Find Eigen values and eigen vectors $\displaystyle F=\begin{pmatrix}1 & 1 & 2\\ 0 & 1 & 0\\ 0 & 0 & 1\end{pmatrix}$
Expanding along $R_3$, I got the equation $(1-\lambda)(\lambda)(\lambda-2)=0$
Eigenvector corresponds to $\lambda=1$: $\displaystyle \begin{pmatrix}1\\-2\\0\end{pmatrix}$.
Got stuck for $\lambda=2$.
$$\begin{pmatrix} -1 & 1 & 2\\0 & -1 & 0\\0 & 0 & -1\end{pmatrix}$$
Cannot obtain the eigenvector $\displaystyle \begin{pmatrix}1\\0\\0\end{pmatrix}$.

The matrix is triangular, hence its eigenvalues are on the diagonal. In this case we have $\lambda=1$ with algebraic multiplicity of $3$.
For $\lambda=1$ we get $(F-\lambda I)v=0\Rightarrow \begin{pmatrix}0&1&2\\0&0&0\\0&0&0\end{pmatrix}\begin{pmatrix}v_1\\v_2\\v_3\end{pmatrix}=0$ And we get $v_2=-2v_3$, thus the solution is of the form $\displaystyle \begin{pmatrix}v_1\\-2v_3\\v_3\end{pmatrix}=v_1\begin{pmatrix}1\\0\\0\end{pmatrix}+v_3\begin{pmatrix}0\\-2\\1\end{pmatrix}$, hence the eigenspace corresponds to $\lambda=1$ is $\displaystyle \left\{\begin{pmatrix}1\\0\\0\end{pmatrix},\begin{pmatrix}0\\-2\\1\end{pmatrix}\right\}$.
As we can see, the geometric multiplicity of $\lambda=1$ is $2$, so this matrix is not diagonalizable.
