Dimension & Rank

When you choose a basis for a subspace, you have many options — there are infinitely many valid bases for most spaces. But no matter which basis you pick, you always count the same number of vectors. This invariant is the dimension, and it is the most fundamental numerical property of a vector space. The entire theory of rank and the rank-nullity theorem rest on this single fact.

Finite-dimensional spaces

A vector space $V$ is finite-dimensional if there exists a finite set of vectors that spans $V$. All of $F^n$, all matrix spaces $M_{m,n}(F)$, and every subspace of a finite-dimensional space are finite-dimensional. A space with no finite spanning set — such as the space of all polynomials over $F$ — is infinite-dimensional.

This checkpoint focuses entirely on finite-dimensional spaces.

All bases have the same size

Linear Subspace introduced bases and stated without proof that all bases of a space have equal size. Here is why that is true.

Lemma (replacement): If $\{v_1, \ldots, v_n\}$ spans $V$ and $\{u_1, \ldots, u_m\}$ is linearly independent in $V$, then $m \le n$.

Proof sketch: You can replace vectors in the spanning set one at a time with vectors from the independent set while maintaining a spanning set at each step. After at most $n$ replacements, the spanning set is exhausted. Because the $u_i$ are independent, none can be “used up” before all $n$ replacement slots are filled — so $m \le n$. $\square$

Theorem: Any two bases of a finite-dimensional vector space $V$ have the same number of elements.

Proof: Let $\mathcal{B} = \{b_1, \ldots, b_n\}$ and $\mathcal{C} = \{c_1, \ldots, c_m\}$ be two bases of $V$. Since $\mathcal{B}$ spans $V$ and $\mathcal{C}$ is linearly independent, the lemma gives $m \le n$. Since $\mathcal{C}$ spans $V$ and $\mathcal{B}$ is linearly independent, the same argument gives $n \le m$. Therefore $m = n$. $\square$

Definition of dimension

Because every basis has the same size, the following is well-defined:

The dimension of a finite-dimensional vector space $V$, written $\dim V$ (or $\dim_F V$ to emphasize the field), is the number of vectors in any basis of $V$.

By convention, the trivial space has $\dim\{\mathbf{0}\} = 0$, taking the empty set as its basis.

Dimensions of common spaces

Space	Standard basis	Dimension
$F^n$	Standard unit vectors $e_1, \ldots, e_n$	$n$
$M_{m,n}(F)$	Matrices $E_{ij}$ (one $1$, rest $0$s)	$mn$
$F[x]_{\le d}$ (polynomials of degree $\le d$)	$1, x, x^2, \ldots, x^d$	$d+1$
$\{\mathbf{0}\}$	$\emptyset$	$0$

The dimension tells you how many independent parameters are needed to describe every element of the space: an element of $F^n$ needs $n$ coordinates, a polynomial of degree $\le d$ needs $d+1$ coefficients, and so on.

Dimension and subspaces

Let $W$ be a subspace of a finite-dimensional space $V$. Then $W$ is also finite-dimensional, and:

1. $\dim W \le \dim V$.
2. $\dim W = \dim V$ if and only if $W = V$.

Point 2 gives a useful shortcut: to prove that a subspace $W$ equals all of $V$, it suffices to find $\dim V$ linearly independent vectors inside $W$. If $W$ contains a linearly independent set of the right size, it must be all of $V$.

Extending and reducing bases

Two practical facts follow from the replacement lemma:

• Any linearly independent set in $V$ can be extended to a basis of $V$ (by adding vectors one at a time).
• Any spanning set of $V$ can be reduced to a basis of $V$ (by removing redundant vectors one at a time).

Together these say: a basis is both a minimal spanning set and a maximal linearly independent set inside $V$.

Rank as a dimension

The rank of a matrix $A$, introduced in Row and Column Spaces, is exactly a dimension:

$\text{rank}(A) = \dim(\text{row}(A)) = \dim(\text{col}(A)).$

The nullity of $A$ is the dimension of its kernel — the solution set of the homogeneous system $Ax = \mathbf{0}$:

$\text{nullity}(A) = \dim(\ker(A)).$

For an $m \times n$ matrix $A$, these two quantities are related by the Rank-Nullity Theorem:

\text{rank}(A) + \text{nullity}(A) = n. \tag{1}

Every column of $A$ is either a pivot column (contributing $1$ to the rank) or a free column (contributing $1$ to the nullity). The $n$ columns are partitioned between the two, with no column counted twice and no column left out.

Summary

• A vector space is finite-dimensional if it has a finite spanning set.
• All bases of a finite-dimensional space have the same number of elements — this follows from the replacement lemma applied in both directions.
• The dimension $\dim V$ is this common basis size: $\dim F^n = n$, $\dim M_{m,n}(F) = mn$, $\dim\{\mathbf{0}\} = 0$.
• For a subspace $W \subseteq V$: $\dim W \le \dim V$, with equality if and only if $W = V$.
• Any linearly independent set in $V$ extends to a basis; any spanning set of $V$ reduces to a basis.
• Rank and nullity are dimensions: $\text{rank}(A) = \dim(\text{col}(A)) = \dim(\text{row}(A))$ and $\text{nullity}(A) = \dim(\ker(A))$, with $\text{rank}(A) + \text{nullity}(A) = n$.