Every matrix comes with two important subspaces. One collects the inputs that $A$ sends to $\mathbf{0}$ (the solutions of a homogeneous system); the other collects all possible outputs.
For an $m \times n$ matrix $A$: the null space $\operatorname{Nul} A$ $= \{\mathbf{x} \text{ in } \mathbb{R}^n : A\mathbf{x}$ $= \mathbf{0}\}$, and the column space $\operatorname{Col} A$ = the span of the columns of $A$ = $\{A\mathbf{x}\}$, a subspace of $\mathbb{R}^m$.
Why it works
$\operatorname{Nul} A$ is a subspace: $A\mathbf{0} = \mathbf{0}$; if $A\mathbf{u} = A\mathbf{v} = \mathbf{0}$, then $A(\mathbf{u} + \mathbf{v}) = \mathbf{0}$ and $A(c\mathbf{u}) = \mathbf{0}$. $\operatorname{Col} A$ is a span, so it is a subspace. Notice they live in different places: null space vectors have $n$ entries (inputs), column space vectors have $m$ entries (outputs).
| Nul A | Col A | |
|---|---|---|
| Lives in | $\mathbb{R}^n$ (inputs) | $\mathbb{R}^m$ (outputs) |
| Described by | a condition: $A\mathbf{x} = \mathbf{0}$ | a list: all combinations of the columns |
| Is $\mathbf{v}$ in it? | Easy: compute $A\mathbf{v}$ | Work: is $A\mathbf{x} = \mathbf{v}$ consistent? |
| Basis from | the general solution of $A\mathbf{x} = \mathbf{0}$ | the pivot columns of $A$ |
Worked example: A basis for Nul A
Find vectors that span $\operatorname{Nul} A$ for $A = \begin{bmatrix} 1 & 2 & 0 & -1 \\ 2 & 4 & 1 & 1 \\ -1 & -2 & 1 & 4 \end{bmatrix}$.
- $R_2 \to R_2 - 2R_1$, $R_3 \to R_3 + R_1$, then $R_3 \to R_3 - R_2$: $\begin{bmatrix} 1 & 2 & 0 & -1 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix}$.
- Pivots in columns 1 and 3; $x_2 = s$ and $x_4 = t$ are free. Then $x_1 = -2s + t$ and $x_3 = -3t$.
- $\mathbf{x} = s\begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \end{bmatrix} + t\begin{bmatrix} 1 \\ 0 \\ -3 \\ 1 \end{bmatrix}$, so $\operatorname{Nul} A$ $= \operatorname{Span}\left\{\begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 0 \\ -3 \\ 1 \end{bmatrix}\right\}$.
- These two are independent (look at entries 2 and 4), so they form a basis. Check: $A\begin{bmatrix} 1 \\ 0 \\ -3 \\ 1 \end{bmatrix} = \mathbf{0}$ ✓.
Answer: Two vectors, one per free variable.
Worked example: A basis for Col A
Find a basis for $\operatorname{Col} A$ for the same $A$.
- The pivot columns are 1 and 3, so take columns 1 and 3 of $A$: $\begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}$.
- The other columns depend on these: column 2 is $2\times$ column 1, and column 4 is $-(\text{column }1) + 3(\text{column }3)$, exactly as the reduced matrix says.
Answer: $\left\{\begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}\right\}$
Why it works
Why the pivot columns of $A$? Row operations do not change the solutions of $A\mathbf{x} = \mathbf{0}$, and those solutions are exactly the dependence relations among the columns. So the columns of $A$ satisfy the same relations as the columns of its reduced form $R$. In $R$ the pivot columns are clearly independent and every other column is a combination of them; therefore the same is true for the corresponding columns of $A$.
Common mistake
Taking the pivot columns of the reduced matrix as a basis for $\operatorname{Col} A$. Row operations keep the relations between columns but change the column space itself. Here $R$ has pivot columns $(1, 0, 0)$ and $(0, 1, 0)$, but $(1, 0, 0)$ is not even in $\operatorname{Col} A$. Use $R$ to locate the pivots, then copy the columns from $A$.
Key idea
$\operatorname{Nul} A$ (in $\mathbb{R}^n$) holds the solutions of $A\mathbf{x} = \mathbf{0}$; its basis comes from the free variables. $\operatorname{Col} A$ (in $\mathbb{R}^m$) is spanned by the columns; its basis is the pivot columns of the original $A$.