# Linear Algebra Problem

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
Homework Helper
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.

Last edited:
• 2 people
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
Homework Helper
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)?

• 1 person
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
Homework Helper
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!

• 1 person
Thank you so much, I really appreciate it!