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I am taking a first course in numerical methods and have passed a course in Linear Algebra. I am curious to know, how norms(a mathematical abstraction of the length of a vector) come about? Are they commonplace in Physics? Do they have interesting applications other than finding limits?

Kind regards,

Quasar.

Quasar
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    Maybe you can find answers for some questions by reading about functional analysis and related concepts as normed spaces, Banach spaces and Hilbert spaces (the Wikipedia's articles can help). – Pedro Apr 15 '20 at 12:25
  • Thank you, I do surely find them cool! – Quasar Apr 15 '20 at 12:27
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    You have a lot of applications of the norm in the field of Riemannian and sub - Riemannian geometry to (which depends of the "norm" defined on the space). – Bruno Apr 15 '20 at 13:45
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    This doesn't completely answer your question, but in my answer here I discuss some of the ways that norms (as applied to matrices) are used in linear algebra – Ben Grossmann Apr 15 '20 at 13:49
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    Norms are important not only for "finding limits", but also for saying what it is that you mean by a "limit" in the first place. Norms are very commonplace in physics. Specifically, norms in Hilbert spaces are often associated with the "energy" associated with a particular vector. – Ben Grossmann Apr 15 '20 at 13:52
  • @Omnomnomnom, cheers for sharing a very informative answer. Thinking out loud; I want to apply this idea to an arbitrary space. I believe, that linear differential operators with constant coefficients $D$, $(D-1)$, $(D+2)^2$ form a vector space, as they satisfy properties of ordinary polynomials. Therefore, I should be able take any function of a differential operator $D$, and show it is continuous at a point $D_0$ , if whenever $\vert \vert D - D_0 \vert \vert < \delta$, I am able to prove that $\vert \vert f(D)-f(D_0)\vert \vert < \epsilon$. The values of $f(D)$ form a continuum. – Quasar Apr 15 '20 at 17:42
  • So, functions of matrices(operators) can be shown to be continuous? – Quasar Apr 15 '20 at 17:50
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    @Quasar Yes, that's right. Some care needs to be taken in defining a topology on differential operators, though, since the derivative map is itself discontinuous – Ben Grossmann Apr 15 '20 at 17:55

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