$A$ 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.

  • $\begingroup$ 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

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 $A$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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