# Question about affine span and symmetric matrices

$$A$$ 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.

• How did you prove that $A$ and $B$ must be symmetric for the other direction? – Berci Nov 4 '18 at 15:55
• 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 $B$. – XYZ Nov 4 '18 at 16:12
• 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 $I+xA+yB$. – Berci Nov 4 '18 at 21:43

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 $$A$$.