Dimension, Rank, and Coordinate Systems

10 min read

Before Dimension counts the vectors in a basis, so this lesson needs bases and independence. Builds on: Null Space, Column Space, and Basis · Linear Independence

Next Coordinates in a new basis return in diagonalization, where a basis of eigenvectors makes a matrix diagonal.

Reference: Lay, Linear Algebra and Its Applications, Section 4.4: Coordinate Systems; Section 4.5: The Dimension of a Vector Space; Section 4.6: Rank

Video: Dr Lynch’s MATH 231 lectures (YouTube playlists)

Before You Start: Prerequisite Check

Can you do these? (Click to reveal self-test)

Test yourself on these prerequisite skills:

Basis size: If $H$ is a subspace and the basis for $H$ has 3 vectors, what is the dimension of $H$?

Check

Dimension is the number of vectors in any basis. If one basis has 3 vectors, then every basis for $H$ has exactly 3 vectors, and $\dim(H) = 3$.
Rank: What does the rank of a matrix $A$ equal?

Check

The rank of $A$ equals the number of pivot columns in any echelon form of $A$, which equals the dimension of $\text{Col}(A)$.
Unique representation: If $\mathcal{B} = \{\mathbf{b}_1, \mathbf{b}_2\}$ is a basis for $H$, and $\mathbf{x} = 3\mathbf{b}_1 - \mathbf{b}_2$, what are the coordinates of $\mathbf{x}$ relative to $\mathcal{B}$?

Check

The $\mathcal{B}$-coordinates are the scalars: $[\mathbf{x}]_\mathcal{B} = \begin{pmatrix}3\\-1\end{pmatrix}$.

If you struggled: Review Null Space, Column Space, and Basis.

Try This First

Let $\mathcal{B} = \left\{\mathbf{b}_1, \mathbf{b}_2\right\} = \left\{\begin{pmatrix}1\\1\end{pmatrix}, \begin{pmatrix}0\\1\end{pmatrix}\right\}$ be a basis for $\mathbb{R}^2$.

Express $\mathbf{x} = \begin{pmatrix}3\\5\end{pmatrix}$ as $c_1\mathbf{b}_1 + c_2\mathbf{b}_2$ by solving the resulting system.

Compare the coordinate vector $[\mathbf{x}]_\mathcal{B}$ with the standard coordinates $\begin{pmatrix}3\\5\end{pmatrix}$. Are they the same?

Show Solution

Solve $c_1\begin{pmatrix}1\\1\end{pmatrix} + c_2\begin{pmatrix}0\\1\end{pmatrix} = \begin{pmatrix}3\\5\end{pmatrix}$:

From entry 1: $c_1 = 3$. From entry 2: $c_1 + c_2 = 5 \implies c_2 = 2$.

\[ [\mathbf{x}]_\mathcal{B} = \begin{pmatrix}3\\2\end{pmatrix} \]

The standard coordinates of $\mathbf{x}$ are $\begin{pmatrix}3\\5\end{pmatrix}$, but the $\mathcal{B}$-coordinates are $\begin{pmatrix}3\\2\end{pmatrix}$. They are different.

The main point: coordinates depend on the basis. The standard coordinates are the $\mathcal{E}$-coordinates relative to the standard basis $\left\{\begin{pmatrix}1\\0\end{pmatrix}, \begin{pmatrix}0\\1\end{pmatrix}\right\}$. A different basis gives different numbers for the same vector.

Three Connected Ideas

Three concepts developed across Lay Sections 4.4-4.6 fit together:

Coordinate systems (4.4): every vector in a space can be described uniquely by its coordinates relative to any basis.
Dimension (4.5): every basis for a given space has the same number of vectors; that number is the dimension of the space.
Rank (4.6): the rank of a matrix is the number of pivots, which equals the dimension of the column space; rank + nullity = number of columns.

Prerequisite Map

Quick Reference

Property	Value
Course	MATH 231
Difficulty	Advanced
Time	~30 minutes
Source	Lay, Linear Algebra and Its Applications, Sections 4.4, 4.5, and 4.6

Rank-Nullity at a Glance

For an $m \times n$ matrix $A$:

Quantity	How to Compute	Formula
rank$(A)$	Number of pivot columns	$\leq \min(m,n)$
nullity$(A)$	Number of free variables	$n - \text{rank}(A)$
dim(Col$(A)$)	rank$(A)$	is in $\mathbb{R}^m$
dim(Nul$(A)$)	nullity$(A)$	is in $\mathbb{R}^n$

Rank Theorem: $\text{rank}(A) + \text{nullity}(A) = n$.

