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A set S is a basis for V if
1. S spans V
2. S is LI.
If S is a basis for V then every vector in V can be written in one and only one way as a linear combo of vectors in S and every set containing more than n vectors is LD.

Basis Test

1. If S is a LI set of vectors in V, then S is a basis for V
2. If S spans V, then S is a basis for V

Change of Basis

P[x]_B' = [x]_B
[x]_B' = P-1 [x]_B
[B B'] -> [ I P-1 ]
[B' B ] -> [ I P ]

Cross Product

if u = u1i + u2j + u3k
v = v1i + v2j + v3k
u x v = (u2v3 - u3v2)i - (u1v3 -u3v1)j + (u1v2 - u2v1)k

Definition of a Vector Space

u + v is within V
u+v = v+u
u+(v+w) = (u+v)+w
u+0 = u
u-u = 0
cu is within V
c(u+v) = cu+cv
(c+d)u = cu+du
c(du) = (cd)u
1*u = u

Diagon­ali­zable Matrices

A is diagon­ali­zable when A is similar to a diagonal matrix.
That is, A is diagon­ali­zable when there exists an invertible matrix P such that P-1AP is a diagonal matrix

Dot Products Etc.

length­/norm ||v|| = sqrt(v­_12 +...+ v_n2
||cv|| = |c| ||v||
v / ||v|| is the unit vector
distance d(u,v) = ||u-v||
Dot product u•v = (u_1v_1 +...+ u_nv_n)
n cos(theta) = u•v / (||u|| ||v||)
u&v are orthagonal when dot(u,v) = 0


The scalar lambda(Y) is called an Eige­nva­lue of A when there is a nonzero vector x such that Ax = Yx.
Vector x is an Eige­nve­ctor of A corres­ponding to Y.
The set of all eigenv­ectors with the zero vector is a subspace of Rn called the Eige­nsp­ace of Y.
1. Find Eigenv­alues: det(YI - A) = 0
2. Find Eigenv­ectors: (YI - A)x = 0
If A is a triangular matrix then its eigenv­alues are on its main diagonal

Gram-S­chmidt Orthon­orm­ali­zation

1. B = {v1, v2, ..., vn}
2. B' = {w1, w2, ..., wn}:
w1 = v1
w2 = v2 - projw1v2
w3 = v3 - projw1v3 - projw2v3
wn = vn - ...
3. B'' = {u1, u2, ..., un}:
ui = wi/||wi||
B'' is an orthon­ormal basis for V
span(B) = span(B'')

Important Vector Spaces

C(-inf, +inf)
C[a, b]

Inner Products

||u|| = sqrt<u­,u>
d(u,v) = ||u-v||
cos(theta) = <u,­v> / (||u|| ||v||)
u&v are orthagonal when <u,­v> = 0
proj_v u = <u,­v>/­<v,­v> * v


For T:V->W The set of all vectors v in V that satisfies T(v)=0 is the kernal of T. ker(T) is a subspace of v.
For T:Rn ->Rm by T(x)=Ax ker(T) = solution space of Ax=0 & Cspace(A) = range(T)

Linear Combo

v is a linear combo of u_1 ... u_n

Linear Indepe­ndence

a set of vectors S is LI if c1v1 +...+ ckvk = 0 has only the trivial solution.
If there are other solutions S is LD. A set S is LI iff one of its vectors can
be written as a combo of other S vectors.

Linear Transf­orm­ation

V & W are Vspaces. T:V->W is a linear transf­orm­ation of V into W if:
1. T(u+v) = T(u) = T(v)
2. T(cu) = cT(u)


If xp is a solution to Ax = b then every solution to the system can be written as x = xp


Nullsp­ace(A) = {x ε Rn : Ax = 0
Nullity(A) = dim(Nu­lls­pac­e(A))
= n - rank(A)

Orthogonal Sets

Set S in V is orthogonal when every pair of vectors in S is orthog­onal. If each vector is a unit vector, then S is orthon­ormal

One-to-One and Onto

T is one-to-one iff ker(T) = {0}
T is onto iff rank(T) = dim(W)
If dim(T) = dim(W) then T is one-to-one iff it is onto

Rank and Nullity of T

nullity(T) = dim(ke­rnal)
rank(T) = dim(range)
range(T) + nullity(T) = n (in m_x n)
dim(do­main) = dim(range) + dim(ke­rnal)

Rank of a Matrix

Rank(A) = dim(Rs­pace) = dim(Cs­pace)

Similar Matrices

For square matrices A and A' of order n, A' is similar to A when there exits an invertible matrix P such that A' = P-1 AP

Spanning Sets

S = {v1...vk} is a subset of vector space V. S spans V if every vector in v can be written as a linear combo of vectors in S.

Test for Subspace

1. u+v are in W
2. cu is in w

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