pivot variable using its RREF row, substitute, and factor out the free variable — exactly the null-space recipe's construction, with $\mathbf y$ 's entries in place of $\mathbf x$ 's.

Example

Row 1: $y_1-y_3=0 \Rightarrow y_1=y_3$ . Row 2: $y_2+y_3=0 \Rightarrow y_2=-y_3$ . So $\mathbf y = y_3\begin{pmatrix}1\\-1\\1\end{pmatrix} \quad\Longrightarrow\quad \boxed{\text{left null space of } C = \operatorname{span}\left\{\begin{pmatrix}1\\-1\\1\end{pmatrix}\right\}}$

Why: the left null space is defined as this exact spanning set — a subspace of $\mathbb R^m$ ( $m=3$ here, $C^{\mathsf T}$ 's own column count), not $\mathbb R^n$ .
4
Compute $\mathbf y^{\mathsf T}A$ by hand (weight each row of $A$ by the matching entry of $\mathbf y$ , then add down each column) and confirm every entry is $0$ — this checks the original equation, not the transposed one.

Example

$\mathbf y^{\mathsf T}C = (1,-1,1)\begin{pmatrix}1&3\\2&1\\1&-2\end{pmatrix} = \big(1(1)-1(2)+1(1),\ \ 1(3)-1(1)+1(-2)\big) = (0,0) \ \checkmark$

Why: Step 4's check confirms the spanning vector solves the equation this section actually asked about, $\mathbf y^{\mathsf T}A=\mathbf 0^{\mathsf T}$ — a slip introduced while transposing in Step 1 wouldn't be caught by re-checking $A^{\mathsf T}\mathbf y=\mathbf 0$ alone, since both sides of that equation were built from the same (possibly mistransposed) matrix.

Common mistakes

Transposing $A$ incorrectly (swapping the wrong pair of indices) and never catching it because the rest of the computation proceeds consistently on the wrong matrix.
Reporting the left null space as a subspace of $\mathbb R^n$ instead of $\mathbb R^m$ — it's built from $A^{\mathsf T}$ 's columns, which number $m$ , not $n$ .
Verifying against $A^{\mathsf T}\mathbf y=\mathbf 0$ only, instead of the original $\mathbf y^{\mathsf T}A=\mathbf 0^{\mathsf T}$ — the whole point of Step 4 is checking the un-transposed statement.
Writing the check as $A\mathbf y$ instead of $\mathbf y^{\mathsf T}A$ — the row vector multiplies from the left, not the right.