Rank counts pivot columns; nullity counts free columns. Together they account for every column, and that simple count connects the size of the outputs to the size of what gets lost. It sharpens what we learned about one-to-one and onto maps.
Key idea
Why it works
The four fundamental subspaces
The row space $\operatorname{Row} A$ is the span of the rows of $A$ (a subspace of $\mathbb{R}^n$). Row operations replace rows by combinations of rows, so they never change the row space, and the nonzero rows of an echelon form are a basis. That gives $\dim \operatorname{Row} A = \operatorname{rank} A$: the row rank equals the column rank. Applying rank–nullity to $A^{T}$ as well gives four subspaces with linked sizes:
| Subspace | Lives in | Dimension |
|---|---|---|
| Column space $\operatorname{Col} A$ | $\mathbb{R}^m$ | $r$ |
| Left null space $\operatorname{Nul} A^{T}$ | $\mathbb{R}^m$ | $m - r$ |
| Row space $\operatorname{Row} A = \operatorname{Col} A^{T}$ | $\mathbb{R}^n$ | $r$ |
| Null space $\operatorname{Nul} A$ | $\mathbb{R}^n$ | $n - r$ |
Worked example: All four dimensions
- $\dim \operatorname{Col} A = \dim \operatorname{Row} A = 3$.
- $\dim \operatorname{Nul} A = 7 - 3 = 4$ (subspace of $\mathbb{R}^7$).
- $A^{T}$ is $7 \times 5$ with rank $3$, so $\dim \operatorname{Nul} A^{T} = 5 - 3 = 2$ (subspace of $\mathbb{R}^5$).
Answer: 3, 3, 4, 2.
Worked example: Is it possible?
- Rank–nullity would need $\operatorname{rank} A = 6 - 1 = 5$.
- But the rank is at most the number of rows, $4$. So the nullity is at least $6 - 4 = 2$.
Answer: No.
Worked example: Reading a matrix
- Row reduce: $\begin{bmatrix} 1 & 3 & 0 & 0 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{bmatrix}$. Two pivots.
- Rank $2$, nullity $4 - 2 = 2$.
- The range has dimension $2 < 3$, so the map is not onto $\mathbb{R}^3$.
Answer: Rank 2, nullity 2, not onto.
Common mistake