Question

Let A, B \(\in M(n, \mathbb C)\)and assume B is invertible. Show that
1) \((B^{-1}AB)^k= B^{-1}A^kB\)
2) Use 1 to show \(e^{tB^{-1}AB}=B^{-1}e^{tA}B\)
Please help

Answer 1

1) is easy.
But I don't know how to do 2
@ganeshie8 @dan815

Answer 2

is it given that A is diagonal matrix ?

Answer 3

Nope, that is all.

Answer 4

invertible meaning non-singular matrix.. there's an inverse.

Answer 5

nvm, I got it. :) Thanks for being here.

Answer 6

for 1)
\((B^{-1}AB)^k = (B^{-1}AB)(B^{-1}AB)........(B^{-1}AB)\)
open parenthesis and \(BB^{-1}=I\) gives us the answer

Answer 7

for 2) by definition \(e^{tA}=\sum_{k=0}^{\infty} \dfrac{t^k}{k!}A^k\)

Answer 8

we replace A by \(B^{-1}AB\) then just simplify it to get the answer.

Answer 9

\(e^{tB^{-1}AB}=\sum_{k=0}^{\infty} \dfrac{t^k}{k!}(B^{-1}AB)^k =\sum_{k=0}^{\infty} \dfrac{t^k}{k!}B^{-1}A^kB \)
?

Answer 10

yeah.... this proof looks familiar. Miracrown helped me with one of them.

Answer 11

how do you conclude

Answer 12

@ganeshie8 sum and t, k are constants, right? hence, we can distribute inside, between \(B^{-1}\) and A

Answer 13

Are you saying we can take \(B^{-1}\) out ?

Answer 14

Yes

Answer 15

\(\sum_{k=0}^{\infty} \dfrac{t^k}{k!}B^{-1}A^kB =B^{-1}\left(\sum_{k=0}^{\infty} \dfrac{t^k}{k!}A^k\right)B \)
? this looks ok intuitively but idk what you're gonna use as justification

Answer 16

Look at the definition of \(e^{tA}\) above

Answer 17

Yes looked.

Answer 18

And we done, right?

Answer 19

I am asking the justification for that distribution

Answer 20

I think its not so obvious, at least for me it is not.

Answer 21

https://en.wikipedia.org/wiki/Matrix_multiplication