### I would like to know if this equation is solvable for $a$ and $\alpha$:


\begin{equation}
\Sigma = \Gamma + a \left( \alpha 1^\top + 1\alpha^\top \right) +a^2 b
\end{equation}




---


1. $\Sigma$ & $\Gamma$ are known. Both are $D\times D$ matrices.
2. $\Sigma$ is symmetric positive definite.
3. $\Gamma$ is symmetric positive semi definite.
4. $\alpha$ is a $D$ vector.
5. $\sum\_{i=0}^D \alpha\_i =0$.
6. $a$ is a scalar.
7. $b$ is a $1$s matrix of $D\times D$.
8. The $1$ is a $D$ vector of $1$s .


If it is solvable, can you explain how?



Two unknowns: one vector, one scalar, one equation
A small rearrangement yields
$$
\Sigma-\Gamma = a(\alpha + a 1 )1^T + a1 \alpha^T.
$$
So for solvability the rank of $\Sigma-\Gamma$ can at most be $2$.


If $\mathrm{rank}\ \Sigma-\Gamma = 0$, then $\Sigma = \Gamma$. Then a solution is $a=0$ and any $\alpha$ with $\alpha^T1=0$. If we assume $a\neq 0$ we equivalently need to solve
$$
0 = \alpha 1^T + 1 \alpha^T + a 11^T
$$
This implies $\alpha$ is a nonzero multiple of $1$ contradicting the requirement $\alpha^T1=0$. So if $\Sigma = \Gamma$ the solution set is
$$
M = \{ (a,\alpha) | a=0, \alpha^T1=0\}.
$$


For the other cases first note that $\mathrm{rank}\ \Sigma-\Gamma >0 $ implies $a\neq 0$. Furthermore, if you multiply the original equation from left and right by $1^T$ and $1$ you obtain that any solution would have to verify the scalar equation
$$
1^T (\Sigma-\Gamma) 1 = a^2 D^2. \tag{$\*$}\label{star}
$$
The other terms vanish because $\alpha^T 1=0$ by assumption. This gives you two potential values for $a$. Also we obtain the next necessary condition that $1^T (\Sigma-\Gamma) 1 \neq 0$ (or even $>0$ if you are looking for real solutions).


When you plug these in the equation is linear in $\alpha$. So for the two potential values of $a$ we need to solve (using $a\neq 0$)
$$
\frac{1}{a}(\Sigma-\Gamma) - a 1 1^T= \alpha 1^T + 1 \alpha^T.
$$
Using $\alpha^T 1 =0$, we get from muliplication by $1$ and division by $D$ that
$$
\frac{1}{aD}(\Sigma-\Gamma)1 - a 1= \alpha.\tag{$\*\*$}\label{starstar}
$$
This at least satisfies $\alpha^T 1 =0$ as from \eqref{star} we get
$$
\alpha^T 1 = \frac{1}{aD}1^T(\Sigma-\Gamma)1 - a D = 0.
$$
So if there is a solution for the two possible $a$'s then it is given by the $\alpha$ from \eqref{starstar}. But note that this does not have to be a solution. This would still depend on $\Sigma - \Gamma$.


