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
 A: What about $B=\left(\begin{array} & 1 & 2 & 1 \\ 2 & 4 & 2 \\ 1 & 2  & 1 \end{array}\right)$ and $C = \left(\begin{array} & -1 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & -1 \end{array}\right)$?
Both matrices have rank 1, their sum is $A = B+C$ and
$BC = \left(\begin{array} & 0 & 0 & 0 \\ 0 & 0 & 0 \\0 & 0 & 0\end{array}\right)$.
The reason why you can do this is the following:
At first, note that all these eigenvectors are orthogonal to each other, meaning that $u^Tv=0$, so using them in $B$ and $C$ will give a result of $BC=0$.
$B$ and $C$ each consist of multiples of the eigenvectors corresponding to eigenvalues $-2$ and $6$. This means we get:
$A\cdot B_i = 6 \cdot B_i$ and $A\cdot C_i = -2 \cdot C_i$ where $B_i$ and $C_i$ denote the $i$-th column. Define $b$ to be the eigenvector to $6$ and $c$ to be the eigenvector to $-2$. As we have rank $1$ matrices, each column has to be a multiple of the corresponding eigen vector. That means our matrices look like this:
$B=\left(\begin{array} &  &  &  \\ \lambda_1 b & \lambda_2 b & \lambda_3 b \\  &  &  \end{array}\right)$ as well as $C=\left(\begin{array} &  &  &  \\ \mu_1 c & \mu_2 c & \mu_3 c \\  &  &  \end{array}\right)$
We can use the information that if a $3\times 3$ matrix has distinct eigenvectors and -values which generate $\mathbb{R}^3$ (which is the case), this matrix can be reconstructed unambiguously given only the eigenvectors and eigenvalues.
So let's look what values for $\lambda_i,\mu_j$ are needed to get the same eigen values as $A$:
First $(B+C)\cdot b = 6\cdot b$. We have $B \cdot b = (\lambda_1+\lambda_2+\lambda_3) b$ so we get $\lambda_1+\lambda_2+\lambda_3 = 6$. Analogously from $(B+C)\cdot c = -2 \cdot c$ we can conclude that $\mu_1+\mu_2+\mu_3 = -2$.
But we also want to have $C\cdot b = B\cdot c = 0$. To make this true the vector $\lambda = \left(\begin{array} & \lambda_1 & \lambda_2 & \lambda_3 \end{array}\right)$ and
$\mu = \left(\begin{array} & \mu_1 & \mu_2 & \mu_3 \end{array}\right)$ must be a multiple of $b$ and $c$. Together with above equations we get our final $B$ and $C$.
Faster approach
As $A$ is symmetric we see that $A = A^T = (B+C)^T = B^T + C^T = B+C$ which means that both $B$ and $C$ should be symmetric. As we already have their eigenvectors, we can just construct $B$ and $C$ by using $B = \lambda b\cdot b^T$ and $C = \mu c\cdot c^T$ and simple experimentation gives $\lambda=\mu=1$
Remark: We cannot use $d$ in our matrix because $d^Tb = d^Tc=d^Td=0$. In addition $(d+b)^Tc=0$ and $(d+c)^Tb=0$ which means that $b$ and $c$ cannot be the correct eigenvectors for the matrices $D+B$ or $D+C$.
A: 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:
$$B=\begin{pmatrix}1&2&1\\ 2&4&2\\ 1&2&1\end{pmatrix},\;C=\begin{pmatrix}-1&0&1\\ 0&0&0\\ 1&0&-1\end{pmatrix}$$
