Linear Algebra (Vector Spaces)

Front 1

The ten axioms to verify Vector Spaces? (u,v,w are vectors m,k are scalars W and V are vector spaces)

Back 1

1.) u +v is in W
2.) u +v = v + u
3.) u +(v +w)= (u +v) + w
4.) There is a 0 so that u + 0 = o +u =u.
5.) there is a -u so that u =(-u)=0
6.)ku is in V
7.) k(u+v)=vu +kv
8.) (k +m)u=ku +mu
9.) k(mu)=-(km) u
10.) 1u =u

Front 2

Theorem 4.1.1 - properties of Zero Vector

Back 2

a. 0u=0
b. k0=0
c. (-1)u= -u
d. if ku=0, the k=0 or u=0

Front 3

Subspace

Back 3

W is a subspace of V is W is a vector space itself and the same rules of addition and multiplication apply in W as they do in V

Front 4

Theorem 4.2.1 - subspace check

Back 4

W is a supspace of V if (u and v are vectors, k is a scalar):
a.) u +v is in W
b.) ku is i W for all k's

Front 5

Span of S

Back 5

The subspace of V that is formed from all possible linear combinations of a non-empty set S is called the span of S and is noted span{w1...w} or Span(S)

Front 6

Theorem 4.2.4 - homogeneous solution sets

Back 6

The solution set of a homogeneous linear system Ax=0 in "n" unknowns is a subspace in Rn

Front 7

Theorem 4.2.5 - equivalent Spans

Back 7

if S and T are nonempty subspace in Vector Space V, then Span(S)=Span(T) if and only if every element in S is an linear combination of the elements in T and vice versa.

Front 8

Linear Independence

Back 8

If S is a nonempty set of vectors in a Vector space V, S is said to be linearly independent if there is only the trivial solution to kS=0

Front 9

Theorem 4.3.1 - Linear independence with 2 or more vectors

Back 9

A set S with two or more vectors is:
a.) Linearly independent if and only if no vectors in S is expressible as a linear combination of other vectors in S.
b.) Lineraly dependent if and only if one or more vectors is expressible as a linear combination of the vectors in the set.

Front 10

Theorem 4.3.2 - Special sets and linear independence

Back 10

a.) Any finite set that contains 0 is dependent
b.) A set with only one vector is linearly independent if and only if that vector is not =0
c.) A set with exactly two vectors is linearly independent if and only if neither vector is a scalar multiple of the other.

Front 11

Theorem 4.3.3 - Dependence in Rn

Back 11

Let S={v1,v2....vr} be a set of vectors in Rn. If r > n then the set is linearly independent.

Front 12

Theorem 4.3.4 - Wronskian explained

Back 12

If the functions f1,f2,fn have fn-1 contiuous derivatives, the wronskian exists. If it is not zero then the set is linearly independent.

Front 13

Wronskian

Back 13

det[[f1,f2,...fn]
[f1ⁱ,f2ⁱ...fnⁱ]
[f1ⁱⁱ,f2ⁱⁱ...fnⁱⁱ]]

Front 14

Basis of a Vector Space

Back 14

S is a basis of V if :
a.) S is linearly independent
b.) S spans V

Front 15

Checking if a set is a Basis

Back 15

set Ax=0 and Ax=b. The matrix of coeffecients will be the same for the two. Theorem 2.3.8 says that if the det≠0 then both systems have the solutions we want and therefore it is a basis.

Front 16

Coordinate Vector

Back 16

Let S={v1,v2....vn} be a basis for vector space V. v=c1v1 +c2v2....+cnvn is the expression for the vector v, then [v]s={c1,c2...cn} is called the coordinate vector of v relative to S.

Front 17

Theorem 4.4.1 -unique basis representation

Back 17

If S={v1,v2....vn} is a basis for Vector space V, then all vectors in v can be expressed as a linear combination of the elements in S in exactly one way.

Front 18

Theorem 4.5.1 - dimension of basis

Back 18

All bases for a finite-dimensional vector space have the same number of vectors.

Front 19

Theorem 4.5.2 - specs on dimension of basis

Back 19

Let V be a finite-dimensional vector space, and let {v1,v2...vn} be any basis:
a.) if a set has more than n vectors, then it is linearly dependent
b.) if a set has few than n vectors, then it does' span V.

Front 20

Dimension

Back 20

The dimension of a finite-dimensional vectorspace V is denotes as dim(V) and represent the # of vectors found in a basis for that vector space.

Front 21

Theorem 4.5.3 - the plus minus theorem

Back 21

Let S be a nonempty set of vectors in a vector space V:
a.) If S is a linearly independent set ,and if v is a vector in V that is outside Span(S), then the set S union v that results from inserting v into is is still linearly independent.
b.) If v is a vector in S that is expressible as a linear combination of other vectors in S, and if S- {v} denotes the set obtained by removing v from S, then S and S - {v} span the same space.

Front 22

Theorem 4.5.4 - Rn basis

Back 22

Let V be a vector space in Rn and let S be a subspace of V with exactly n vectors. Then S is a basis for V if and only if S spans V or S is linearly independent

Hint 22

You only have to check one of them instead of both

Front 23

Theorem 4.5.5

Back 23

Let S be a finite set of vectors in a finite-dimensional vector space V: a.) If S spans V but is not a basis for V, then S can be reduced to a basis for C by removing dependent vectors from S.
b.) If S is linearly independent set that is not already a basis for V, then S can be enlarged to a basis for V by inserting linearly independent vectors into S.

Front 24

Transition matrix

Back 24

The matrix that, when multiplied on the right by a coordinate vector relative to a basis B returns the coordinate vector relative to basis S.

Front 25

How to find the transition matrix

Back 25

PB to Bⁱ is found by forming the matrix [Bⁱ|B] and doing gauss Jordan until you end up with [I|PB to Bⁱ]

Hint 25

[new basis|old basis]

Front 26

Theorem 4.6.2

Back 26

Let Bⁱ={u1...un} ad S={e1,...en}, then the transition matrix from Bⁱ to S is simply to vectors u1,...un written as columns like so [u1|u2|un]

Front 27

Dot product

Back 27

|v||u| cos (ø)

Front 28

Cross Product

Back 28

|v||u| sin (ø)

Hint 28

Only valid in R3 and the cross product is orthogonal to both original vectors.

Front 29

Orthogonality

Back 29

Two vectors are orthogonal if the result of their dot product is zero.

Front 30

Triple Scalar Product

Back 30

v dot u x w

Hint 30

The scalar triple product yields the volume of the parallelepipied created by the three vectors. Also, that quantity is zero, they three vectors all lie in the same plane.