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30 Cards in this Set

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  • Back
  • 3rd side (hint)
The ten axioms to verify Vector Spaces? (u,v,w are vectors m,k are scalars W and V are vector spaces)
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
Theorem 4.1.1 - properties of Zero Vector
a. 0u=0
b. k0=0
c. (-1)u= -u
d. if ku=0, the k=0 or u=0
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
Theorem 4.2.1 - subspace check
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
Span of S
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)
Theorem 4.2.4 - homogeneous solution sets
The solution set of a homogeneous linear system Ax=0 in "n" unknowns is a subspace in Rn
Theorem 4.2.5 - equivalent Spans
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.
Linear Independence
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
Theorem 4.3.1 - Linear independence with 2 or more vectors
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.
Theorem 4.3.2 - Special sets and linear independence
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.
Theorem 4.3.3 - Dependence in Rn
Let S={v1,v2....vr} be a set of vectors in Rn. If r > n then the set is linearly independent.
Theorem 4.3.4 - Wronskian explained
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.
Basis of a Vector Space
S is a basis of V if :
a.) S is linearly independent
b.) S spans V
Checking if a set is a Basis
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.
Coordinate Vector
Let S={v1,} be a basis for vector space V. v=c1v1 +c2v2....+cnvn is the expression for the vector v, then [v]s={c1,} is called the coordinate vector of v relative to S.
Theorem 4.4.1 -unique basis representation
If S={v1,} 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.
Theorem 4.5.1 - dimension of basis
All bases for a finite-dimensional vector space have the same number of vectors.
Theorem 4.5.2 - specs on dimension of basis
Let V be a finite-dimensional vector space, and let {v1,} 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.
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.
Theorem 4.5.3 - the plus minus theorem
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.
Theorem 4.5.4 - Rn basis
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
You only have to check one of them instead of both
Theorem 4.5.5
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.
Transition matrix
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.
How to find the transition matrix
PB to Bⁱ is found by forming the matrix [Bⁱ|B] and doing gauss Jordan until you end up with [I|PB to Bⁱ]
[new basis|old basis]
Theorem 4.6.2
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]
Dot product
|v||u| cos (ø)
Cross Product
|v||u| sin (ø)
Only valid in R3 and the cross product is orthogonal to both original vectors.
Two vectors are orthogonal if the result of their dot product is zero.
Triple Scalar Product
v dot u x w
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.