In Example DEMS5 the matrix \begin{equation*} H= \begin{bmatrix} 15 & 18 & -8 & 6 & -5\\ 5 & 3 & 1 & -1 & -3\\ 0 & -4 & 5 & -4 & -2\\ -43 & -46 & 17 & -14 & 15\\ 26 & 30 & -12 & 8 & -10 \end{bmatrix} \end{equation*} has characteristic polynomial \begin{equation*} \charpoly{H}{x}=x(x-2)(x-1)(x+1)(x+3) \end{equation*} and so is a $5\times 5$ matrix with 5 distinct eigenvalues.
By Theorem DED we know $H$ must be diagonalizable. But just for practice, we exhibit a diagonalization. The matrix $S$ contains eigenvectors of $H$ as columns, one from each eigenspace, guaranteeing linear independent columns and thus the nonsingularity of $S$. Notice that we are using the versions of the eigenvectors from Example DEMS5 that have integer entries. The diagonal matrix has the eigenvalues of $H$ in the same order that their respective eigenvectors appear as the columns of $S$. With these matrices, verify computationally that $\similar{H}{S}=D$. \begin{align*} S&= \begin{bmatrix} 2 & 1 & -1 & 1 & 1\\ -1 & 0 & 2 & 0 & -1\\ -2 & 0 & 2 & -1 & -2\\ -4 & -1 & 0 & -2 & -1\\ 2 & 2 & 1 & 2 & 1 \end{bmatrix} &D&= \begin{bmatrix} -3 & 0 & 0 & 0 & 0\\ 0 & -1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 2 \end{bmatrix} \end{align*} Note that there are many different ways to diagonalize $H$. We could replace eigenvectors by nonzero scalar multiples, or we could rearrange the order of the eigenvectors as the columns of $S$ (which would subsequently reorder the eigenvalues along the diagonal of $D$).