Answer to Question #219721 in Linear Algebra for Kailash

Question #219721
What do you mean by Norm of an Inner Product Space
1
Expert's answer
2021-07-22T17:55:44-0400

Any inner product induces a norm given by


"\\|v\\|=\\sqrt{\\langle v, v\\rangle}"

Proof. The axioms for norms mostly follow directly from those for inner products.

If "u, v \u2208 V" and "\u03b1 \u2208 F," then

(i)

"\\|v\\|=\\sqrt{\\langle v, v\\rangle}\\geq0,"

since "\\langle v, v\\rangle\\geq0" with equality if and only if "v = 0."


(ii)

"\\|\\alpha v\\|=\\sqrt{\\langle \\alpha v,\\alpha v\\rangle}=\\sqrt{|\\alpha|^2\\langle v, v\\rangle}"

"=|\\alpha|\\sqrt{\\langle v, v\\rangle}=|\\alpha|\\|a v\\|"

(iii) The triangle inequality


Cauchy-Schwarz inequality

If "V" is an inner product space, then


"|\\langle u, u\\rangle|\\leq\\|u\\|\\|v\\|"

for all "u, v \u2208 V ." Equality holds exactly when "u" and "v" are linearly dependent.


Using the Cauchy-Schwarz inequality,


"\\|u+ v\\|^2=\\langle u+v, u+v\\rangle"

"=\\langle u, u\\rangle+\\langle u, v\\rangle+\\langle v, u\\rangle+\\langle v, v\\rangle"

"=\\|u\\|^2+\\langle u, v\\rangle+\\overline{\\langle u, v\\rangle}+\\|v\\|^2"

"=\\|u\\|^2+2Re\\langle u, v\\rangle+\\|v\\|^2"

"\\leq\\|u\\|^2+2|\\langle u, v\\rangle|+\\|v\\|^2"

"\\leq\\|u\\|^2+2\\|u\\|\\|v\\|+\\|v\\|^2"

"=(\\|u\\|+\\|v\\|)^2"

Taking square roots yields


"\\|u+ v\\|\\leq\\|u\\|+\\|v\\|,"

since both sides are nonnegative.


Therefore "\\|v\\|=\\sqrt{\\langle v, v\\rangle}" is a Norm.



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