A plane has many different bases, but every one of them has two vectors. That number, the same for every basis, measures the size of a subspace.
Key idea
Why it works
| Space | Dimension |
|---|---|
| $\mathbb{R}^n$ | $n$ |
| A line through the origin | $1$ |
| A plane through the origin | $2$ |
| $\{\mathbf{0}\}$ | $0$ (the empty set is its basis) |
| $\mathbb{P}_n$, polynomials of degree at most $n$ | $n + 1$ |
| $m \times n$ matrices | $mn$ |
Worked example: Rank and nullity of a matrix
- Row reduce: $\begin{bmatrix} 1 & 0 & -3 & -4 \\ 0 & 1 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{bmatrix}$.
- Two pivot columns: $\operatorname{rank} A = 2$.
- Four columns, two pivots, so two free variables: $\dim \operatorname{Nul} A = 2$.
Answer: Rank 2, nullity 2.
Worked example: Dimension of a span
- These are the columns of the matrix in the first example, which has 2 pivots.
- So the span is 2-dimensional: a plane in $\mathbb{R}^3$, with basis $\{(1, 0, 1), (2, 1, 3)\}$ (the pivot columns).
Answer: $2$
Worked example: Dimension of a solution space
- It is the null space of $\begin{bmatrix} 1 & 0 & -1 & 0 & 2 \\ 0 & 1 & 0 & 1 & 0 \end{bmatrix}$, already in RREF.
- Two pivots, five columns: three free variables, so dimension $3$.
Answer: $3$
For a subspace given as a span, the dimension is the rank of the matrix whose columns are the spanning vectors. For a subspace given by equations, it is the number of free variables. These are the two ways subspaces usually appear, and both reduce to counting pivots.
Two consequences you can use right away. In a $p$-dimensional space, any $p$ independent vectors automatically span (so they form a basis), and any $p$ vectors that span are automatically independent. And a subspace of $\mathbb{R}^n$ never has dimension bigger than $n$.