Basis and Dimension of a Vector Space

Basis — two properties in one

A basis of a vector space $V$ is a set of vectors that is:

1. Linearly independent (no redundancy), and
2. Spans $V$ (reaches every vector via linear combinations).

Every vector in $V$ has a unique representation as a linear combination of basis vectors.

Dimension

The dimension of $V$, written $\dim V$, is the number of vectors in any basis. Remarkably, all bases of a given $V$ have the same size — dimension is well-defined.

Examples:

• $\dim \mathbb{R}^n = n$, with standard basis $\{\mathbf{e}_1, \mathbf{e}_2, \ldots, \mathbf{e}_n\}$.
• $\dim \{\mathbf{0}\} = 0$.
• $\dim P_n$ (polynomials of degree $\le n$) $= n + 1$; basis $\{1, x, x^2, \ldots, x^n\}$.
• $\dim M_{m \times n}(\mathbb{R}) = m \cdot n$.

Finding a basis for the column space

The column space $\operatorname{Col}(A)$ is the span of the columns of $A$. Its basis consists of the pivot columns of the ORIGINAL matrix (not the RREF columns themselves — the pivot positions of the RREF tell you which original columns form the basis).

Step-by-step

$A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 6 & 8 \end{pmatrix}$

Step 1 — Row reduce to RREF.

$R_2 \to R_2 - 2 R_1$: $(2, 4, 5) - 2(1, 2, 3) = (0, 0, -1)$. $R_3 \to R_3 - 3 R_1$: $(3, 6, 8) - 3(1, 2, 3) = (0, 0, -1)$.

$\begin{pmatrix} 1 & 2 & 3 \\ 0 & 0 & -1 \\ 0 & 0 & -1 \end{pmatrix}$

$R_3 \to R_3 - R_2$: $(0, 0, -1) - (0, 0, -1) = (0, 0, 0)$.

$\begin{pmatrix} 1 & 2 & 3 \\ 0 & 0 & -1 \\ 0 & 0 & 0 \end{pmatrix}$

$R_2 \to -R_2$: $(0, 0, 1)$. Then $R_1 \to R_1 - 3R_2$: $(1, 2, 0)$.

$\text{RREF}(A) = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}$

Step 2 — Identify pivot columns. Pivots appear in columns 1 and 3.

Step 3 — The basis is the corresponding columns of the ORIGINAL $A$ (not the RREF): $\text{Basis for } \operatorname{Col}(A) = \left\{ \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}, \begin{pmatrix} 3 \\ 5 \\ 8 \end{pmatrix} \right\}$

So $\dim \operatorname{Col}(A) = 2 = \operatorname{rank}(A)$.

Why we use original columns, not RREF columns

Row reduction changes the column space — RREF columns span a different set! But row reduction preserves the linear relations among columns. So the pivot positions identify which original columns are independent; the pivot columns of RREF do not themselves form a basis of $\operatorname{Col}(A)$.

Column 2 as a dependency

In the RREF, column 2 has no pivot. The relation is visible: column 2 = 2 · column 1. In the original matrix, this corresponds to $\mathbf{a}_2 = 2 \mathbf{a}_1$ (check: $(2, 4, 6) = 2 \cdot (1, 2, 3)$ ✓). So $\mathbf{a}_2$ is redundant in the basis.

Common mistakes

• Using RREF columns instead of original columns. The pivot positions are right, but the actual basis vectors come from $A$.
• Confusing column space with row space. They have the same dimension (both equal rank) but live in different ambient spaces.
• Counting vectors that aren't independent. A basis has no redundancy.

Try it in the visualization

Each column of $A$ is drawn in 3D. Pivot columns pulse gold; non-pivot columns fade toward gray and show their linear combination back to the pivots. Dimension updates live as you change entries.