Independence, Span, Basis & Dimension

Here is the word that crossed the border from probability — independence — now meaning something about arrows. Pick a handful of them. How few can you keep and still reach every point of a space by adding and scaling? An arrow that points in a genuinely new direction earns its place; one that is secretly a stretch of the others is dead weight.

Make that concrete before the algebra arrives. On a flat plane, an arrow pointing east and an arrow pointing north are already enough to reach anywhere: walk some distance east, some distance north, done. Hand yourself a third arrow pointing north-east and you have gained nothing, because east-plus-north could already take you there. Spotting that kind of passenger is the skill this lesson builds — and it opens four tightly linked ideas, independence, span, basis, dimension, which GATE tests as a cluster.

Independence — no redundant directions

Vectors v₁, …, vₙ are linearly independent when no one of them is a linear combination of the others — where a linear combination just means “scale each arrow by some number and add them up”. The crisp algebraic test: the only way to make

c₁v₁ + c₂v₂ + … + cₙvₙ = 0

is to take every coefficient cᵢ = 0. If some non-zero choice of coefficients also gives zero, the vectors are dependent — at least one is redundant, addable from the rest.

That definition is stated in the negative, which is why it reads backwards on first meeting. It is not saying “you cannot combine these arrows to reach zero” — of course you can. Set every cᵢ to zero and any set of arrows sums to nothing; that recipe always works and says nothing about anybody. The claim is that this is the only recipe.

Dependence, then, is the existence of a second way to reach zero — one that actually uses some of the arrows. And any such recipe can be rearranged into a confession. For (1,2) and (2,4), the combination 2·(1,2) − 1·(2,4) = (0,0) reaches zero with non-zero coefficients; move one term across and it reads (2,4) = 2·(1,2). The algebraic test and the intuitive “that one’s a passenger” are the same statement, written twice.

Drag the two arrows below. Point them different ways and every combination fills out the plane — independence in action. Now lay one on top of (or opposite) the other and the combinations collapse onto a single line — the hallmark of dependence.

Span, basis, dimension

The span of a set of vectors is everything you can build from them by adding and scaling — all their linear combinations. Two independent vectors in R² span the whole plane; one non-zero vector spans only a line.

A basis of a space is a set that is both independent and spanning — just enough arrows to reach everything, with none redundant. The dimension is simply the number of vectors in a basis (every basis of a given space has the same count). So R² has dimension 2, R³ has dimension 3. And a space has many valid bases: any n independent vectors that span an n-dimensional space will do — there is nothing special about the “standard” one.

Orthonormal — perpendicular and unit length

A set is orthonormal when its vectors are mutually perpendicular (every pair has dot product 0) and each has length 1. The standard basis (1,0), (0,1) is orthonormal. A fact GATE tests: an orthonormal set is automatically independent — perpendicular directions can never be combinations of one another. But the converse is weaker: independent vectors need not be perpendicular.

A worked example — independent or dependent?

(1) Are (1, 0) and (1, 1) independent, and do they form a basis of R²? (2) Are (1, 2) and (2, 4) independent?

For two vectors in R², the quickest test is the determinant of the matrix they form — non-zero means independent. The reason is worth carrying: for two vectors the determinant measures, up to a sign, the area of the parallelogram they stretch out between them. Point them different ways and that parallelogram has real area; lay one along the other and it collapses to a flat line of zero area. So det = 0 is precisely the algebra reporting “these two have squashed onto one direction”.

det [1 1]  = 1·1 − 0·1 = 1  ≠ 0   →  (1,0),(1,1) independent → a basis of R²
    [0 1]

det [1 2]  = 1·4 − 2·2 = 0        →  (1,2),(2,4) dependent
    [2 4]

The first pair are independent, and two independent vectors in the 2-dimensional R² automatically span it, so they are a basis — a different basis from the standard (1,0), (0,1). The second pair are dependent because (2, 4) = 2·(1, 2): the second adds no new direction, so they span only a line.

A question to carry forward

For a clutch of arrows we just asked “how many point in genuinely new directions?”. A matrix is exactly a stack of column arrows — so the same question applies to it. Here is the thread onward: how many independent directions does a matrix hold, what do we call that number, and what does it instantly tell us about the equation Ax = b?

In one breath

• Independent ⇔ c₁v₁+…+cₙvₙ = 0 forces all cᵢ = 0 (no vector is a combination of the others); a non-zero solution means dependent (a redundant direction).
• Span = all linear combinations; basis = independent and spanning; dimension = number of basis vectors (every basis has the same count; Rⁿ → n).
• A space has many bases — any n independent vectors that span.
• Orthonormal (perpendicular + unit length) ⇒ automatically independent, but independent ⇏ orthogonal ((1,0),(1,1) are independent, not perpendicular).
• Fast check for n vectors in Rⁿ: form a matrix — det ≠ 0 ⇔ independent (a basis). Any n+1 vectors in Rⁿ are always dependent.

Practice