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 $D = PTP^{-1}$?

  • $\begingroup$ 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

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.


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