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Proposition 13.4 of Alexander Barvinok's A Course in Convexity shows the existence of the following result:

Let $n\ge 3$. For two $n\times n$ symmetric matrices $A$ and $B$, and a PSD matrix $X$ with $\mbox{trace}(X) = 1$, there exists a unit vector $x$ such that $x^T A x = \mbox{trace}(AX)$ and $x^T B x = \mbox{trace}(BX)$.

This book does not show how to construct such a vector!

In Boyd & Vandenberghe's Convex Optimization, on page 656, there exists a constructive method for a looser version of above (basically, no constraints on the trace of $X$ and consequently, the magnitude of $x$).

I could not so far use their proof for the above stronger version result and basically construct such $x$. Any proof, idea, or help?

Rodrigo de Azevedo
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Sam
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  • The matrix you are looking for is $X = x x^T$. By cyclicity of the trace this equals the LHS. – Rammus Jun 24 '20 at 06:59
  • That works if $\text{rank}(X) = 1$, but if $X$ has a larger rank, there is not a vector $x$ such that $X = xx^T$. – JimmyK4542 Jun 24 '20 at 07:05
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    There are probably some existing algorithms in the literature of joint numerical range. – user1551 Jun 24 '20 at 07:23
  • @JimmyK4542 You're right, I read the question as given $x$ find $X$ and not the other way around. – Rammus Jun 24 '20 at 08:05
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    This result is equivalent to the convexity of the joint numerical range of a pair of symmetric matrices. A proof for arbitrary tuples of Hermitian matrices (so a stronger result) is given here. – Ben Grossmann Jun 24 '20 at 08:40
  • Thank you all for your kind and useful comments: @Rammus, JimmyK4542, user1551, Omnomnomnom, and Rodrigo de Azevedo. Still, I have not found a constructive method for my question but we'll keep looking! – Sam Jun 24 '20 at 15:38
  • @Sam you should say in your question that your question is how to construct such an $x$ – mathworker21 Jun 26 '20 at 19:55
  • @mathworker21 thanks and done! – Sam Jun 27 '20 at 06:58
  • For $\mathrm{rank}(X) \ge 3$, one may construct $x$ based on some decomposition of $X$ (omitted). I am interested in how to construct $x$ when $\mathrm{rank}(X)=2$ based on some decomposition of $X$. – River Li Jul 13 '20 at 03:44

1 Answers1

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Some thoughts

Let $\alpha = \mathrm{Tr}(AX)$ and $\beta = \mathrm{Tr}(BX)$. We need to find $x$ such that \begin{align} x^\mathsf{T} A x &= \alpha, \\ x^\mathsf{T} B x &= \beta, \\ x^\mathsf{T} x &= 1. \end{align} We use the Newton-Raphson method to solve the above system of three quadratic equations. The Newton-Raphson map is $$x \mapsto x - J^\mathsf{T}(JJ^\mathsf{T})^{-1} \left( \begin{array}{l} x^\mathsf{T} A x - \alpha \\ x^\mathsf{T} B x - \beta \\ x^\mathsf{T} x - 1 \end{array} \right) $$ where $J$ is the Jacobian matrix given by $$J = \left( \begin{array}{l} 2x^\mathsf{T}A \\ 2x^\mathsf{T}B \\ 2x^\mathsf{T} \end{array} \right). $$

Example. Let $n=5$, $$A = \left(\begin{array}{rrrrr} -6 & -2 & 0 & -1 & 3\\ -2 & 2 & 2 & -2 & -4\\ 0 & 2 & 2 & -5 & 2\\ -1 & -2 & -5 & 4 & -2\\ 3 & -4 & 2 & -2 & 0 \end{array}\right),$$ $$B = \left(\begin{array}{rrrrr} 2 & -1 & -3 & -2 & 2\\ -1 & -6 & -4 & -1 & 0\\ -3 & -4 & 0 & -5 & -1\\ -2 & -1 & -5 & -6 & -4\\ 2 & 0 & -1 & -4 & 6 \end{array}\right)$$ and $$X = \frac{1}{127}\left(\begin{array}{rrrrr} 29 & -9 & -20 & 16 & -1\\ -9 & 35 & 9 & -20 & 6\\ -20 & 9 & 26 & -22 & -8\\ 16 & -20 & -22 & 24 & 5\\ -1 & 6 & -8 & 5 & 13 \end{array}\right).$$ Randomly choose an initial $x_0$. After a few iterations, we get $x \approx [0.5829, 0.5326, 0.2344, -0.4843, 0.2951]^\mathsf{T}$.

River Li
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  • Thank you so much for your time and effort @River Li. This numerical method definitely makes sense as there is proof showing the existence of at least one solution for the nonlinear systems that you wrote. Nevertheless, I am interested in constructing such a vector given matrices $A, B$, and $X$. Please take a look at page 656 of the Convex Optimization book to see what I mean. I am voting up though for sure. Thanks again for a useful answer. – Sam Jul 02 '20 at 05:08
  • @Sam You are welcome. Hope to see a solution as you desired. – River Li Jul 02 '20 at 11:30
  • Do you mind explaining how you got Newton's updates you have in your answer, please? I do not clearly see how you applied Newton's method on a system with exactly 3 equations and n (possibly strictly bigger 3) variables as the standard Newton's method requires the same number of equations and variables. Thank you so much @River Li. – Sam Jul 15 '20 at 02:07
  • @Sam It is the Moore-Penrose pseudoinverse of Jacobian. See: A. Ben-Israel, A Newton-Raphson method for the solution of systems of equations. http://benisrael.net/NEWTON-GI-2.pdf – River Li Jul 15 '20 at 02:26
  • @Sam By the way, you can go to find the algorithms based on decomposition of $X$ rather than Newton's method. – River Li Jul 15 '20 at 02:28
  • Sorry for my late response and thank you so much for your useful comments. I was doing research on them. – Sam Jul 19 '20 at 07:44