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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!

2 Answers 2

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As linear operator $\nabla f$ has a so-called operator norm. In your case this is $L$

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    @Norbert Thanks.2012-10-16
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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.