$$A = \begin{pmatrix} 2 & 0 & 3 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{pmatrix}$$ Prove that $A$ is diagonalizable, find the bases for the eigenspaces, the diagonalizing matrix $P$, and compute $P^{-1} A P$.
Then finding the eigenvectors, we look at the null space of $A-I\lambda$ for the two eigenvalues.
For $\lambda=2$, we find the nullspace of
$\begin{bmatrix}0&0&3\\0&1&0\\0&0&1 \end{bmatrix}$, which eyeballing we can see is the vector $\begin{bmatrix}1\\0\\0\end{bmatrix}$
or $\lambda=3$, we find the nullspace of
$\begin{bmatrix}-1&0&3\\0&0&0\\0&0&0 \end{bmatrix}$ which we can eyeball one easily as $\begin{bmatrix}0\\1\\0\end{bmatrix}$.
Then how to find the third eigenvector?
Best Answer
By eyeballing, we can see that the nullspace of $\begin{bmatrix}-1&0&3\\0&0&0\\0&0&0 \end{bmatrix}$ also contains $\begin{bmatrix}3\\0\\1\end{bmatrix}$.