Solve Linear Algebra Problem with Invertible Matrix - Finding Matrix B(t)

[motherh]

Hey guys, I'm having problems with a question.

Let P be an invertible matrix and assume that A = PMP^{-1}. Where M is

M = [{3,1,0}{0,3,0}{0,0,2}]

Find a matrix B(t) such that e^{tA} = PB(t)P^{-1}.

Now this might be an easy problem, but I really have no idea what to do because my lecturer is so bad and the book for the course doesn't cover this material.

I have seen something about A= PBP^{-1} implying e^{tA} = Pe^{tB}P^{-1} so I have tried computing the exponential of M, but to no avail. Any advice is much appreciated.

 

[Dick]

Hey guys, I'm having problems with a question.

Let P be an invertible matrix and assume that A = PMP^{-1}. Where M is

M = [{3,1,0}{0,3,0}{0,0,2}]

Find a matrix B(t) such that e^{tA} = PB(t)P^{-1}.

Now this might be an easy problem, but I really have no idea what to do because my lecturer is so bad and the book for the course doesn't cover this material.

I have seen something about A= PBP^{-1} implying e^{tA} = Pe^{tB}P^{-1} so I have tried computing the exponential of M, but to no avail. Any advice is much appreciated.

Yes, matrix exponential. This problem is fairly easy because you can split Mt into the sum of a diagonal matrix D and an offdiagonal matrix N which is nilpotent. And they commute with each other. So you can use exp(D+N)=exp(D)exp(N). Finding the exponential of each matrix is pretty easy.

 

[motherh]

Thank you so much, you hero! Just to check I haven't gone completely wrong, should I have

B(t) = e^{2t}*[{e^{t},e^{t},0}{0,e^{t},0}{0,0,1}]?

 

[Dick]

Thank you so much, you hero! Just to check I haven't gone completely wrong, should I have

B(t) = e^{2t}*[{e^{t},e^{t},0}{0,e^{t},0}{0,0,1}]?

Close, but no cigar. That's only correct for t=1. Your nilpotent matrix is N={[0,t,0],[0,0,0],[0,0,0]}. What's exp(N)?

 

[motherh]

Aha, of course! I think I see where I went wrong. Is it

B(t) = e^{2t}*[{e^{t},t*e^{t},0}{0,e^{t},0}{0,0,1}]?

 

[Dick]

Aha, of course! I think I see where I went wrong. Is it

B(t) = e^{2t}*[{e^{t},t*e^{t},0}{0,e^{t},0}{0,0,1}]?

Yup!

 

[motherh]

Thank you so much, I really appreciate it!