Question

@Marki Ok so let's do something quick and fun!

Answer 1

yayy

Answer 2

\[\Large \det|e^{A}| = e^{tr(A)}\]

Answer 3

A is a matrix =P

Answer 4

haha

Answer 5

rt constant ?

Answer 6

Ah, nope that's the trace, which is just the sum of the diagonal on the matrix.

Answer 7

oh

Answer 8

Ok so I guess let's show that this is true XD I guess I should say words haha

Answer 9

na its ok
im trying something

Answer 10

im used to
e^tA for t variable :)

Answer 11

hehe

Answer 12

\( \large e^A=I +\sum _{m=0}^{\infty}\frac{k^m }{k!}\)

Answer 13

hmm

Answer 14

sound something wrong with this definition xD

Answer 15

is it true like this ?

Answer 16

Almost, just use the power series like replace x with the matrix A

Answer 17

oh got it

Answer 18

lol sorry k was a typo xD

Answer 19

\(\Huge e^A=I +\sum _{k=1}^{\infty}\frac{A^k }{k!}\)

Answer 20

im too sleeping :(
but have class at 8 AM :( means 1 and 40 mnts left :(

Answer 21

Wnat me to give you more hints cause you're tired so we can get to the answer quicker?

Answer 22

well so far its diagonal , right ?

Answer 23

Yeah, we're going to assume that it's diagonalizable

Answer 24

hehe , well it is :) since we are assuming det not zero :P

Answer 25

\(det(A)=(1+\sum _{k=1 }^{\infty}\frac{A^k }{k!})^n \) for nxn matrix

Answer 26

eh sounds complicated , show what ur got
and we have
\(\large e^{tr(A)}= 1+\sum _{k=1 }^{\infty}\frac{tr(A)^k }{k!} \) hmmm

Answer 27

Well I wrote it as: \[\Large e^A =\sum_{n=0}^\infty \frac{1}{n!}A^n\] and since A is diagonalizable, I wrote it as \[\Large A = P^{-1}DP\] So if we plug it in we have: \[\Large e^A =\sum_{n=0}^\infty \frac{1}{n!}(P^{-1}DP)^n = P^{-1} \left( \sum_{n=0}^\infty \frac{1}{n!}D^n \right) P\]

Answer 28

Since 1/n! is a scalar and exponentiating a diagonal matrix is the same as raising every element to that power, we are really just doing e^x on each of the eigenvalues along the diagonal in the middle, so we can write it as e^D to mean that.
\[\Large \det |P^{-1}e^D P| = e^{tr(P^{-1}DP)}\] So now maybe it's better? Its yours now good luck. =P

Answer 29

:O

Answer 30

that was nice , thanks for sharing !

Answer 31

Yeah, now if you can find a use for it, that would be awesome haha.

Answer 32

:O
i'll show u some other time
what in my mind is solving some ODE

Answer 33

Yeah I had heard of that but not really sure how to use this. But this equation is awesome cause the side with the trace makes it easy, I love it.

Answer 34

I want to make a theorem, "All the best formulas have e^x in them" lol

Answer 35

lol