# Non-symmetric Positive Definite/Semidefinite Matrix in Quadratic Program

A necessary condition in any quadratic programming to be convex is the matrix $$\mathbf{Q}$$ in the formulation $$x^\intercal \mathbf{Q}x$$ to be positive definite or positive semidefinite. Positive definiteness (PD) or semidefiniteness (PSD) requires the eigen values of the matrix either to be $$> 0$$ or $$\geq 0$$ respectively. Is the symmetry of the matrix $$\mathbf{Q}$$ a necessary condition for the matrix to be PD or PSD?

This link in Matlab documentations checks for the symmetry of the matrix before finding the eigen values. I can't find this information explicitly anywhere in any reference.

Yes, a real PSD matrix $$M$$ is a symmetric matrix with $$x^TMx\ge 0$$ for any $$x$$ (see e.g. https://en.wikipedia.org/wiki/Definite_matrix).

However, this is not a real restriction. (We have two meanings of "real" here). We can form $$M' = \frac{M+M^T}{2}$$ Now $$M'$$ is symmetric and we have $$x^TM'x = x^TMx$$ for any $$x$$. So you can make $$M$$ symmetric by preprocessing it without affecting the solution.

• Thanks sir for this detailed answer. Consider this matrix: $$\mathbf{Q} = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \end{bmatrix}$$ Now if I apply the equation you kindly provided, the eigen values I get are (-0.5000, -0.5000, 0.5000, 0.5000) which turns the matrix to be indefinite, am I right? Jun 27 at 18:56
• Yes, the "symmetrized" matrix is indefinite. Jun 27 at 21:16