linear independence, basis, subspace

vectors v1, v2, …, vr are linearly independent iff

the only way that c1v1 + … + crvr = 0→ is if all the ci are zero

in Rⁿ, no set can have more than

n linearly independent vectors; if m > n, then any set of m vectors in Rⁿ must be linearly dependent

c1v1 + … + cnvn = 0 is trivial if

c1 = c2 = … = cn = 0

1 vector is linearly independent iff

it is not the zero vector

2 vectors are linearly independent iff

they are not on the same line

3 vectors are linearly independent iff

they are not on the same plane

span of 1 linearly independent vector is

a line

span of 2 linearly independent vectors is

a plane

span of 3 linearly independent vectors is

a 3d space

smallest span of vectors is

zero

the vectors v2, v2, …, vk are said to be a basis of the set v (with v being the span of some vectors u1, u1, .. uj) if

they span v (span(v1, v2, …, vk) = span (u1,…,uj)

they are linearly independent

a basis of v is

the minimal set of vectors needed to span all of v

parametric equation of a plane

p = p₀ + tu + sv , where p₀ is a point in the plane and u, v are two noncollinear vectors on the plane

subspace of Rⁿ

a set of vectors in Rⁿ that can be described as a span of vectors

given n vectors in Rⁿ, how can you tell whether the set of them is a basis of Rⁿ

check linear independence; need n leading ones in rref of [ v1 v2 … vn | 0 ], and also that there is a leading one in each row so they also span all of Rⁿ

if {x1, x2, …, xm} and {y1, y2, …, yk} are basis of a subspace of Rⁿ, then

m = k

a set V of vectors in Rⁿ is a subspace if

the set includes the zero vector; if x and y are in the set then x + y is in the set; and if x is in the set then ax is in the set for every real number a

vectors are a basis of a subspace V if

they span V and are linearly independent

a basis of a subspace V is the

minimal set of vectors needed to span all of V

dimension of a subspace V is

number of vectors in a basis of V

why does it not span R³

one of the entries is zero → the first vector is only 2D