Consider the matrix \begin{equation*} E= \begin{bmatrix} 29 & 14 & 2 & 6 & -9\\ -47 & -22 & -1 & -11 & 13\\ 19 & 10 & 5 & 4 & -8\\ -19 & -10 & -3 & -2 & 8\\ 7 & 4 & 3 & 1 & -3 \end{bmatrix} \end{equation*} then \begin{equation*} \charpoly{E}{x}=-16+16x+8x^2-16x^3+7x^4-x^5=-(x-2)^4(x+1) \end{equation*} So the eigenvalues are $\lambda=2,\,-1$ with algebraic multiplicities $\algmult{E}{2}=4$ and $\algmult{E}{-1}=1$.
Computing eigenvectors, \begin{align*} \lambda&=2\\ E-2I_5&= \begin{bmatrix} 27 & 14 & 2 & 6 & -9\\ -47 & -24 & -1 & -11 & 13\\ 19 & 10 & 3 & 4 & -8\\ -19 & -10 & -3 & -4 & 8\\ 7 & 4 & 3 & 1 & -5 \end{bmatrix} \rref \begin{bmatrix} \leading{1} & 0 & 0 & 1 & 0\\ 0 & \leading{1} & 0 & -\frac{3}{2} & -\frac{1}{2}\\ 0 & 0 & \leading{1} & 0 & -1\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}\\ \eigenspace{E}{2}&=\nsp{E-2I_5} =\spn{\set{\colvector{-1\\\frac{3}{2}\\0\\1\\0},\,\colvector{0\\\frac{1}{2}\\1\\0\\1}}} =\spn{\set{\colvector{-2\\3\\0\\2\\0},\,\colvector{0\\1\\2\\0\\2}}}\\ \lambda&=-1\\ E+1I_5&= \begin{bmatrix} 30 & 14 & 2 & 6 & -9\\ -47 & -21 & -1 & -11 & 13\\ 19 & 10 & 6 & 4 & -8\\ -19 & -10 & -3 & -1 & 8\\ 7 & 4 & 3 & 1 & -2 \end{bmatrix} \rref \begin{bmatrix} \leading{1} & 0 & 0 & 2 & 0\\ 0 & \leading{1} & 0 & -4 & 0\\ 0 & 0 & \leading{1} & 1 & 0\\ 0 & 0 & 0 & 0 & \leading{1}\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}\\ \eigenspace{E}{-1}&=\nsp{E+1I_5}=\spn{\set{\colvector{-2\\4\\-1\\1\\0}}}\\ \end{align*}
So the eigenspace dimensions yield geometric multiplicities $\geomult{E}{2}=2$ and $\geomult{E}{-1}=1$. This example is of interest because $\lambda=2$ has such a large algebraic multiplicity, which is also not equal to its geometric multiplicity.