0
$\begingroup$

I am having a hard time trying to figure out the order eigenvalues and eigenvectors result in when trying to diagonalize a $2\times 2$ matrix: $$\left[\begin{matrix} 3 & 1\\ 1 & 3 \end{matrix}\right]$$

$\lambda_1 = 4, \lambda_2 = 2$ with basis $b_1 = (1, 1), b_2 = (-1,1)$.

When diagonalizing to matrix with $P = \left[\begin{matrix} 1 & 1\\ -1 & 1\end{matrix}\right]$ the diagonal matrix is $D = \left[\begin{matrix} 2 & 0\\ 0 & 4\end{matrix}\right]$, and the eigenvalues are in a different order and I'm not entirely sure why.

Should the equation be $D = P^{-1}TP$ instead of $D = PTP^{-1}$?

$\endgroup$
  • $\begingroup$ If you want to switch the order of the eigenvalues, switch the order of the columns of $P$. $\endgroup$ – Git Gud Jun 4 '13 at 6:19
  • 2
    $\begingroup$ The eigenvectors go into columns of $P$, not rows. $\endgroup$ – Erick Wong Jun 4 '13 at 6:30
  • $\begingroup$ @ErickWong Well caught! Didn't notice what the OP was doing. $\endgroup$ – Git Gud Jun 4 '13 at 6:31
3
$\begingroup$

Note that the first column of $P$ is an eigenvector for the eigenvalue $\lambda_2 = 2$, and the second column of $P$ is an eigenvector for the eigenvalue $\lambda_1 = 4$. That is why $2$ appears on the diagonal before $4$.

As Git Gud mentioned, if you swap the two columns of $P$, the two diagonal elements will switch places.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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