Thread: numerical analysis (norm)

prove that if |.| is a vector norm, and if we define |x|'=a .|x| (a is a positive constant), then |.|' is a norm.

* numerical analysis

There are three properties needed to be checked to make |.|' a vector norm.

(I) |cv|' = a|cv| = a(c|v|) = c(a|v|) = c|v|'.
(II) |v + w|' = a|v + w| <= a(|v| + |w|) = a|v| + a|w| = |v|' + |w|'.
(III) if |v|' = 0, then a|v| = 0,|v| = 0, and v = 0 [this last 0, the zero vector.]

In each case, the norm properties of |.| have been used to show that |.|' satisfy them as well.

Hope this helps.

thanks : )
yea it works.