Consider matrix $A=\left[\begin{matrix}1 & -2\\-2 & 1\\\end{matrix}\right]$, with characteristic equation $\begin{vmatrix}1-\lambda & -2\\-2 & 1-\lambda\\\end{vmatrix}=(\lambda+1)(\lambda-3)$
$\lambda_1=-1$ gives eigenvector $\begin{bmatrix}1\\1\\\end{bmatrix}$
and $\lambda_2=3$ gives eigenvector $\begin{bmatrix}-1\\1\\\end{bmatrix}$, resulting in $P=\begin{bmatrix}1&-1\\1&1\\\end{bmatrix}$
$D=P^TAP=\begin{bmatrix}1&1\\-1&1\\\end{bmatrix}\begin{bmatrix}1&-2\\-2&1\\\end{bmatrix}\begin{bmatrix}1&-1\\1&1\\\end{bmatrix}$ gives me $D=\begin{bmatrix}-1&0\\0&3\\\end{bmatrix}$
But, the answer tells me it's supposed to be $D=\begin{bmatrix}3&0\\0&-1\\\end{bmatrix}$
Does the order of eigenvalues and eigenvectors make a difference? If so, how?
Or is both of the resulting $D$ "equal" and correct? Also, maybe I made a mistake somewhere…
Thank you!
Best Answer
Your answer is fine too, but it doesn't only change $D$, it's connected with $P$:
The order of the eigenvalues on the diagonal in $D$, match the order of the corresponding eigenvectors in the columns of $P$ : if eigenvector $\vec v_i$ is in the $i$-th column of $P$, then the corresponding eigenvalue $\lambda_i$ is on the $ i$-th (diagonal) position in $D$ and vice versa.