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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.


marked as duplicate by amd, mrp, MathOverview, Daniel W. Farlow, JonMark Perry Oct 12 '16 at 2:26

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    $\begingroup$ Mostly because you get the same quadratic form if you replace $M$ by its symmetric part $(M+M^T) / 2.$ Put another way, the quadratic form remains the same if we subtract off the anti-symmetric part $(M-M^T) / 2.$ $\endgroup$ – Will Jagy Oct 11 '16 at 16:54

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.


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 $\mathrm A$.

  • $\begingroup$ Brilliant, thanks! $\endgroup$ – Elias Strehle Oct 13 '16 at 11:39

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?


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