Find Two Rank 1 Matrices I'm looking for rank one matrices B and C such that A = B + C and BC = 0. Where
$$A = \begin{bmatrix}
0 & 2 & 2\\
2 & 4 & 2\\
2 & 2 & 0
\end{bmatrix}$$
with eigenvalues 0, -2, 6 and eigenvectors \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}
Sorry, I don't really know how to format

You can diagonalize $A$ as $D_A=P^{-1}AP$ with:
$$P=
    \begin{pmatrix}
    1 & -1 & 1 \\
    -1 & 0 & 2 \\
    1 & 1 & 1 \\
    \end{pmatrix} \text{, }
D_A=
    \begin{pmatrix}
    0 & 0 & 0 \\
    0 & -2 & 0 \\
    0 & 0 & 6 \\
    \end{pmatrix}
=\text{diag}(0,-2,6)
$$
We try to find $B$ and $C$ with the same eigenvectors as $A$ (it's easier, but with other conditions it might not be possible). In that case the matrices are simultaneously diagonizable and the EVs can be added and multiplied componentwhise when the matrices are added an multiplied. So the conditions translate to the eigenvalues $(b_1,b_2,b_3)$ of $B$ and $(c_1,c_2,c_3)$ of $C$ as follows:

*

*From $BC=0$ we get: $b_1*c_1=0,\: b_2*c_2=0, \:b_3*c_3=0$.

*From $A=B+C$ we get: $b_1+c_1=0$, $b_2+c_2=-2$, $b_3+c_3=6$.

*Since $B$, $C$ are rank 1 neither can have EVs $(0,0,0)$
The equations can be resolved for $D_B=\text{diag}(0,0,6) \text{ and } D_C=\text{diag}(0,-2,0)$ and we have $D_A=D_B+D_C$ and $D_B*D_C=0$.
Now your conditions are fulfilled for $B=PD_BP^{-1}$ and  $A=PD_AP^{-1}$. That results in:
$$B=\begin{pmatrix}1&2&1\\ 2&4&2\\ 1&2&1\end{pmatrix},\;C=\begin{pmatrix}-1&0&1\\ 0&0&0\\ 1&0&-1\end{pmatrix}$$
