For what values of K is the matrix diagonalizable?

Question

Can someone please help with this problem. I have tried it several times but can't get the answer as k*0 = 0 makes it hard to work with k.

Answer 1

Hint: What are the eigenvalues? Do the eigenvectors span $\mathbb{R}^2$?

Solution:

Since $\mathbf{B}$ is upper triangular, its eigenvalues are $6$ and $9$ for all $k\in\mathbb{R}$. Eigenvectors corresponding to distinct eigenvalues are linearly independent, so it is possible to construct a basis for $\mathbb{R}^2$ consisting of eigenvectors. Thus, $\mathbf{B}$ is diagonalizable for all $k\in\mathbb{R}$.