4.5.15 · D5Linear Algebra (Full)
Question bank — Linear independence — formal definition, testing
Before we start, one reminder of the vocabulary so no symbol is unearned:
- A combination means "scale each vector by a number and add them up."
- The trivial (boring) combination is all — that always gives , for any vectors. It proves nothing.
- Independent = the only way to reach is the boring way. Dependent = there is an exciting way (at least one ).
True or false — justify
Each item is a statement. Decide true/false, then give the reason, never just the verdict.
Two vectors are independent as long as they are not equal.
False — they must not be scalar multiples. and are unequal yet dependent since .
If a set of 3 vectors is independent, then every subset of 2 of them is also independent.
True — a dependency among a subset would immediately be a dependency in the whole set (pad the missing coefficients with ). So independence is inherited downward.
If every pair of vectors in a set is independent, the whole set is independent.
False — pairwise independence is weaker. are independent in every pair, yet makes the trio dependent.
A set of independent vectors can still fail to span the space.
True — independence and spanning are separate. are independent in but only span a plane; you need a basis for both.
Adding a vector to an independent set can never make it independent-er, only possibly dependent.
True — you either add a genuinely new direction (still independent) or add a redundant one (becomes dependent). Independence can only be preserved or broken, never "improved."
If for a square matrix whose columns are your vectors, the vectors are dependent.
True for square — zero determinant means the columns collapse into a lower-dimensional space, so a nontrivial combination hits .
Independent columns of guarantee is invertible.
False in general — only if is square. A matrix can have independent columns yet isn't square, so the invertibility question doesn't even apply.
The empty set of vectors is linearly dependent.
False — the empty set is independent by convention, since there is no nontrivial combination to even write down (vacuously, only the trivial one exists).
If has only the trivial solution, the columns of are independent.
True — that is the definition restated: the homogeneous system having a trivial-only null space is exactly independence.
Spot the error
Each item contains a plausible-sounding argument with a hidden flaw. Name the flaw.
" are three distinct non-parallel vectors, so they're independent."
Flaw: distinctness and non-parallelism don't prevent a three-term dependency. Here the third is second first, so they're dependent — you must solve the full system, not eyeball pairs.
"I computed a combination and it wasn't zero, so the set is independent."
Flaw: independence requires every nontrivial combination to be nonzero, not one lucky choice. You must show no nontrivial combo reaches , i.e. solve fully.
"These are 4 vectors in , so they must be independent."
Flaw: matching count and dimension is necessary but not sufficient. Four vectors in can still be dependent (e.g. two equal ones); you need or full rank.
"The matrix isn't square, but I'll take its determinant to test dependence."
Flaw: the determinant is defined only for square matrices. For a non-square use row reduction and count pivots instead.
"There's a zero row after row reduction, but the vectors are the rows, and one row went to zero — that zero row is just a leftover, so ignore it."
Flaw: a zero row after reduction is precisely the signal that rank number of vectors — it proves a dependency exists, it is not to be ignored.
"Set contains plus two independent vectors, so is independent because the two real vectors are."
Flaw: any set containing is dependent, because is nontrivial (). One passenger poisons the whole set.
" and has 2 columns, but I found a free variable, so it's dependent."
Flaw: contradiction — if rank equals the number of columns () there are no free variables. The "free variable" was a computational slip; recheck the reduction.
Why questions
These ask for the mechanism, not a fact.
Why does a nontrivial solution to mean one vector is redundant?
If some , divide by it and solve , expressing as a combination of the rest — a passenger.
Why does the trivial solution never prove independence on its own?
Because times anything is , so the all-zero combination works for every set, dependent or not. It carries zero information.
Why must more than vectors in always be dependent?
Stacking them as columns gives a matrix with more columns than rows; rank column count forces a free variable, hence a nontrivial null-space solution.
Why do we phrase the test as a homogeneous system ?
Because is the linear combination by definition of matrix–vector product, so "combination equals " becomes "solve ."
Why does independence guarantee unique coordinates for every vector in the span?
If a vector had two representations, subtracting them gives a nontrivial combination equal to — impossible for an independent set. So the representation is one-of-a-kind.
Why is "pivot in every column" the right rank condition?
A pivotless (free) column corresponds to a free variable in , which produces nontrivial solutions; every column having a pivot rules that out, forcing independence.
Why does work only in the square case, and what does it see geometrically?
Determinant measures signed volume of the parallelepiped spanned by the columns; it exists only for vectors in . Zero volume means the columns lie in a lower dimension — dependent.
Edge cases
The degenerate and boundary scenarios people forget.
Is a single nonzero vector independent?
Yes — with forces , the only (trivial) solution, so a lone nonzero vector is always independent.
Is the single zero vector independent?
No — holds for every , including , a nontrivial solution. So is dependent.
Can a set of just 2 vectors be dependent without either being zero?
Yes — if one is a nonzero scalar multiple of the other, e.g. and : nontrivially, dependent.
If vectors live in with , can they still be dependent?
Yes — fewer vectors than the dimension does not force independence; e.g. and in are two vectors, still dependent (parallel).
What is the rank of a matrix whose columns are a maximal independent set of size ?
Exactly — every column is a pivot column, so rank equals the number of vectors, the borderline where any added vector would create dependence.
Are the standard basis vectors of dependent for any ?
Never — forces each directly, so they are independent for every .
Recall One-line summary of every trap
Independence is decided by one honest question — is there a nontrivial combination hitting ? Not by distinctness, not by one lucky combination, not by matching counts, not by pairwise checks. Solve and count pivots.
Connections
- Linear independence — formal definition, testing — the parent this bank drills.
- Span and spanning sets — independence + spanning together give a basis.
- Basis and dimension — independence is one half of the basis definition.
- Rank of a matrix — pivot count decides independence.
- Determinant — the square-case shortcut.
- Homogeneous systems and null space — the algebraic home of the test.
- Invertible matrix theorem — independent square columns ⇔ invertible.