Matrix Diagonalisation

Question 1

caley hamilton theorem

Answer 1

let B be a square matrix with characteristic polynomial Pb(t) then
Pb(b) = 0 as a matrix eq
find Pb(t) then sub in B for t and times constant no matrix part by I

Question 2

equivalence relation

Answer 2

~ on a set X is a bianry relation (between 2 elements) with the following properties
- a ~a (reflexivity)
- if a ~ b then b ~ a (symmetry)
- if a ~ b and b ~ c then a ~ c (transitivity)

Question 3

similar

Answer 3

2 square matrices A and B of same size are similar if there exists an invertible matrix M such that A = MBM^-1

Question 4

similarity and equivalence

Answer 4

similarity is an equivalence relation A~B if A and B are similar

Question 5

similarity and eigenvalues

Answer 5

similar matrices have the same eigenvalues
(but not the other way round - same eigenvalues dont mean similar)

Question 6

diagonalisable

Answer 6

a square matrix A is diagonalisable if A is similar to a diagonal matrix D
A = MDM^-1
D has λs diagonal and 0s elsewhere

Question 7

facts about diagonalisable

Answer 7

• if A and B same eigenvalues and both diagonalisable then A~B
• if A is diagonalisable and B not then are not similar
• if eigenvalues of A are distinct then A is diagonalisable

Question 8

diagonalisable and eigenvectors

Answer 8

A is diagonalisable iff it has n linearly independent eigenvectors

Question 9

diagonalisable and geometric multiplicity

Answer 9

dimV = k for A to be diagonalisable
for a matrix to be diagonalisable for each λ we must have dimV = k in this case
∑ ki = ∑dimVλi = n
V = Vλ1 ⊕ Vλ2 ⊕…

Question 10

solving a system of differential eq using diagonalisation

Answer 10

d/dt v = Av
d/dt M^-1v = M^-1AMM^-1v
d/dt w = D w
as D = M^-1AM and w = M^-1v
can then solve easily and v = Mw so convert back

Question 11

eigenvectors eigenvalues and LI

Answer 11

eigenvectors that correspond to distinct eigenvalues of A are LI
if λ1 ≠ λ2 ≠ …≠ λn
then {v1,v2..vn} are LI

Question 12

eigenvalues and diagonalisation

Answer 12

if all eigenvalues are distinct the matrix A is diagonalisable

Question 13

remark about diagonalisation and complex

Answer 13

• if A real may be possible to diagonalise A over C but not over R
• the standard basis in C^n is the same as R^n
  eg z = z1 e1 + z2 e2 + …+ zn en