Let $A$ be the matrix $$ \begin{pmatrix} -1 & -2 & 1 & 2\\ -1 & 2 & 1 & 0\\ -4 & -3 & 3 & 3\\ -4 & -2 & 2 & 4 \end{pmatrix} $$
How can I show that $A$ has a single eigenvalue with geometric multiplicity 2 using the trace of $A$?
I know that $A$ has the eigenvalue $2$ with algebraic multiplicity 4, and I also know that $$tr(A) = \sum_i \lambda_i$$ where $\lambda_i$ is the $i$'th eigenvalue.
Edit: The whole exercise is as follows:
Consider $$ A = \begin{pmatrix} -1 & -2 & 1 & 2\\ -1 & 2 & 1 & 0\\ -4 & -3 & 3 & 3\\ -4 & -2 & 2 & 4 \end{pmatrix} ,\quad \begin{pmatrix} {\bf v}_2 & {\bf v}_1 & {\bf v}_0 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 0\\ 1 & 0 & 1\\ 1 & 3 & 0\\ 2 & 2 & 2\\ \end{pmatrix} $$
Show that $\begin{pmatrix}{\bf v}_2 & {\bf v}_1 & {\bf v}_0\end{pmatrix}$ is a Jordan chain for $A$ and compute the corresponding eigenvalue $\lambda$. Show using the trace or otherwise that $A$ has only one eigenvalue and show it has geometric multiplicity 2.
