Consider the matrix \begin{equation*} C= \begin{bmatrix} 1 & 0 & 1 & 1\\ 0 & 1 & 1 & 1\\ 1 & 1 & 1 & 0\\ 1 & 1 & 0 & 1 \end{bmatrix} \end{equation*} then \begin{equation*} \charpoly{C}{x}=-3+4x+2x^2-4x^3+x^4=(x-3)(x-1)^2(x+1) \end{equation*} So the eigenvalues are $\lambda=3,\,1,\,-1$ with algebraic multiplicities $\algmult{C}{3}=1$, $\algmult{C}{1}=2$ and $\algmult{C}{-1}=1$.
Computing eigenvectors, \begin{align*} \lambda&=3&C- 3I_4&= \begin{bmatrix} -2 & 0 & 1 & 1\\ 0 & -2 & 1 & 1\\ 1 & 1 & -2 & 0\\ 1 & 1 & 0 & -2 \end{bmatrix} \rref \begin{bmatrix} \leading{1} & 0 & 0 & -1\\ 0 & \leading{1} & 0 & -1\\ 0 & 0 & \leading{1} & -1\\ 0 & 0 & 0 & 0 \end{bmatrix}\\ &&\eigenspace{C}{3}&=\nsp{C-3I_4} =\spn{\set{\colvector{1\\1\\1\\1}}}\\ \lambda&=1&C-1I_4&= \begin{bmatrix} 0 & 0 & 1 & 1\\ 0 & 0 & 1 & 1\\ 1 & 1 & 0 & 0\\ 1 & 1 & 0 & 0 \end{bmatrix} \rref \begin{bmatrix} \leading{1} & 1 & 0 & 0\\ 0 & 0 & \leading{1} & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}\\ &&\eigenspace{C}{1}&=\nsp{C-1I_4} =\spn{\set{\colvector{-1\\1\\0\\0},\,\colvector{0\\0\\-1\\1}}}\\ \lambda&=-1&C+1I_4&= \begin{bmatrix} 2 & 0 & 1 & 1\\ 0 & 2 & 1 & 1\\ 1 & 1 & 2 & 0\\ 1 & 1 & 0 & 2 \end{bmatrix} \rref \begin{bmatrix} \leading{1} & 0 & 0 & 1\\ 0 & \leading{1} & 0 & 1\\ 0 & 0 & \leading{1} & -1\\ 0 & 0 & 0 & 0 \end{bmatrix}\\ &&\eigenspace{C}{-1}&=\nsp{C+1I_4} =\spn{\set{\colvector{-1\\-1\\1\\1}}}\\ \end{align*}
So the eigenspace dimensions yield geometric multiplicities $\geomult{C}{3}=1$, $\geomult{C}{1}=2$ and $\geomult{C}{-1}=1$, the same as for the algebraic multiplicities. This example is of interest because $A$ is a symmetric matrix, and will be the subject of Theorem HMRE.