Consider 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*} then \begin{equation*} \charpoly{H}{x}=-6x+x^2+7x^3-x^4-x^5=x(x-2)(x-1)(x+1)(x+3) \end{equation*} So the eigenvalues are $\lambda=2,\,1,\,0,\,-1,\,-3$ with algebraic multiplicities $\algmult{H}{2}=1$, $\algmult{H}{1}=1$, $\algmult{H}{0}=1$, $\algmult{H}{-1}=1$ and $\algmult{H}{-3}=1$.
Computing eigenvectors, \begin{align*} \lambda&=2\\&H-2I_5= \begin{bmatrix} 13 & 18 & -8 & 6 & -5\\ 5 & 1 & 1 & -1 & -3\\ 0 & -4 & 3 & -4 & -2\\ -43 & -46 & 17 & -16 & 15\\ 26 & 30 & -12 & 8 & -12 \end{bmatrix} \rref \begin{bmatrix} \leading{1} & 0 & 0 & 0 & -1\\ 0 & \leading{1} & 0 & 0 & 1\\ 0 & 0 & \leading{1} & 0 & 2\\ 0 & 0 & 0 & \leading{1} & 1\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}\\ &\eigenspace{H}{2}=\nsp{H-2I_5} =\spn{\set{\colvector{1\\-1\\-2\\-1\\1}}} \end{align*} \begin{align*} \lambda&=1\\&H-1I_5= \begin{bmatrix} 14 & 18 & -8 & 6 & -5\\ 5 & 2 & 1 & -1 & -3\\ 0 & -4 & 4 & -4 & -2\\ -43 & -46 & 17 & -15 & 15\\ 26 & 30 & -12 & 8 & -11 \end{bmatrix} \rref \begin{bmatrix} \leading{1} & 0 & 0 & 0 & -\frac{1}{2}\\ 0 & \leading{1} & 0 & 0 & 0\\ 0 & 0 & \leading{1} & 0 & \frac{1}{2}\\ 0 & 0 & 0 & \leading{1} & 1\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}\\ &\eigenspace{H}{1}=\nsp{H-1I_5} =\spn{\set{\colvector{\frac{1}{2}\\0\\-\frac{1}{2}\\-1\\1}}} =\spn{\set{\colvector{1\\0\\-1\\-2\\2}}} \end{align*} \begin{align*} \lambda&=0\\&H-0I_5= \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} \rref \begin{bmatrix} \leading{1} & 0 & 0 & 0 & 1\\ 0 & \leading{1} & 0 & 0 & -2\\ 0 & 0 & \leading{1} & 0 & -2\\ 0 & 0 & 0 & \leading{1} & 0\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}\\ &\eigenspace{H}{0}=\nsp{H-0I_5} =\spn{\set{\colvector{-1\\2\\2\\0\\1}}} \end{align*} \begin{align*} \lambda&=-1\\&H+1I_5= \begin{bmatrix} 16 & 18 & -8 & 6 & -5\\ 5 & 4 & 1 & -1 & -3\\ 0 & -4 & 6 & -4 & -2\\ -43 & -46 & 17 & -13 & 15\\ 26 & 30 & -12 & 8 & -9 \end{bmatrix} \rref \begin{bmatrix} \leading{1} & 0 & 0 & 0 & -1/2\\ 0 & \leading{1} & 0 & 0 & 0\\ 0 & 0 & \leading{1} & 0 & 0\\ 0 & 0 & 0 & \leading{1} & 1/2\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}\\ &\eigenspace{H}{-1}=\nsp{H+1I_5} =\spn{\set{\colvector{\frac{1}{2}\\0\\0\\-\frac{1}{2}\\1}}} =\spn{\set{\colvector{1\\0\\0\\-1\\2}}} \end{align*} \begin{align*} \lambda&=-3\\&H+3I_5= \begin{bmatrix} 18 & 18 & -8 & 6 & -5\\ 5 & 6 & 1 & -1 & -3\\ 0 & -4 & 8 & -4 & -2\\ -43 & -46 & 17 & -11 & 15\\ 26 & 30 & -12 & 8 & -7 \end{bmatrix} \rref \begin{bmatrix} \leading{1} & 0 & 0 & 0 & -1\\ 0 & \leading{1} & 0 & 0 & \frac{1}{2}\\ 0 & 0 & \leading{1} & 0 & 1\\ 0 & 0 & 0 & \leading{1} & 2\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}\\ &\eigenspace{H}{-3}=\nsp{H+3I_5} =\spn{\set{\colvector{1\\-\frac{1}{2}\\-1\\-2\\1}}} =\spn{\set{\colvector{-2\\1\\2\\4\\-2}}} \end{align*}
So the eigenspace dimensions yield geometric multiplicities $\geomult{H}{2}=1$, $\geomult{H}{1}=1$, $\geomult{H}{0}=1$, $\geomult{H}{-1}=1$ and $\geomult{H}{-3}=1$, identical to the algebraic multiplicities. This example is of interest for two reasons. First, $\lambda=0$ is an eigenvalue, illustrating the upcoming Theorem SMZE. Second, all the eigenvalues are distinct, yielding algebraic and geometric multiplicities of 1 for each eigenvalue, illustrating Theorem DED.