Positive Definite Matrices

Certify x^T A x > 0 via eigenvalues, pivots, and minors.

The idea

A symmetric matrix $A$ is positive definite when $x^{T} A x > 0 \quad \text{for every nonzero } x.$ The scalar $x^{T} A x$ is the energy of $A$ at $x$: a quadratic expression in the coordinates of $x$, the matrix analogue of $a x^{2}$ in one variable. Just as $a x^{2} > 0$ for every $x \neq 0$ says $a > 0$, positive definiteness says the energy is positive in every direction. Geometrically the energy is a bowl, zero at the origin and curving upward every way from it. By contrast, a symmetric matrix $B$ whose energy $x^{T} B x$ is positive along some directions and negative along others has a saddle for its surface — rising one way, falling another.

The definition cannot be verified one vector at a time, since there are infinitely many $x$ to check. What makes it usable is that several finite tests are equivalent to it: a symmetric $A$ is positive definite if and only if all of its eigenvalues are positive; if and only if all of its pivots from elimination are positive; and if and only if every leading principal minor — the determinant of the top-left $k \times k$ block, for each $k$ — is positive. Any one of these certifies the whole infinite family of inequalities at once.

Ways to work on it

• Walkthrough. What makes a symmetric matrix positive definite, and the tests that detect it.
• Practice. Apply the leading-minor test to a symmetric matrix.
• Hardest. Find the definiteness threshold of a parametrized matrix.

Not sure where to start? Take the ten-question placement test.