Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Problem: Find the transition matrix P such that $P^{-1}AP=B$ where: $$A=\begin{bmatrix} 3 & -1 & 0 \\ -1 & 0 & -1 \\ 0 & 1 & 1 \end{bmatrix} \quad\text{and}\quad B=\begin{bmatrix} \frac{2}{\sqrt{3}} & 0 & 0 \\ \frac{4}{\sqrt{2}} & \frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{2}{\sqrt{6}} & \frac{-3}{\sqrt{6}} & \frac{-3}{\sqrt{6}} \end{bmatrix}$$

So I am unsure how to do this because we have not discussed it. I was just wondering If somebody could give me some ideas about how to find transition matrices and what exactly they are. Any help is appreciated. Thanks

share|cite|improve this question
It's quite all right. I can tell you off the top of my head that transition matrices have all non-negative real entries, and that the entries in each column (or row, or both) sum to 1. You need such a matrix $P$, which will be $3\times 3$, and satisfy $AP=PB$. That probably isn't the most elegant way to go about it, though. Have you discussed eigenvalues and eigenvectors, yet? – Cameron Buie May 15 '12 at 2:37
Yea we studied them for awhile a couple weeks ago – Mathstudent May 15 '12 at 2:39
Actually, now that I think about it, I'm not sure that will help you, since $B$ isn't diagonal. I'm blanking, here. Long day. Go ahead with the one approach I gave, if you like (it isn't too laborious). Eventually someone else will come along and give you better advice. – Cameron Buie May 15 '12 at 2:41
Ok Thanks a lot! – Mathstudent May 15 '12 at 2:42
Something's odd here: $\,A\,,\,B\,$ don't have the same trace so there can't exist a matrix $\,P\,$ a required... – DonAntonio May 15 '12 at 3:40

This is impossible. $A$ has a real eigenvalue greater than 3. $B$ has a real eigenvalue less than 2. $B$'s other two eigenvalues are not real.

(I just evaluated the characteristic polynomial of $A$ at 1,2,3. The form of $B$ shows one real eigenvalue and a complex pair.)

share|cite|improve this answer
More simply, as Don Antonio notes in the comments, $A$ and $B$ don't have the same trace. – Gerry Myerson May 15 '12 at 3:57
@GerryMyerson: My default is brute force... – copper.hat May 15 '12 at 3:58


  1. $A, B$ have the same eigenvalues... why?

  2. Recall what eigendecomposition is..

  3. What does the eigendecomposition on both $A, B$ in $P^{-1} A P = B$ would look like?

share|cite|improve this answer
$A,B$ should have the same eigenvalues...but apparently they don't! – Gerry Myerson May 15 '12 at 3:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.