Basis
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 |
AND |
v = v1i + v2j + v3k |
THEN |
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 |
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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 |
Eigenshit
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}: |
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w1 = v1 |
w2 = v2 - projw1v2 |
w3 = v3 - projw1v3 - projw2v3 |
wn = vn - ... |
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3. B'' = {u1, u2, ..., un}: |
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ui = wi/||wi|| |
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B'' is an orthonormal basis for V |
span(B) = span(B'') |
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Important Vector Spaces
Rn |
C(-inf, +inf) |
C[a, b] |
P |
P_n |
M_m,n |
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 |
Kernal
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) |
Non-Homogeny
If xp is a solution to Ax = b then every solution to the system can be written as x = xp |
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Nullity
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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