Key Concepts

Coordinate Systems

Let $\mathcal{B} = \{\mathbf{b}_1, \ldots, \mathbf{b}_p\}$ be a basis for a subspace $H$.

Unique Representation Theorem: For every $\mathbf{x}$ in $H$, there exists a unique set of scalars $c_1, \ldots, c_p$ such that $\mathbf{x} = c_1\mathbf{b}_1 + \cdots + c_p\mathbf{b}_p$.

The coordinate vector of $\mathbf{x}$ relative to $\mathcal{B}$ is: \[ [\mathbf{x}]_\mathcal{B} = \begin{pmatrix}c_1\\\vdots\\c_p\end{pmatrix} \]

To find $[\mathbf{x}]_\mathcal{B}$: Solve $P_\mathcal{B}\mathbf{c} = \mathbf{x}$ where $P_\mathcal{B} = [\mathbf{b}_1 \; \cdots \; \mathbf{b}_p]$ is the change-of-basis matrix.

Source: Lay, Linear Algebra and Its Applications, Section 4.4: Coordinate Systems.

Dimension

The dimension of a nonzero subspace $H$, written $\dim(H)$, is the number of vectors in any basis for $H$. (The zero subspace has dimension 0.)

Key theorems (Lay, Section 4.5):

Let $H$ be a subspace of dimension $p$ in a vector space $V$.

Any linearly independent set with exactly $p$ vectors in $H$ is automatically a basis for $H$.
Any set of exactly $p$ vectors that spans $H$ is automatically a basis for $H$.
Any linearly independent set in $H$ can be extended to a basis for $H$.
Any spanning set for $H$ can be reduced to a basis for $H$.

Consequence: to verify a set of $p$ vectors is a basis for $H$, you only need to verify ONE of the two conditions (independent OR spanning), provided you know $\dim(H) = p$.

Common dimensions:

Space	Dimension
$\{mathbf{0}\}$	0
$\mathbb{R}^n$	$n$
Line through origin in $\mathbb{R}^n$	1
Plane through origin in $\mathbb{R}^n$	2

Source: Lay, Linear Algebra and Its Applications, Section 4.5: The Dimension of a Vector Space.

The Rank Theorem

Definition: The rank of $A$ is the dimension of the column space of $A$: $\text{rank}(A) = \dim(\text{Col}(A))$.

Equivalently, rank$(A)$ = number of pivot columns = number of pivot rows = dim(Row$(A)$).

Rank Theorem: If $A$ is $m \times n$, then: \[ \text{rank}(A) + \text{nullity}(A) = n \]

where nullity$(A) = \dim(\text{Nul}(A))$ = number of free variables.

Intuition: The $n$ columns of $A$ split into pivot columns (which produce rank linearly independent vectors) and free columns (which produce nullity-many degrees of freedom). Pivot + free = total columns.

Source: Lay, Linear Algebra and Its Applications, Section 4.6: Rank.

Worked Example

Let $A = \begin{pmatrix}1&2&0&4\\-1&-2&1&-3\\2&4&1&5\end{pmatrix}$.

From earlier work, the RREF of $A$ is $\begin{pmatrix}1&2&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}$ with pivot columns 1, 3, 4.

rank$(A) = 3$ (three pivot columns).
nullity$(A) = n - \text{rank} = 4 - 3 = 1$.
dim(Col$(A)$) = 3 (a 3-dimensional subspace of $\mathbb{R}^3$, so Col$(A)$ = $\mathbb{R}^3$).
dim(Nul$(A)$) = 1 (a line through the origin in $\mathbb{R}^4$).

Check: $3 + 1 = 4 = n$. Rank Theorem satisfied.

Coordinate example:

Using basis $\mathcal{B} = \left\{\begin{pmatrix}1\\0\\0\end{pmatrix}, \begin{pmatrix}1\\1\\0\end{pmatrix}, \begin{pmatrix}1\\1\\1\end{pmatrix}\right\}$ for $\mathbb{R}^3$, find $[\mathbf{x}]_\mathcal{B}$ for $\mathbf{x} = \begin{pmatrix}4\\3\\1\end{pmatrix}$.

