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Math Help - prove f(x) is a norm

  1. #1
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    prove f(x) is a norm

    Hi!,
    i'm having trouble to prove that f(x) is a vector norm if and only if A is positive definite, in the property that says  ||x + y|| \leq ||x|| + ||y|| .

    <br />
f(x) = \frac{(x^t Ax)^\frac{1}{2}}{2}<br />

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by redalert View Post
    I'm having trouble to prove that f(x) is a vector norm if and only if A is positive definite, in the property that says  ||x + y|| \leq ||x|| + ||y|| .

    <br />
f(x) = \frac{(x^t Ax)^\frac{1}{2}}{2}<br />
    If A is not positive definite then x^{\textsc t}Ax need not be positive, so f(x) will not be positive-valued and hence will not be a norm.

    If A is positive definite then it has a positive square root, say A = S^{\textsc t}S for some positive definite matrix S. Then f(x) = \tfrac12\|Sx\| (where \|Sx\| means the euclidean norm of the vector Sx), and it is easy to check that this satisfies the properties for a norm.
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