Linear Algebra · step 4 of 13

Vector spaces, span & linear independence

The span of a set is all its linear combinations; the set is linearly independent when the only combination giving 0 is the trivial one. Testing independence is always the same computation: set a linear combination equal to zero, form the homogeneous system, and row reduce — a free variable means dependence, and the reduction exhibits the relation.

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Method: how to approach it

The order below is what examiners expect to see, and each step carries its own marks.

Write the vectors as columnsForm a matrix and row reduce it. This single computation answers both independence and span questions.
Independence from pivotsThe set is independent iff every column has a pivot. A pivot-free column is a dependent vector.
Read off the dependence relationFree-variable columns express dependent vectors as combinations of the pivot ones.
Check the subspace axiomsA subspace must contain 0 and be closed under addition and scalar multiplication. Failing to contain 0 is the quickest disproof.

Worked example

Are (1,2,3), (2,4,6) and (1,0,1) linearly independent in ℝ³?

Note (2,4,6) = 2·(1,2,3) immediately.
So a non-trivial combination gives zero: 2·(1,2,3) − 1·(2,4,6) + 0·(1,0,1) = 0.
The definition of independence is violated.
The span is therefore only 2-dimensional, spanned by (1,2,3) and (1,0,1).

Answer. Not independent — the second vector is twice the first. The span is a plane in ℝ³, not all of ℝ³.

Where marks get dropped

These are the specific errors that cost credit on vector spaces, span & linear independence questions — QED's rubric penalises each of them separately.

Testing independence pairwise. Three vectors can be pairwise non-parallel and still dependent, as any three coplanar vectors show.
Concluding independence because no obvious multiple appears. Only the row reduction settles it.
Forgetting that any set containing the zero vector is automatically dependent.

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Vector spaces, span & linear independence — frequently asked questions

How many vectors can be independent in ℝⁿ?

At most n. Any n+1 vectors in ℝⁿ are dependent, since the homogeneous system has more unknowns than equations and therefore a free variable.

What is the span of the empty set?

The zero subspace {0}, by the convention that an empty sum is 0. This makes the dimension formulas work uniformly.

How do I show something is a subspace?

Check it contains 0, and is closed under addition and scalar multiplication. Solution sets of homogeneous systems always are; solution sets of inhomogeneous ones never are.

The rest of Linear Algebra

Matrices, systems, determinants, eigenvalues. Each subtopic below has its own method, worked example and mark-losing traps.

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