The Dimension of a Vector Space and Rank

Coordinates showed that an $n$-vector basis makes a space behave like ${\mathbb{R}}^n$. The next question is whether that $n$ is a property of the space or an accident of the basis. It is a property: every basis of a given space has the same number of vectors. That count is the dimension, and applied to the column space of a matrix it becomes the rank — the single number that ties together a matrix's rows, columns, and null space.¹

Dimension is well defined

The key fact is a counting bound: a basis of $n$ vectors caps the size of any independent set.

The proof runs through coordinates. If $\{u_1,\dots ,u_p\}$ has $p>n$ vectors, their coordinate vectors $[u_1{]}_{\mathcal{B}},\dots ,[u_p{]}_{\mathcal{B}}$ are $p$ vectors in ${\mathbb{R}}^n$, more vectors than entries, so they are dependent. The coordinate mapping is an isomorphism, so a dependence relation among the coordinate vectors is a dependence relation among the $u_i$ themselves. From this, equal size of all bases follows at once.

Given two bases ${\mathcal{B}}_1$ of $n$ vectors and ${\mathcal{B}}_2$, independence of ${\mathcal{B}}_2$ against the basis ${\mathcal{B}}_1$ forces $|{\mathcal{B}}_2|\le n$, and the symmetric argument forces $|{\mathcal{B}}_2|\ge n$. So $|{\mathcal{B}}_2|=n$, and the count is unambiguous.

A few dimensions to anchor the definition:

• $\dim {\mathbb{R}}^n=n$, from the standard basis $\{e_1,\dots ,e_n\}$.
• $\dim {\mathbb{P}}_n=n+1$, from $\{1,t,\dots ,t^n\}$.
• $\mathbb{P}$, the space of all polynomials, is infinite-dimensional, and so is the space $C(\mathbb{R})$ of continuous functions.

Subspaces of ${\mathbb{R}}^3$ by dimension

Dimension classifies the subspaces of ${\mathbb{R}}^3$ completely. The value of $\dim$ names the geometric type.

Building and recognizing bases

Two theorems make dimension a practical tool: independent sets can always be grown to a basis, and once the dimension is known, either half of the basis definition suffices.

Adding vectors outside the current span keeps a set independent (by the predecessor test), and the theorem bounding independent sets by a basis caps how far the process can run, so it terminates at a basis.

With the right number of vectors, spanning and independence each imply the other. This shortcut is decisive in applied settings — solutions of a differential or difference equation — where independence is easy to verify but spanning is not.

Dimensions of the null and column spaces

The basis-finding procedures for a matrix immediately give the two dimensions.

The null-space method produces one independent vector per free variable, and the pivot columns are a basis for the column space. So both dimensions are read off a single echelon form.

Rank and the row space

The rows of $A$ span a subspace in their own right. If $A$ is $m\times n$, each row has $n$ entries, so the row space $\mathrm{Row}A$ is a subspace of ${\mathbb{R}}^n$, equal to $\mathrm{Col}{A}^{\top }$.

Row operations replace rows by linear combinations of rows, which cannot enlarge or shrink their span, and they are reversible. The nonzero rows of an echelon form are independent because no such row is a combination of those below it. Unlike the column-space basis, this basis uses rows of the echelon form, which are valid representatives since the row space is preserved.

Because $\mathrm{Row}A=\mathrm{Col}{A}^{\top }$, the dimension of the row space is $\mathrm{rank}{A}^{\top }$. The central theorem says these two ranks coincide.

Every pivot contributes one pivot column (a column-space basis vector) and one nonzero echelon row (a row-space basis vector), so $\dim \mathrm{Col}A=\dim \mathrm{Row}A=(\text{number of pivots})$. The columns split into pivot and nonpivot, and the nonpivot columns correspond to the free variables counted by $\dim \mathrm{Nul}A$:

The theorem is a fast reasoning tool.

The same counting settles questions phrased without any linear-algebra vocabulary.

The four fundamental subspaces

A matrix $A$ and its transpose together carry four subspaces. Two live in the domain ${\mathbb{R}}^n$ and two in the codomain ${\mathbb{R}}^m$, and their dimensions are fixed by the rank $r$.

Within each ambient space the two subspaces are more than a partition of dimension: $\mathrm{Row}A$ and $\mathrm{Nul}A$ meet only at $0$ and are in fact orthogonal, as are $\mathrm{Col}A$ and $\mathrm{Nul}{A}^{\top }$. That orthogonality is the subject of inner products and orthogonality; here it is enough that the dimensions add to $n$ and to $m$ respectively.

Rank in the Invertible Matrix Theorem

For a square $n\times n$ matrix, full rank is equivalent to invertibility. The following statements join the Invertible Matrix Theorem, each equivalent to $A$ being invertible.

• (m) The columns of $A$ form a basis of ${\mathbb{R}}^n$.
• (n) $\mathrm{Col}A={\mathbb{R}}^n$.
• (o) $\dim \mathrm{Col}A=n$.
• (p) $\mathrm{rank}A=n$.
• (q) $\mathrm{Nul}A=\{0\}$.
• (r) $\dim \mathrm{Nul}A=0$.

These follow from the rank theorem and the earlier invertibility conditions by a short chain: full column span gives $\dim \mathrm{Col}A=n$, hence rank $n$, hence $\dim \mathrm{Nul}A=0$ by $\mathrm{rank}A+\dim \mathrm{Nul}A=n$, hence only the trivial solution to $Ax=0$.