Ryan B. answered • 03/31/22

6+ years tutoring Linear Algebra

First, a bit of vocabulary:

A **linear combination** of the vectors v_{1},...,v_{n} ∈ V is a_{1}v_{1} + ... + a_{n}v_{n}, where a_{1},...,a_{n} ∈ R

Note that a linear combinations of the vectors v_{1},...,v_{n} ∈ V is itself a vector ∈ V

An **equation of dependency** of the vectors v_{1},...,v_{n} ∈ V is an equation where the left-hand side is a linear combination of v_{1},...,v_{n} ∈ V, and the right-hand side is the zero vector: a_{1}v_{1} + ... + a_{n}v_{n} = **0**

Clearly, this equation is true if a_{1} = ... = a_{n} = 0; this is called the **trivial solution**.

The **span** of the vectors v_{1},...v_{n} ∈ V is the set of all linear combinations of those vectors.

The notation: *span*{v_{1},...,v_{n}} = {a_{1}v_{1} + ... + a_{n}v_{n} : a_{1},...,a_{n} ∈ R}

We say that {v_{1},...,v_{n}} is a **spanning set** for V if V = *span*{v_{1},...,v_{n}}

Putting this together, for B = {v_{1},...,v_{n}}, we have:

(a) "B spans V" means "{v_{1},...,v_{n}} is a spanning set for V"

(b) "B is linearly independent" means "The only solution to an equation of dependency for v_{1},...,v_{n} is the trivial solution"

(c) "B is a basis for V" means "{v_{1},...,v_{n}} is a linearly independent spanning set for V"