Question

Matrix mulitipication

Answer 1

Can someone tell my why this is true:

\[\sum_{n=1}^{∞}\frac{ (t(QDQ^{-1}))^n }{ n! }=\sum_{n=1}^{∞}\frac{ Q(tD)^nQ^{-1} }{ n! }\]
Where Q and D is at matrix.

Answer 2

t is scalar ?

Answer 3

t ?

Answer 4

Yes

Answer 5

D is a diagonal if its matters.

Answer 6

yes its true
because
(QDQ−1)^2=(QDQ−1)(QDQ−1)=(QD^2Q−1)
by induction
(QDQ−1)^n=(QD^nQ−1)
t is scalar so
t^n (QD^nQ−1)=(Qt^nD^nQ−1)
=(Q(tD)^nQ−1)

Answer 7

d is diagonal
and Q is jordan form

Answer 8

what course is that you are in ?

Answer 9

It is stochastic processes. Are you a expert in this? :)

Answer 10

my research is in ode , pde
but i had a stochastic diff equation 2 years ago

Answer 11

Thank you @amoodarya