Consider the matrix \begin{equation*} B= \begin{bmatrix} -2 & 1 & -2 & -4\\ 12 & 1 & 4 & 9\\ 6 & 5 & -2 & -4\\ 3 & -4 & 5 & 10 \end{bmatrix} \end{equation*} then \begin{equation*} \charpoly{B}{x}=8-20x+18x^2-7x^3+x^4=(x-1)(x-2)^3 \end{equation*} So the eigenvalues are $\lambda=1,\,2$ with algebraic multiplicities $\algmult{B}{1}=1$ and $\algmult{B}{2}=3$.
Computing eigenvectors, \begin{align*} \lambda&=1&B- 1I_4&= \begin{bmatrix} -3 & 1 & -2 & -4\\ 12 & 0 & 4 & 9\\ 6 & 5 & -3 & -4\\ 3 & -4 & 5 & 9 \end{bmatrix} \rref \begin{bmatrix} \leading{1} & 0 & \frac{1}{3} & 0\\ 0 & \leading{1} & -1 & 0\\ 0 & 0 & 0 & \leading{1}\\ 0 & 0 & 0 & 0 \end{bmatrix}\\ &&\eigenspace{B}{1}&=\nsp{B-1I_4} =\spn{\set{\colvector{-\frac{1}{3}\\1\\1\\0}}} =\spn{\set{\colvector{-1\\3\\3\\0}}}\\ \lambda&=2&B-2I_4&= \begin{bmatrix} -4 & 1 & -2 & -4\\ 12 & -1 & 4 & 9\\ 6 & 5 & -4 & -4\\ 3 & -4 & 5 & 8 \end{bmatrix} \rref \begin{bmatrix} \leading{1} & 0 & 0 & 1/2\\ 0 & \leading{1} & 0 & -1\\ 0 & 0 & \leading{1} & 1/2\\ 0 & 0 & 0 & 0 \end{bmatrix}\\ &&\eigenspace{B}{2}&=\nsp{B-2I_4} =\spn{\set{\colvector{-\frac{1}{2}\\1\\-\frac{1}{2}\\1}}} =\spn{\set{\colvector{-1\\2\\-1\\2}}}\\ \end{align*}
So each eigenspace has dimension 1 and so $\geomult{B}{1}=1$ and $\geomult{B}{2}=1$. This example is of interest because of the discrepancy between the two multiplicities for $\lambda=2$. In many of our examples the algebraic and geometric multiplicities will be equal for all of the eigenvalues (as it was for $\lambda=1$ in this example), so keep this example in mind. We will have some explanations for this phenomenon later (see Example NDMS4).