Why do positive definite matrices have to be symmetric?

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.

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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 $\mathrm A$.

Positive definite matrices do not have to be symmetric it is just rather common to add this restriction for examples and worksheet questions.

Though this restriction may seem a little severe, there are a number of important applications, which include some classes of partial differential equations and some classes of least squares problems. The advantage of this restriction is that the number of operations to do Gaussian elimination can be cut in half.

You might find this previous question regarding non-symmetric positive definite matrices worthwhile reading: Does non-symmetric positive definite matrix have positive eigenvalues?