Algebra question types

is V a vector space?

find if summ of 2 points in the vector space satisfy V, same for multiplying by any scalar and should also satisfy V

proof a linear map is a subspace

same as Vector space: T(v)= T(u)+T(w) because v=u+w. and same as vector space. T(a*v)= a*T(v) which is a l.d form bc multilplying by a scalar

how many l.i vectors are there?

see if there is a solution to matrix (columns=vectors) if only solution is (0,0,0) then they are l.i. [0] is always in the vector space. so if rank=#rows then only solution is [0] vector.

rank of a matrix

row reduce the matrix A, the rank is only the number of non-zero rows (aka pivots)

dim of sets of vectors

basis of a subspace

the basis of a subspace are all l.i vectors included in subspcace. If F defined by v1,v2,v3 but only v1,v3 are l.i. then the basis is v1,v3. check by null(f), if solution not [0], then check for other vectors/columns of A

corrdinates from a basis to another

for u, v arrival basis

Yc=AXc

Xc=V* x(b)

Yc= U* y(b)

combining we get —> U* y(b) = A* V* x(b)

y(b)= U^(-1)* A* V* x(b)

grassman formula

dim(f+g)= dim f+ dim g - dim (f inter g)

how to determine dim of subspace

number of vectors of the basis are wht define the dim!!

1: line

2: plane

3: 3d

operations with subspaces

intersection and sum of 2 subspcaces

direct intersection

when F inter G = [0]

intersection of F and G when they span the whole space

if F+G spans the whole space then F and G are complimentary: then the expansion of f, g vectors are all linearly independant and spans E so dim(g inter f)=0

change vector basis (1—> 2)

w= A(b) = C(d) | same vector in different basis

C= A(d) * (d)^(-1)

to obtain vector C in basis d, which is known

diagonalization

for f matrix in B basis made of v1,v2,…, vn vectors

then f(vi) in b= lambda(i)*vi

orthogonal spaces

when F inter Ft = [0] and F+Ft = E

Nullspace (F) =

some values of xyz (or simply a vector x such that Fx=0.

simply a function f such that f(v)=0 for f generated by v vectors

Image of F

dim of null(f)

dim E vector space - rank A

endomorphism

if f function goes from E—> H and E=H

endomorphisms are just multiplication of square matrix A

n*f(x)=A^n*x

diagonalizing a matrix A:

1. define charracteristic polynomial of A: det(A-lambda*Id)

2. find roots (= eigenvalues)

3. for each eigenvalue, find the corresponding eigenvector with null(A-lambda*I) for specific value of lambda

diagonalizing matrix

D is diagonalizing matrix for [(lambda 1, 0), (0, lambda 2)]

We diagonalize A by finding D:

A= VDV^-1

D= V^-1 AV

othogonal vector spaces

if every v has unit length 1

and if l.i (ie orthogonal to each other with scalar product =0)

check if Q is orthogonal

Q*Qt= Id.

isotropy principle

let U= a1v1, … , anvn

and W = b1v1, … , bnvn

then scalar product of U*W = a1b1+ … + anbn

orthogonal projection for dim 1

(<u,v> / ||u||² )* u

orthogonal projection for dim=1 into perpendicular subspace

pi into G(v) = v - pi into Gt(v)

orthogonal projections in general

At* v = At*A*alpha

we get solution as alpha 1(v ) + alpha 2 (v)

where v are vectors generators of A (being a subspace).