Is this matrix diagonalizable for every value of $k$?

Question

If I have a matrix

$$\begin{pmatrix} 2 & -2 & 0\\ 0 & 2 & k \\ 0 & 0 & 3 \end{pmatrix}$$

finding the eigen values and eigenvectors:

with eigenvalues 2 and 3 and eigenvectors $\ v =(-2k,k,1)$ and $\ v1 = (1,0,0)$, is the matrix diagonalizable?

Answer 1

Let $J = \begin{pmatrix} 2&0&0 \\ 0&3&1 \\ 0&0&3 \end{pmatrix}$ and let $V = (v_1,v_2,v_3),$ where $v_3$ is chosen such that $V$ has linearly independent columns. Then the matrix $A = VJV^{-1}$ has exactly two eigenpairs, $(2, v_1)$, and $(3,v_2)$, but is not diagonalizable as it does not have a third eigenvector so that the eigenbasis spans the entire space.

Edit: This answer was given for a previous version of the question about whether the existence of two eigenpairs implies diagonalizability in $\mathbb{R}^3$.

Answer 2

For any $k$, your eigenvalues $2$ and $3$ and associated eigenspaces (generated respectively by $(1,0,0)$ and $(-2k,k,1)$) are correct, and sufficient to conclude that the matrix is not diagonalizable, since the sum of the dimensions of the eigenspaces, $1+1=2$, is strictly smaller than the dimension of the matrix, $3$.