# Is the positive semidefinite cone still self-dual in the larger vector space of all real $n \times n$ matrices?

In section 2.24 of Boyd and Vandenberghe's Convex Optimization, it is shown that the set of positive semidefinite matrices is self-dual within the vector space of symmetric matrices.

Is it still self-dual in the larger vector space of all real $n \times n$ matrices?

• In this context, does positive definite imply symmetric? – Omnomnomnom Jul 16 '18 at 11:04
• I was thinking of it as meaning symmetric with positive eigenvalues. – user3281410 Jul 16 '18 at 11:23
• A nonsymmetric matrix $X$ is positive semidefinite if its symmetric part $(X+X^T)/2$ is. In many contexts, semidefiniteness implies symmetry, but in this particular context, it would be good to be clear about that. – Michael Grant Jul 16 '18 at 20:42

## 2 Answers

The dual cone would include non-symmetric matrices that give non-negative quadratic forms.

Follow the same proof. Assume that $q^TYq\geq0$ for all vectors $q$, but $Y$ not necessarily symmetric. Let $X$ a (symmetric) positive semi-definite matrix, i.e. any element of the set that we are computing the cone of. Write it as $X=\sum_i \lambda_i q_iq_i^T$, with $\lambda_i\geq0$.

Then $$tr(YX)=\sum_i \lambda_i tr(Yq_iq_i^T)=\sum_i \lambda_i tr(q_i^TYq_i)\geq0$$

Therefore, $Y$ belongs to the dual cone of the positive semi-definite matrices.

This shows that the dual cone of the (symmetric) positive semi-define matrices, when considered in the space of all matrices, will be strictly larger.

It is not. For instance, note that any skew-symmetric matrix would be a (non positive semidefinite) element of the dual cone to the positive semidefinite matrices.