Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

I'm trying to find the value of this matrix, $A = \begin{pmatrix} 1 & 4 \\ 3 & 2\end{pmatrix}$ to the power of $10$.

I've determined (and confirmed on Wolfram) that the eigenvalues are $5$, and $-2$.
I started looking for P by using the eigenvalue $5$, and found the eigenvector $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$.
I did the same for the eigenvalue $-2$, and found eigenvector $\begin{bmatrix} -4 \\ 3 \end{bmatrix}$

I then put these together to get $P: \begin{bmatrix} 1 & -4\\ 1 & 3 \end{bmatrix}$. I followed that by finding $P^{-1}: \dfrac{1}{7} \begin{bmatrix} 3 & 4\\ -1 & 1 \end{bmatrix}$.

The problem is, that $PAP^{-1}$ is not diagonalizing. I checked Wolfram, and I found that the reason is because my $P$ should have been $\begin{bmatrix} -4 & 1\\ 3 & 1 \end{bmatrix}$.

My question is, what is the general rule about what eigenvalue to plug in first and how to arrange the columns of $P$?

share|cite|improve this question
You have and $P$ and $P^{-1}$ swapped, should be $P^{-1}AP$. – Amzoti Dec 9 '12 at 22:17
If $\Pi$ is a permutation matrix and $\Lambda$ is diagonal, then $\Pi^{-1} \Lambda \Pi$ is also diagonal (with the same, rearranged entries). Hence if $P^{-1} A P = \Lambda$, then $(P \Pi)^{-1} A P \Pi = \Pi^{-1} \Lambda \Pi$, the rearranged diagonal matrix. – copper.hat Dec 9 '12 at 22:40

The problem does not lie in the order of the eigenvalues or the order of the columns of $P$. What should be diagonal is $P^{-1}AP$, not $PAP^{-1}$. The order of the eigenvalues or eigenvectors do not matter, as long as the ordering of eigenvalues is consistent with the ordering of eigenvectors.

share|cite|improve this answer
Ahaa!! Silly mistake!! Gratefulness I have to you, thank for showing me. – Phee Dec 9 '12 at 22:20
I made this kind of mistakes all the time. – user1551 Dec 9 '12 at 22:21

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.