TITLE: Why do positive definite matrices have to be symmetric?
QUESTION [16 upvotes]: Definitions of positive definiteness usually look like this:

A symmetric matrix $M$ is positive definite if $x^T M x > 0$ for all vectors $x \neq 0$.

Why must $M$ be symmetric? The definition seems to make sense for general square matrices.

REPLY [17 votes]: Let quadratic form $f$ be defined by
$$f (\mathrm x) := \mathrm x^\top \mathrm A \,\mathrm x$$
where $\mathrm A \in \mathbb{R}^{n \times n}$. Since $\mathrm x^\top \mathrm A \,\mathrm x$ is a scalar, then $(\mathrm x^\top \mathrm A \,\mathrm x)^\top = \mathrm x^\top \mathrm A \,\mathrm x$, i.e., $\mathrm x^\top \mathrm A^\top \mathrm x = \mathrm x^\top \mathrm A \,\mathrm x$. Hence,
$$\mathrm x^\top \left(\frac{\mathrm A - \mathrm A^\top}{2}\right) \mathrm x = 0$$
Thus, the skew-symmetric part of matrix $\mathrm A$ does not contribute anything to the quadratic form. What is left is, then, the symmetric part
$$\frac{\mathrm A + \mathrm A^\top}{2}$$
which is diagonalizable and has real eigenvalues and orthogonal eigenvectors, all nice properties.

Addendum
Taking affine combinations of $\mathrm A$ and $\mathrm A^\top$, we obtain
$$\mathrm x^\top (\gamma \mathrm A + (1-\gamma) \mathrm A^\top) \mathrm x = f (\mathrm x)$$
which yields $f$ for all $\gamma \in \mathbb{R}$. Choosing $\gamma = \frac{1}{2}$, we obtain the symmetric part of $\mathrm A$.