<s>[INST] 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
 [/INST] 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 $D+C$.
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