1
$\begingroup$

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?

$\endgroup$
8
  • $\begingroup$ Is the question whether a solution exists, whether a solution exists and is unique, whether there is a "closed form" solution (whatever that means), whether there is an efficient algorithm to find the/a solution, or some combination? $\endgroup$
    – LSpice
    Sep 28, 2022 at 20:15
  • $\begingroup$ you have $D+1$ equations with $D+1$ unknowns, so this should generically give you a solution; I doubt more can be said in full generality. $\endgroup$ Sep 28, 2022 at 20:18
  • 1
    $\begingroup$ @CarloBeenakker, re, isn't it $\frac{D(D + 1)}2$ scalar equations? Is there some further obvious redundancy than the symmetry? $\endgroup$
    – LSpice
    Sep 28, 2022 at 20:45
  • $\begingroup$ @LSpice I believe the question is whether a solution exists, whether a solution exists and is unique between each variable. Yes, for a closed form solution (so not infinite sum or limit etc). I'm not asking for an efficient algorithm. Just basically how can you solve for both variables $\endgroup$
    – Bayesian
    Sep 29, 2022 at 0:57
  • 2
    $\begingroup$ Please have pity on an aged mathematician with failing eyesight, and avoid using both “$a$” and “$\alpha$” in the same formula, and above all next to each other. $\endgroup$
    – Lubin
    Sep 30, 2022 at 2:12

1 Answer 1

3
$\begingroup$

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 $\Sigma - \Gamma$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.