Solve $P_\mathcal{B}\mathbf{c} = \mathbf{x}$: \[ \left[\begin{array}{rrr|r}1&1&1&4\\0&1&1&3\\0&0&1&1\end{array}\right] \]

Back-substitute: $c_3 = 1$, $c_2 = 3-1 = 2$, $c_1 = 4-2-1 = 1$.

$[\mathbf{x}]_\mathcal{B} = \begin{pmatrix}1\\2\\1\end{pmatrix}$.

Verify: $1\begin{pmatrix}1\\0\\0\end{pmatrix} + 2\begin{pmatrix}1\\1\\0\end{pmatrix} + 1\begin{pmatrix}1\\1\\1\end{pmatrix} = \begin{pmatrix}1+2+1\\0+2+1\\0+0+1\end{pmatrix} = \begin{pmatrix}4\\3\\1\end{pmatrix}$.

Common Pitfalls

Mistake	Why It Is Wrong	Correct Approach
Saying rank$(A)$ + nullity$(A)$ $= m$ (rows)	The Rank Theorem says rank + nullity $= n$ (number of COLUMNS)	Count columns: rank + nullity $= n$ always
Assuming coordinates are the same in every basis	The same vector has different coordinates in different bases	Coordinates depend on the choice of basis; solve $P_\mathcal{B}\mathbf{c} = \mathbf{x}$ to find the correct ones
Claiming rank can exceed $\min(m,n)$	Rank equals the number of pivots, which cannot exceed the number of rows OR the number of columns	$\text{rank}(A) \leq \min(m, n)$ always

Practice Problems

Level 1 State the Dimension

(a) A basis for a subspace $H$ of $\mathbb{R}^5$ has 3 vectors. What is $\dim(H)$?

(b) A $4 \times 7$ matrix has rank 3. What is the nullity? What is the dimension of the column space?

(c) Can a subspace of $\mathbb{R}^4$ have dimension 5? Explain.

Thought Process

For (a): dimension equals the number of vectors in any basis.

For (b): apply rank + nullity = n (number of columns).

For (c): any $p$ linearly independent vectors in $\mathbb{R}^4$ can have $p \leq 4$ (no more than $n$ independent vectors can exist in $\mathbb{R}^n$).

Show Answer

(a) $\dim(H) = 3$.

(b) $n = 7$ columns; rank $= 3$; nullity $= 7 - 3 = 4$. The column space has dimension 3 (same as rank).

(c) No. In $\mathbb{R}^4$, any set of more than 4 vectors is linearly dependent (by the “more than $n$ vectors” theorem). So no subspace of $\mathbb{R}^4$ can have a basis larger than 4, and no subspace of $\mathbb{R}^4$ can have dimension exceeding 4.

Level 2 Find Coordinates Relative to a Basis

Let $\mathcal{B} = \left\{\begin{pmatrix}2\\1\end{pmatrix}, \begin{pmatrix}3\\2\end{pmatrix}\right\}$ be a basis for $\mathbb{R}^2$.

Find $[\mathbf{x}]_\mathcal{B}$ for $\mathbf{x} = \begin{pmatrix}7\\4\end{pmatrix}$.

Thought Process

Solve $c_1\begin{pmatrix}2\\1\end{pmatrix} + c_2\begin{pmatrix}3\\2\end{pmatrix} = \begin{pmatrix}7\\4\end{pmatrix}$ by row-reducing $[P_\mathcal{B} \mid \mathbf{x}]$.

Show Answer

\[ \left[\begin{array}{rr|r}2&3&7\\1&2&4\end{array}\right] \]

$R_1 \leftrightarrow R_2$, then $R_2 - 2R_1$: \[ \left[\begin{array}{rr|r}1&2&4\\0&-1&-1\end{array}\right] \]

$c_2 = 1$, $c_1 = 4 - 2(1) = 2$.

\[ [\mathbf{x}]_\mathcal{B} = \begin{pmatrix}2\\1\end{pmatrix} \]

Verify: $2\begin{pmatrix}2\\1\end{pmatrix} + 1\begin{pmatrix}3\\2\end{pmatrix} = \begin{pmatrix}4+3\\2+2\end{pmatrix} = \begin{pmatrix}7\\4\end{pmatrix}$ (correct).

Level 3 Verify the Rank Theorem

For $A = \begin{pmatrix}1&2&-1&0\\2&4&0&2\\-1&-2&3&2\end{pmatrix}$:

(a) Find rank$(A)$ and nullity$(A)$ by row-reducing $A$.

