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Lecture 5 — May 22, 2026
Administrative
- Reflection 1: After submitting your first attempt you should see your score (how many correct) but not which questions you answered correctly — the goal is for you to remain doubtful of all your answers and identify your own mistakes. Some students reported not seeing any score at all; this is unintentional and will be fixed for Reflection 2. Correct answers are revealed automatically once the deadline passes.
- Solutions to reflections will be posted after each deadline.
- Tutorials: the instructor will no longer be attending tutorials regularly. If your tutorial section has issues, reach out early — even one or two unsuccessful tutorials can quickly erode engagement for an entire section.
1. Review: parametric solutions are spans
Last lecture established that the solution set of any homogeneous linear system is a subspace of
For example, the single equation
This is exactly
The shift for today. We move from the arithmetic world of specific linear systems to a more algebraic one: given a collection of vectors in an abstract vector space, what does their span look like, and which collections describe their span efficiently, without redundancy?
2. Span examples in
We work in
One vector. The span of any single nonzero vector is a line through the origin — the set of all scalar multiples. In
Two vectors in independent directions. Consider
To reach an arbitrary
Two vectors in the same direction. Now consider
Three vectors, one redundant.
3. Spanning as solvability
Whether a set
has at least one solution
The only obstruction is a “false equation”: a row reducing to
Example. Does
The second row reads
When does a consistent system fail to span?
For
vectors in , the spanning condition requires every row of the RREF of the coefficient matrix to contain a pivot. When , there are fewer columns than rows, so at least one row must reduce to zero — meaning the span cannot be all of . You cannot span with fewer than vectors. Conversely, when , you are guaranteed a free variable (more unknowns than equations), so the system always has solutions for any right-hand side — but those solutions are non-unique, which is a problem of a different kind.
4. Non-unique descriptions and why they are bad
When a set contains more vectors than necessary to span, the same vector in the span can be expressed in multiple ways as a linear combination. This is the core problem the next definition is designed to address.
Example. We know
Subtracting one from the other:
This is a linear combination of the three vectors equalling
Why is this bad? When we parametrize a solution set, we want the parameters to identify solutions uniquely — one set of parameter values, one solution. Redundant vectors give redundant parameters: a single point on the solution line can be reached by many different combinations. We want to use the fewest vectors that still span, so that the description is unique and efficient.
5. Linear dependence
Definition
A set
of vectors in a vector space is called linearly dependent if there exist and scalars , not all zero, such that Such a relation is called a linear dependence relation, or a non-trivial linear combination equal to
. The word non-trivial refers to the “not all zero” condition: setting all always gives from any set of vectors, so that case carries no information.
Examples.
-
: linearly dependent, since with coefficients . -
: linearly dependent, since .
In both cases, the dependence relation directly reflects the geometric redundancy — one vector can be written as a combination of the others, and removing it would not shrink the span.
Dependence means one vector is in the span of the others
If
with , then dividing by : so
. Removing from the set leaves the span unchanged. Dependence always witnesses redundancy in exactly this form: one vector is already reachable from the others.
6. Linear independence
Definition
A set
of vectors in a vector space is called linearly independent if the only linear combination of vectors from that equals is the trivial one. That is: whenever and it follows that
.
Examples.
-
: independent. forces . -
: independent. From we get and , forcing as well. -
Any set containing
is linearly dependent: is a non-trivial dependence relation. -
A set
with is always linearly independent: with forces .
Independence and unique representations
is linearly independent if and only if every vector in has a unique representation as a linear combination of elements of . This is precisely the property that makes a set of vectors a good coordinate system: each point in the span is identified by exactly one tuple of coefficients.
7. Basis
We now combine spanning and linear independence into one of the most important definitions in the course.
Definition
Let
be a vector space. A basis of is a set of vectors that is:
- linearly independent, and
- spanning:
.
A basis is the most efficient description of
Examples in
-
— the standard basis. Linearly independent and spans . ✓ -
— also a basis for : independent (verified above) and spanning (verified in §2). ✓ -
— spans but linearly dependent. Not a basis. ✗ -
— linearly independent but does not span . Not a basis. ✗
The number of vectors in any basis of
Bases are not unique
There is no single “correct” basis for a vector space — there are infinitely many. The standard basis
for is the most convenient for computation, but other bases can be far more natural for a given problem. Choosing a basis is choosing a coordinate system; different coordinate systems suit different purposes.
8. The zero vector space
One edge case deserves attention. The zero vector space is
What is a basis for
- Is
linearly independent? Vacuously yes — there are no vectors in it, so the implication “if some combination equals then all coefficients are zero” holds with no cases to check. - Does
span ? Yes: by convention.
Therefore:
This space is zero-dimensional: there are no directions to go, so no vectors are needed to span it.
Recovering a vector from a scalar multiple
A small algebraic fact appeared at the end of lecture: if
and , then . This uses associativity of scalar multiplication: . Nonzero scalars are invertible in , so multiplication by is always reversible. This is the vector space axiom at work.
Looking ahead
The key theorem ahead: any two bases of the same vector space have the same number of elements. This number is the dimension. The proof rests on one lemma — sometimes called the Steinitz exchange lemma — which says that a linearly independent set can never have more vectors than a spanning set. Establishing this will be the main goal of the next lecture.
Textbook references
| Topic | Axler (2nd ed.) | Hefferon (4th ed.) | Treil |
|---|---|---|---|
| Parametric solutions as spans | — | Ch. 1, §I.3 | Ch. 1, §2 |
| Span examples in | Ch. 2, “Span and Linear Independence” | Ch. 2, §I.2 | Ch. 2, §1 |
| Spanning as solvability of a linear system | — | Ch. 1, §I.1–2 | Ch. 1, §1 |
| Linear dependence — definition | Ch. 2, “Linear Independence” | Ch. 2, §II.1, Def. 1.1 | Ch. 2, §2 |
| Dependent ⟹ one vector in span of others | Ch. 2, 2.17 | Ch. 2, §II.1 | Ch. 2, §2 |
| Linear independence — definition | Ch. 2, “Linear Independence” | Ch. 2, §II.1, Def. 1.3 | Ch. 2, §2 |
| Basis — definition | Ch. 2, “Bases” | Ch. 2, §I.1, Def. 1.4 | Ch. 2, §3 |
| Zero vector space and empty basis | Ch. 1, “Definition of Vector Space” | Ch. 2, §I.1 | — |