Decide whether matrix $A$ is diagonalizable. If so, find $P$ such that $P^{-1}AP$ is diagonal.
We are given: $A = \begin{bmatrix}1 & 0 & 0 \\ 1 & 2 & 1 \\ 0 & 0 & 1\end{bmatrix}$
We set up and and solve: $|A - \lambda I| = 0$, which yields:
$$\left|\begin{matrix}1-\lambda & 0 & 0 \\ -1 & 2-\lambda & -1 \\ 0 & 0 & 1-\lambda\end{matrix}\right| = 0$$
This yields a characteristic polynomial and eigenvalues as:
$$-\lambda^3 + 4 \lambda^2 - 5 \lambda + 2 = (1-\lambda)^2 (2-\lambda) = 0 ~~~\rightarrow ~~~ \lambda_1 = 1, \lambda_{2,3} = 2$$
We have multiplicities of $2$ and $1$ for those eigenvalues.
So, for $\lambda_1 = 1$, we have:
$[A- I]v_1 = \begin{bmatrix}0 & 0 & 0 \\ -1 & 1 & -1 \\ 0 & 0 & 0\end{bmatrix}v_1 = 0$
Thus, $v_1 = (1,1,0)$ and $v_2 = (0,1,1)$ .
Repeating this same process for the second eigenvalue, we have as RREF:
$\begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}v_3 = 0$
This yields an eigenvector of $v_3 = (0,1,0)$.
Putting all of this together, we have the eigenvalue/eigenvector pairs:
- $\lambda_1 = 1, v_1 = (1,1,0)$
- $\lambda_2 = 1, v_2 = (0,1,1)$
- $\lambda_3 = 2, v_3 = (0,2,0)$
So P=\begin{bmatrix}1 & 0 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 0\end{bmatrix}
Is this right? Do I need to show more work? Do I need to take any additional steps after I find $P$?