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$A$ and $B$ are real $n\times n$ matrices and $S=\{(x,y)\in \mathbb{R^2}$ : $I+xA+yB$ is positive semidefinite matrix $\}$.

Prove that $\mathbb{R^2}= Aff(S)$ if and only if $A$ and $B$ are symmetric. Here $Aff(S)$ stands for affine span of $S$. I have proved that if $\mathbb{R^2}= Aff(S)$ then $A$ and $B$ are symmetric. I have problem with proving other direction. Any ideas would be helpful.

Since $S \subset \mathbb{R^2}$ then $Aff(S) \subset \mathbb{R^2}$. It is obvious since $\mathbb{R^2}$ is affine subspace. I am not sure how to prove other inclusion if it is known that matrices $A$ and $B$ are symmetric.

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  • $\begingroup$ How did you prove that $A$ and $B$ must be symmetric for the other direction? $\endgroup$ – Berci Nov 4 '18 at 15:55
  • $\begingroup$ Well I do not know if It is correct but I took an element in $\mathbb{R^2}$ which is also element of affine span of $S$ so it must be affine combination of elements in S. For each element of $S$ we have that matrix $I+x_iA+y_iB$ is semi definite so it is symmetric. Symmetric matrix is equal to its transpose matrix. From there I have got that transpose of $A$ is $A$ and the same for matrix $B$. $\endgroup$ – XYZ Nov 4 '18 at 16:12
  • $\begingroup$ It can be correct. The main thing is that linear combinations of symmetric matrices are symmetric, and thus $A$ and $B$ can be expressed by several instances of $I+xA+yB$. $\endgroup$ – Berci Nov 4 '18 at 21:43
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Hint: It's enough to show $(0,0),\ (x,0),\ (0,y)\in S$ for some nonzero $x, y$.
Then find such an $x$ using the eigenvalues of $A$.

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