Symmetric matrix: Difference between revisions

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m Symmetric matrices represent self-adjoint operators only if the basis is orthonormal.
Hessian: there was nothing mathematically wrong with was written but it gave the impression that a sufficient condition was actually necessay when it is not the case
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== Hessian ==
Symmetric <math>n \times n</math> matrices of real functions appear as the [[Hessian matrix|Hessians]] of twice continuously differentiable functions of <math>n</math> real variables (the continuity of the second derivative is not needed, despite common belief to the opposite<ref>{{Cite book |last=Dieudonné |first=Jean A. |title=Foundations of Modern Analysis |publisher=Academic Press |year=1969 |isbn=978-1443724265 |edition=Enlarged and Corrected printing |pages=Theorem (8.12.2), p. 180 |language=en}}</ref>).
 
Every [[quadratic form]] <math>q</math> on <math>\mathbb{R}^n</math> can be uniquely written in the form <math>q(\mathbf{x}) = \mathbf{x}^\textsf{T} A \mathbf{x}</math> with a symmetric <math>n \times n</math> matrix <math>A</math>. Because of the above spectral theorem, one can then say that every quadratic form, up to the choice of an orthonormal basis of <math>\R^n</math>, "looks like"