(b) Verify the Rank Theorem: rank $+$ nullity $= n$.

Thought Process

Row-reduce $A$ to find pivot columns and free columns. Count both.

Show Answer

(a) Row-reduce $A$:

$R_2 - 2R_1$: $[0, 0, 2, 2]$; $R_3 + R_1$: $[0, 0, 2, 2]$.

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

$R_3 - R_2$: $[0,0,0,0]$. $R_2 \to \frac{1}{2}R_2$: $[0,0,1,1]$.

RREF: $\begin{pmatrix}1&2&0&1\\0&0&1&1\\0&0&0&0\end{pmatrix}$

Pivot columns: 1, 3. rank$(A) = 2$.

$n = 4$ columns; nullity$(A) = 4 - 2 = 2$.

(b) rank $+$ nullity $= 2 + 2 = 4 = n$. Rank Theorem confirmed.

(c) Free variables: $x_2 = s$, $x_4 = t$.

From RREF: $x_3 = -t$, $x_1 = -2s - t$.

\[ \text{Basis for Nul}(A): \left\{\begin{pmatrix}-2\\1\\0\\0\end{pmatrix}, \begin{pmatrix}-1\\0\\-1\\1\end{pmatrix}\right\} \]

Level 4 Use Rank-Nullity Without Full Row Reduction

A $3 \times 5$ matrix $A$ has rank 2.

(a) What is the nullity of $A$?

(b) Is $A\mathbf{x} = \mathbf{b}$ consistent for every $\mathbf{b} \in \mathbb{R}^3$? Why or why not?

(c) If $A\mathbf{x} = \mathbf{b}$ is consistent, how many solutions does it have?

(d) Can the columns of $A$ be linearly independent? Explain.

Thought Process

For (a): rank + nullity = 5.

For (b): for every $\mathbf{b}$ in $\mathbb{R}^3$, the system is consistent iff every row of $A$ contains a pivot. With rank 2 and 3 rows, at least one row has no pivot.

For (c): nullity $> 0$ means free variables exist, so infinitely many solutions if any exist.

For (d): linear independence of columns requires rank $= 5$ (all columns are pivot columns). With rank 2, this is impossible.

Show Answer

(a) nullity $= n - \text{rank} = 5 - 2 = 3$.

(b) No. Rank 2 with 3 rows means at most 2 pivot rows. The third row has no pivot, so there exist right-hand sides $\mathbf{b} \in \mathbb{R}^3$ that produce a contradiction row. The system is not consistent for every $\mathbf{b}$.

(c) When consistent: nullity $= 3 > 0$, so 3 free variables exist. There are infinitely many solutions (a 3-parameter family).

(d) The columns cannot be linearly independent. Linear independence requires rank $= 5$ (a pivot in every column). But rank $= 2 < 5$: only 2 pivot columns and 3 free columns. The 5 columns are linearly dependent.

Level 5 One Condition Suffices for a Basis

Let $\mathcal{B} = \left\{\begin{pmatrix}1\\2\\1\end{pmatrix}, \begin{pmatrix}0\\1\\2\end{pmatrix}, \begin{pmatrix}3\\5\\1\end{pmatrix}\right\}$ be a set of three vectors in $\mathbb{R}^3$.

(a) Row-reduce the matrix $A = [\mathbf{b}_1 \; \mathbf{b}_2 \; \mathbf{b}_3]$ and find the rank.

(b) Is $\mathcal{B}$ linearly independent? Does $\mathcal{B}$ span $\mathbb{R}^3$?

(c) The dimension of $\mathbb{R}^3$ is 3 and $|\mathcal{B}| = 3$. State the theorem that says checking EITHER independence OR spanning is sufficient to confirm $\mathcal{B}$ is a basis.

Thought Process

Row-reduce $A$ and count pivots. 3 pivots means rank 3, which means both independent and spanning (for a 3x3 matrix).

For (c): quote the relevant theorem from Lay Section 4.5.

Show Answer

(a) \[ A = \begin{pmatrix}1&0&3\\2&1&5\\1&2&1\end{pmatrix} \]

$R_2 - 2R_1$: $[0,1,-1]$; $R_3 - R_1$: $[0,2,-2]$.

$R_3 - 2R_2$: $[0,0,0]$.

\[ \text{REF}: \begin{pmatrix}1&0&3\\0&1&-1\\0&0&0\end{pmatrix} \]

Rank $= 2$ (only 2 pivot columns). The third column is a free column.

