Answer
The matrix is not diagonalizable.
Work Step by Step
Consider the matrix A = $\begin{pmatrix} 5 & 1 \\ 0 & 5 \end{pmatrix}$. We begin by finding the respective eigenvalues:
$\lambda_n$ = $det(A - \lambda I) = \begin{vmatrix} 5 - \lambda & 1 \\ 0 & 5 - \lambda \end{vmatrix} = (5 - \lambda)(5 - \lambda)$
$\therefore \lambda = 5$
Now, we use the fact that $\lambda = 5$ to find the eigenvectors for $A$:
$Nul(A - \lambda I) = Nul(A - 5 I) = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}$
$\therefore Nul(A - \lambda I) = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$
As the number of eigenvectors is smaller than $n$, they do not form a linearly independent basis for $\mathbb{R}^n$, therefore A is not diagonalizable.
