I came across the following:
"Let f be continuously differentiable with Lipschitz gradient, i.e.,
$ ||\nabla f(\mathbf{x}) - \nabla f(\mathbf{y})|| \leq L||(\mathbf{x} - \mathbf{y})|| $
where L is the modulus of the Hessian (if exists)."
What is the modulus of a matrix? Is it the same thing as the determinant?
Thanks!
$\endgroup$2 Answers
$\begingroup$As linear operator $\nabla f$ has a so-called operator norm. In your case this is $L$
$\endgroup$ 1 $\begingroup$A modulus of a matrix A = sqrt(A'A). Since A'A is always positive semidefinite, we can talk about the square root of it. Here, the square root of a positive semidefinite matrix is defined as the unique X which is also positive semidefinite and X^2 = A'A.
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