Chapter 4 Study Guide

calendar_today August 9, 2026 article 805 words access_time 6 min replay Modified August 9, 2026

Table of Contents

Chapter 4 — Vector Spaces

Covers §4.1, 4.2, 4.3, 4.5 (§4.4 and §4.6 not assigned). Full problems and recorded answers in 2160 Compendium.

Weighting

67 questions, 13 true/false (19%). This is the abstraction step of the course: everything from Chapters 1–2 gets restated without reference to $\mathbb{R}^n$. §4.3 alone carries 23 questions, the largest single section in the course.

§4.1 Vector Spaces and Subspaces

Definition 1 (Subspace).

A subset $H$ of a vector space $V$ is a subspace if:

$\mathbf{0} \in H$
$H$ is closed under addition — $\mathbf{u},\mathbf{v}\in H \Rightarrow \mathbf{u}+\mathbf{v}\in H$
$H$ is closed under scalar multiplication — $\mathbf{u}\in H, c$ scalar $\Rightarrow c\mathbf{u}\in H$

All three must hold. To disprove a subspace, a single counterexample to any one is enough.

The fastest disqualifiers

Does the set contain $\mathbf{0}$? If not, done — not a subspace.
Any nonlinear condition ($x^2$, $xy$, $|x|$) or inequality ($\le$, $\ge$) almost always breaks closure. A set defined by $6x^2+y^2 \le 4$ fails scalar multiplication immediately — scale a point up and you leave the region.
A condition like $x_1 + x_2 = 1$ (nonzero constant) excludes $\mathbf{0}$.

These dispose of 4.1.3, 4.1.5, 4.1.7, 4.1.8 on sight.

The reliable way to prove a subspace

Show the set is a span. Theorem 1: $\text{Span}\{\mathbf{v}_1,\dots,\mathbf{v}_p\}$ is always a subspace. So write the general element in terms of its free parameters and factor:

$$\begin{bmatrix} a+b \\ b-c \\ a \end{bmatrix} = a\begin{bmatrix}1\\0\\1\end{bmatrix} + b\begin{bmatrix}1\\1\\0\end{bmatrix} + c\begin{bmatrix}0\\-1\\0\end{bmatrix}$$

This is the intended method for 4.1.10, 4.1.12, 4.1.15, 4.1.17.

Tested as — subspace or not 4.1.3–4.1.8 · show a set is a subspace 4.1.9–4.1.17 · membership in a span 4.1.13, 4.1.14 · true/false 4.1.24–4.1.30

§4.2 Null Spaces, Column Spaces, and Row Spaces

Definition 2 (The three spaces).

Space	Definition	Lives in	Found by
$\text{Nul}\,A$	$\{\mathbf{x} : A\mathbf{x}=\mathbf{0}\}$	$\mathbb{R}^{n}$ (cols)	solve $A\mathbf{x}=\mathbf{0}$, parametric form
$\text{Col}\,A$	span of the columns of $A$	$\mathbb{R}^{m}$ (rows)	pivot columns of original $A$
$\text{Row}\,A$	span of the rows of $A$	$\mathbb{R}^{n}$	nonzero rows of the echelon form

The two traps

$\text{Nul}\,A$ and $\text{Col}\,A$ live in different spaces unless $A$ is square. For a $3\times5$ matrix, $\text{Nul}\,A \subseteq \mathbb{R}^5$ but $\text{Col}\,A \subseteq \mathbb{R}^3$.

Testing membership differs fundamentally:

Is $\mathbf{w}\in\text{Nul}\,A$? — just compute $A\mathbf{w}$ and check it is $\mathbf{0}$.
Is $\mathbf{w}\in\text{Col}\,A$? — solve $A\mathbf{x}=\mathbf{w}$ and check consistency.

That contrast is precisely what 4.2.23 and 4.2.24 ask.

Tested as — membership in $\text{Nul}\,A$ 4.2.1 · spanning set for $\text{Nul}\,A$ 4.2.3, 4.2.5, 4.2.6 · is $W$ a vector space 4.2.8–4.2.13 · find $A$ with a given $\text{Col}\,A$ 4.2.15 · vectors in each space 4.2.21

§4.3 Linearly Independent Sets; Bases

Definition 3 (Basis).

$\mathcal{B}=\{\mathbf{b}_1,\dots,\mathbf{b}_p\}$ is a basis for subspace $H$ if it is linearly independent and spans $H$. Both conditions — a spanning set that is dependent is not a basis, and an independent set that is too small is not either.

Finding bases — three different procedures

Basis for $\text{Nul}\,A$: solve $A\mathbf{x}=\mathbf{0}$, write in parametric vector form; the vectors attached to the free variables form the basis.

Basis for $\text{Col}\,A$: row reduce to find which columns are pivot columns, then take those columns of the original $A$ — never the reduced ones. Row reduction preserves which columns are pivotal, not the column space itself.

Basis for $\text{Row}\,A$: take the nonzero rows of the echelon form. Here the reduced version is correct, because row operations preserve the row space.

The asymmetry between $\text{Col}$ and $\text{Row}$ is the most commonly tested subtlety in the chapter — 4.3.13, 4.3.14, 4.3.28, 4.3.30 all turn on it.

"Describe the set" problems

Eight problems open with “Describe the set $\{\mathbf{v}_1,\mathbf{v}_2,\dots\}$”. They are asking for the geometric object spanned: a line through the origin, a plane through the origin, or all of $\mathbb{R}^n$. Count the independent vectors — that is the dimension of the span.

Tested as — describe a span 4.3.1–4.3.8 · basis for $\text{Nul}\,A$ 4.3.9, 4.3.10 · basis for a plane/line 4.3.11, 4.3.12 · bases for all three spaces 4.3.13, 4.3.14 · basis for a spanned space 4.3.16 · function and polynomial spaces 4.3.36, 4.3.44

§4.5 The Dimension of a Vector Space

Definition 4 (Dimension, rank, nullity).

$\dim H$ is the number of vectors in any basis for $H$ (all bases have the same size).

$$\operatorname{rank}A=\dim\text{Col}\,A=\dim\text{Row}\,A=\#\text{pivots}$$$$\operatorname{nullity}A=\dim\text{Nul}\,A=\#\text{free variables}$$

The Rank–Nullity Theorem

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

$$\operatorname{rank}A+\dim\text{Nul}\,A=n \quad (\text{the number of \textbf{columns}})$$

Every §4.5 dimension problem is an application. Note it is $n$ — the columns — not $m$.

Also: $\dim\text{Row}\,A=\dim\text{Col}\,A$ always, even for non-square $A$. Row rank equals column rank.

Worked instance

If $A$ is $5\times7$ with $\dim\text{Nul}\,A=3$, then $\operatorname{rank}A=7-3=4$, so $\dim\text{Col}\,A=\dim\text{Row}\,A=4$. That is 4.5.37 in one line.

Tested as — basis and dimension of a subspace 4.5.6, 4.5.8 · dimension of a span 4.5.9, 4.5.10 · all three dimensions from a matrix 4.5.11–4.5.15 · rank–nullity 4.5.37

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