(b) Rank $< 3$: the columns are linearly dependent (the set is NOT independent). Since rank $= 2 < 3 = m$, the columns do not span all of $\mathbb{R}^3$ ($A$ does not have a pivot in every row). $\mathcal{B}$ does NOT span $\mathbb{R}^3$.

$\mathcal{B}$ is not a basis for $\mathbb{R}^3$: it fails both conditions.

(c) Lay Section 4.5 (Basis Theorem): If $\dim(H) = p$, then any set of exactly $p$ vectors in $H$ that either spans $H$ OR is linearly independent automatically satisfies both conditions and is a basis for $H$.

In this problem, we found rank $2$, so only 2 independent columns exist. With $|\mathcal{B}| = 3 > 2$, the set has the wrong number of independent vectors for $\mathbb{R}^3$: the theorem confirms that for 3 vectors to form a basis for $\mathbb{R}^3$, they must be independent (which would force rank 3, which would force spanning as well).

Common Misconceptions

Common misconception

the Rank Theorem says rank + nullity equals the number of rows $m$. The theorem counts COLUMNS, not rows: rank$(A)$ + nullity$(A)$ $= n$ where $n$ is the number of columns of $A$. The number of rows $m$ affects whether the system is consistent but does not appear in the Rank Theorem. Remember: nullity counts free columns (non-pivot columns), and free + pivot = total columns = $n$.

Common misconception

the coordinate vector of $\mathbf{x}$ is the same regardless of which basis you use. Coordinates are entirely determined by the chosen basis. The same vector has completely different coordinates in different bases: the standard-basis coordinates record how much of each standard axis direction; the $\mathcal{B}$-coordinates record how much of each basis vector. Different bases are different “ruler systems” for the same space.

Common misconception

rank can exceed the smaller of $m$ and $n$. The rank of an $m \times n$ matrix equals the number of pivot positions, which cannot exceed the number of rows (only $m$ rows can each have one pivot) or the number of columns (only $n$ columns can each have a pivot). Therefore rank$(A) \leq \min(m, n)$.

Mastery Checklist

Novice (Level 1-2):

I can state the definition of dimension as the number of vectors in any basis.
I can find coordinates of a vector relative to a given basis by solving a system.
I can apply rank + nullity $= n$ to find one quantity given the other.

Competent (Level 3-4):

I can compute rank and nullity from a row reduction and verify the Rank Theorem.
I can determine consistency and solution-count structure from rank and nullity alone.
I can determine whether columns of a rectangular matrix can be independent based on rank constraints.

Proficient (Level 5):

I can apply the Basis Theorem to conclude a set is a basis from a single condition.
I can recognize when a set of $p$ vectors fails to be a basis (not enough independent members, or not spanning).
I can connect rank to the IMT: an $n \times n$ matrix is invertible iff rank $= n$.

Mental Model

The Bookkeeping

The Rank Theorem is bookkeeping for the $n$ columns of $A$. Each column is either a pivot column or a free column; together they account for all $n$ columns:

\[ \underbrace{\text{rank}}_{\text{pivot columns}} + \underbrace{\text{nullity}}_{\text{free columns}} = \underbrace{n}_{\text{total columns}} \]

Pivot columns contribute independent information to the column space; free columns are “explained by” the pivot columns and contribute to the null space. The split is clean and exhaustive.

Coordinates are the analogous bookkeeping for vectors: every vector in a $p$-dimensional space is uniquely determined by $p$ numbers (its coordinates in any basis). Different bases are different accounting systems for the same underlying data.

Connections

Looking ahead in MATH 231:

Eigenvalues and Eigenvectors: the rank of $(A - \lambda I)$ drops below $n$ precisely when $\lambda$ is an eigenvalue
Diagonalization: a matrix is diagonalizable iff it has enough eigenvectors to form a basis; dimension language describes this precisely

Real-world connections:

In data analysis, the rank of a data matrix is the “true dimensionality” of the dataset, often much smaller than the number of observed features. Compression exploits this.
In engineering, the rank-nullity relationship governs controllability and observability of a linear system.

Source: Lay, Linear Algebra and Its Applications, Section 4.4: Coordinate Systems; Section 4.5: The Dimension of a Vector Space; Section 4.6: Rank.

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Null Space, Column Space, and Basis	MATH231 Skills Index	Eigenvalues and Eigenvectors

Last updated: 2026-06-16

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