# Thread: prove f(x) is a norm

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

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

$\displaystyle f(x) = \frac{(x^t Ax)^\frac{1}{2}}{2}$

I'm having trouble to prove that $\displaystyle f(x)$ is a vector norm if and only if A is positive definite, in the property that says $\displaystyle ||x + y|| \leq ||x|| + ||y||$.
$\displaystyle f(x) = \frac{(x^t Ax)^\frac{1}{2}}{2}$
If A is not positive definite then $\displaystyle x^{\textsc t}Ax$ need not be positive, so $\displaystyle f(x)$ 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 $\displaystyle A = S^{\textsc t}S$ for some positive definite matrix S. Then $\displaystyle f(x) = \tfrac12\|Sx\|$ (where $\displaystyle \|Sx\|$ means the euclidean norm of the vector Sx), and it is easy to check that this satisfies the properties for a norm.