It is known that a 3 by 3 real symmetric matrix $A$ has an eigendecomposition

$$ A = Q E Q^T $$

where $Q$ is an orthogonal matrix and $E$ is a diagonal matrix whose elements, $E_{11}$, $E_{22}$ and $E_{33}$, are the eigenvalues of $A$.

Moreover, if those eigenvalues are non-negative then $A$ is positive-semidefinite.

The question is: if those eigenvalues are not only non-negative but also verify the triangle inequalities $$ \begin{aligned} E_{11} + E_{22} &\geq E_{33}\\ E_{11} + E_{33} &\geq E_{22}\\ E_{22} + E_{33} &\geq E_{11} \end{aligned} $$ is there anything special about the structure of $A$ besides the fact that it is positive-semidefinite?

Can those extra triangle inequalities constraints be written as functions of the elements of $A$, just like the positive-semidefinite constraints can be written down using the Sylvester's criterion?

Can those extra constraints be somehow represented in a semidefinite programming / linear matrix inequalities framework?

  • $\begingroup$ using the fact that the lengths of a triangle are of the form $x+y,y+z,z+x$ you can see that $A$ is a sum of three matrices which are symmetric, simultaneous diagonalizable with eigenvalues $(x,x,0),(y,y,0),(z,z,0)$. I'm not sure if this helps... $\endgroup$ – Beni Bogosel Jun 18 '16 at 11:51

Condition $$E_{11}+E_{22}\ge E_{33}$$ is equivalent to $$E_{11}+E_{22}+E_{33}\ge 2 E_{33}$$ or $$E_{33} \le \frac12 tr\ A$$ since the sum of eigenvalues is equal to the trace. Combining with the other two conditions gives $$\lambda_{max}(A) \le \frac12 tr\ A$$ which is semidefinite representable as $$A \preceq \frac12(A_{11}+A_{22}+A_{33}) I$$

  • $\begingroup$ Wow, that is great. Thank you very much. Let me ask you, did you find this solution out of thin air and a lot of knowledge, or is this something that appears in some problems? $\endgroup$ – Cristóvão D. Sousa Jun 20 '16 at 9:57

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