Let $u= \begin{pmatrix} 1\\ -1\\ 2 \end{pmatrix}$ and $v= \begin{pmatrix} 2\\ 1\\ -1 \end{pmatrix}$
$E=uv^T= \begin{pmatrix} 2 & 1 & -1\\ -2 & -1 &1 \\ 4 & 2 & -2 \end{pmatrix} $
What would be a good strategy to find a Schur decomposition here ? I know that $u$ has eigenvalue $v^Tu=-1$ and the orthognal complement of $u$ must have eigenvalue $0$. I've been trying to do some gram smith but it all turns out very ugly, and I think I'm supposed to something smarter. Any ideas ?
Let $e_1=(1,0,0)^T$. If $Q$ is a real orthogonal matrix $Q$ that has $u/\|u\|$ as its first column (i.e. $Qe_1=u$), then $Q^TEQ=(Q^Tu)(v^TQ)=e_1(v^TQ)$ is upper triangular (its first row is $v^TQ$ and the other two rows are zero). To construct $Q$, you may use a Householder matrix: $$ Q = I-\frac{2ww^T}{\|w\|^2},\ w=u-\|u\|e_1. $$