Find $P$ such that $A= PJP^{-1}$

Question

$A$ = $\begin{bmatrix}2 & 3 \cr 3 & 2\end{bmatrix}$

I have found the characteristic polynomial and minimal polynomial to be: $(x+1)(x-5)$

Also the JCF to be $J$ =$\begin{bmatrix}-1 & \cr &5 \end{bmatrix}$ because A has two distinct eigenvalues and is diagonalizable. All examples of finding $P$ I have found only have one eignevalue and are not diagonlizable so I'm not sure how to find $P$ exactly. Using $P$ to find $A^n = PJ^nP^{-1}$

Answer 1

The eigenspaces are generated by the vectors $\begin{bmatrix}1\\-1\end{bmatrix}$ (eigenvalue $-1$) and $\begin{bmatrix}1\\1\end{bmatrix}$ (eigenvalue $5$). Hence, if we take $P=\begin{bmatrix}1&1\\-1&1\end{bmatrix}$, we have $$J=P^{-1}AP\iff A=PJP^{-1}.$$

Answer 2

Since the matrix is diagonalizable, the Jordan form will exactly coincide with the diagonalization. The Jordan form is only different in the case that the matrix is non-diagonalizable.

As a side note, $A$ is real-symmetric, hence orthogonally diagonalizable.

1. Find eigenvectors $\vec{v_1}$ and $\vec{v_2}$ associated with the eigenvalues of $\lambda_1 = -1$ and $\lambda_2 = 5$.
2. Form $P = \begin{bmatrix} \vec{v_1} & \vec{v_2} \end{bmatrix}$. Since $\vec{v_1}$ and $\vec{v_2}$ are from distinct eigenvalues, they are automatically linearly independent. Thus $P$ is invertible.
3. $A = P \begin{bmatrix} -1 & \\ 0 & 5 \end{bmatrix} P^{-1}$
4. $A^n = P \begin{bmatrix} -1 & \\ 0 & 5 \end{bmatrix}^n P^{-1}$