3
$\begingroup$

There is a lot of info on them; also several ways to check for the property.

What I am missing is the application. Why is it important to know whether a matrix (a linear map) is positive definite?

$\endgroup$
  • 2
    $\begingroup$ An example: en.wikipedia.org/wiki/Hessian_matrix#Second_derivative_test $\endgroup$ – Michael Hoppe Aug 29 '17 at 10:50
  • $\begingroup$ @Michael Hoppe Thanks -- so it seems quite useful in gradient-based optimisation, to check if a point is a minimum (sounds like a stopping criterion). $\endgroup$ – A.L. Verminburger Aug 29 '17 at 10:57
  • 1
    $\begingroup$ If $A$ is any matrix with real entries, then $A^T A$ is positive semidefinite. Furthermore, if $C$ is a positive semidefinite matrix (for example, a diagonal matrix with positive entries), then $A^T C A$ is positive semidefinite. This pattern $A^T C A$ tends to recur throughout math. Even the Laplace operator has this form. The ubiquity of $A^T C A$ in applied math is emphasized by Gilbert Strang in his textbooks. He gives many applications and examples where $A^T C A$ appears. Pos. definite matrices have nice properties, e.g., they are orthogonally diagonalizable with positive eigenvalues. $\endgroup$ – littleO Aug 29 '17 at 11:01
0
$\begingroup$

Let $B$ be a symmetric bilinear form defined in $\mathbb{R}^n\times\mathbb{R}^n$ and let $M_B$ be the matrix of $B$ with respect to the standard basis (that is, the entries of $M_B$ are the $B(e_i,e_j)$'s). Then $B$ is an inner product if and only if $M_B$ is positive definite.

$\endgroup$
0
$\begingroup$

If one has function $f$ on $\mathbb R$ and wants to know if it is the Fourier transform of a finite measure, a very convenient test is that all the matrices $M=(M_{i,j})$ with entries of form $M_{i,j} = f(x_i-x_j)$ must be positive definite.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.