MAST10007 Chap.6 Span and Linear Independence
Span and Linear Independence
Chapter 6 develops span and linear independence for University of Melbourne MAST10007. The span of vectors is the set of every linear combination they can generate. Independence asks whether the zero vector has only the trivial representation. These questions become systems: put candidate generators in the columns of a matrix, then row-reduce.
A pivot in every generator column means independence; a consistent augmented system means the target lies in the span. It pairs exact definitions with a new completed example, two concept diagrams, diagnostic contrasts and answered retrieval prompts.
Span and independence ask different questions of the same column matrix.
A membership test augments the generators with a target and asks whether the resulting system is consistent. An independence test sets the target to zero and asks whether any nontrivial coefficient vector survives. A dependence relation is useful evidence, not merely failure: it identifies a redundant generator that can be removed without changing the span.
Keeping the coefficient interpretation visible prevents a pivot calculation from becoming detached from the vector statement it is meant to prove.
What this chapter covers
- 01
Linear combinations
- 02
Span as all combinations
- 03
Geometric pictures of span
- 04
Testing membership in a span
- 05
Independence definition
- 06
Dependence relations
- 07
Pivot tests for columns
- 08
Redundant generators
- 09
Building and pruning spanning sets
- 10
Proof strategy
- 11
Reading coefficients from a span equation
- 12
Using a dependence relation to remove redundancy
- 13
Comparing target-membership and independence systems
Span and Linear Independence worked example
- stepSolve av1+bv2=v3, producing equations a+2b=3, a=1, and 2b=2.
- stepThe second and third equations give a=1 and b=1.
- stepThe first equation is then 1+2=3, so the system is consistent.
- stepTherefore v3=v1+v2. The same equality is a nontrivial dependence relation among v1,v2,v3 if all three are considered together.
Key terms
- Linear combinations
- A linear combination c1v1+⋯+ckvk uses scalar coefficients to build a vector. The order of a list does not change its span, but it will later affect coordinate vectors. Write the ambient space so coefficient and dimension claims have a home.
- Span as all combinations
- Span{v1,…,vk} is every linear combination of the listed vectors and is automatically a subspace. To test whether b belongs, solve Ac=b with the generators as columns. Consistency is exactly membership.
- Geometric pictures of span
- One nonzero vector spans a line through the origin; two nonparallel vectors in R3 span a plane through the origin; three suitable vectors may span all of R3. The picture predicts dimension but row reduction certifies it.
- Testing membership in a span
- Membership in a span is an augmented-system question. A contradiction row proves the target is outside the span. A consistent system provides coefficients, and those coefficients should be substituted into the combination as a direct check.
- Independence definition
- A list is linearly independent when c1v1+⋯+ckvk=0 has only the trivial solution. The zero vector cannot appear in an independent list because assigning it a nonzero coefficient creates a dependence immediately.
- Dependence relations
- A nontrivial dependence relation identifies redundancy. If one coefficient is nonzero, solve for that vector as a combination of the others. Removing it preserves the span; adding it back cannot increase dimension.
- Redundant generator
- A generator is redundant when it lies in the span of the remaining generators. Removing it leaves the span unchanged. A nontrivial dependence relation shows exactly how to express that vector using the others and therefore explains the redundancy.
- Pivot test for generator columns
- Place the candidate vectors as columns and reduce the matrix. A pivot in every column means the only solution of the associated homogeneous coefficient equation is trivial, so the original vectors are linearly independent. A free column signals dependence.
Span and Linear Independence FAQ
How do I check linear combinations?
A linear combination c1v1+⋯+ckvk uses scalar coefficients to build a vector. The order of a list does not change its span, but it will later affect coordinate vectors. Write the ambient space so coefficient and dimension claims have a home. No; the third vector is the sum of the first two.
What is the main trap in span and linear independence?
Spanning and independence are different tests. Including the zero vector makes a list dependent. More vectors than the ambient dimension forces dependence, but fewer vectors does not guarantee independence.
Can AI help with this MAST10007 topic?
Yes. Ask Sia to explain one step, create a fresh practice problem, or check your reasoning. Use it to learn rather than complete graded assessment, and follow University of Melbourne academic-integrity rules.
Why does a consistent augmented system prove span membership?
The unknowns in that system are the coefficients of a linear combination of the generator columns. Consistency means at least one coefficient list produces the target. The final answer should state those coefficients and reconstruct the target, not merely report that reduction succeeded.
What should I do after finding a nontrivial dependence relation?
Interpret the coefficients. Choose a vector with a nonzero coefficient and solve for it as a combination of the others. This identifies a removable generator and explains why the original list was redundant while preserving the same span.
Exam move
Use the varied 11-page chapter as a dependency map: define each object, reproduce the fresh worked example, explain both diagrams, answer the two retrieval prompts, and perform an independent check. For span and linear independence, revisit the first line where your representation, dimensions, or theorem conditions diverge.
Before reducing, write whether the unknown coefficients are being used to reach a target or to represent zero. After reducing, translate pivot and free-variable information back into a sentence about the original vectors. Reconstruct a target for membership problems and extract a dependence relation for redundancy problems. Practise converting that relation into a smaller spanning set without changing the generated space.
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