$A =\begin{bmatrix} -14 & 12 \\ -20 & 17 \\ \end{bmatrix} $
I did this by first calculating the eigenvalues which turn out to be $λ = 2$ and $λ = 1$
Then I calculated the eigenvectors by first substituting $λ = 2$ and then $λ = 1$ in the $(Iλ – A)$ matrix and reducing both matrices before calculating the eigenvectors.
E.g for $λ = 2$
$\begin{bmatrix} 1 & \frac{-3}{4} \\ 0 & 0 \\ \end{bmatrix}\begin{bmatrix} x_1 \\ x_2\end{bmatrix} =\begin{bmatrix} 0 \\ 0\end{bmatrix} $
In both cases I made $x_2 = t$
which resulted in the first eigenvector being $t \begin{bmatrix} \frac{3}{4} \\ 1\end{bmatrix}$ and the second one $t \begin{bmatrix} \frac{4}{5} \\ 1\end{bmatrix}$
So I made $t = 4$ in the first one and $t = 5$ in the second one to get $ \begin{bmatrix} 3 \\ 4\end{bmatrix}$ and $\begin{bmatrix} 4 \\ 5\end{bmatrix}$ which represent the $P_1$ and $P_2$ columns of the matrix $P = \begin{bmatrix} 3 & 4 \\ 4 & 5 \\ \end{bmatrix}$ which gives: $$P^{-1} AP = \begin{bmatrix} 2 & 0 \\ 0 & 1 \\ \end{bmatrix}$$
In my textbook solution they made $x_1 = t$ for both cases, instead of $x_2$ like I did. Which makes $P_1 = \begin{bmatrix} 4 \\ 5\end{bmatrix}$ and $P_2 = \begin{bmatrix} 3 \\ 4\end{bmatrix}$, so same $P$ but in different order $P = \begin{bmatrix} 4 & 5 \\ 3 & 4 \\ \end{bmatrix}$. When doing $P^{-1} AP$ with this new matrix $P$ the diagonal matrix turns out to be: $$P^{-1} AP = \begin{bmatrix} 1 & 0 \\ 0 & 2 \\ \end{bmatrix}$$
Would this be the same transformation just that in a different basis (represented by the eigenvectors in different order) and is equally valid as a diagonalized matrix $A$?
Best Answer
It is all okay. You just calculated the eigenvalues base in other order. In fact, those two matrices are equivalent.
It is (and will be for you) a well known fact that if the eigenvalues are real and all distinct, then your matrix is diagonalizable and the diagonal matrix is the diagonal matrix with the eigenvalues in the diagonal. (edit: in the ground field, but I said that because I assumed that you were working over the reals)