Finding the third eigenvector of a diagonalizing matrix.

Question

$$A = \begin{pmatrix} 2 & 0 & 3 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{pmatrix}$$ 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?

Answer 1

By eyeballing, we can see that the nullspace of $\begin{bmatrix}-1&0&3\\0&0&0\\0&0&0 \end{bmatrix}$ also contains $\begin{bmatrix}3\\0\\1\end{bmatrix}$.

Answer 2

Your eigenspace for $\lambda=3$ is two dimensional: Solving(not eyeballing)

$$\begin{pmatrix}-1 &0 &3\\0 &0 &0\\0 &0 &0\end{pmatrix}x=\mathbf0$$

yields the following parametric solution:

$$x=c(3,0,1)^\top+c'(0,1,0)^\top$$

Thus, you have two linearly independent eigenvectors(spanning the corresponding eigenspace) corresponding to the eigenvalue $\lambda=3$ and one eigenvector(spanning the corresponding eigenspace) corresponding to $\lambda=2$. These now form a basis s.t. your matrix is diagonalizable.

Answer 3

The matrix $A-3I$ is one step away from being in row-reduced echelon form: just multiply the first row by $-1$. You can then read a basis for its null space directly from the rref using the method described in this answer, which yields the vectors $(0,1,0)^T$ and $(3,0,1)^T$.