f|AC|t that any sum of squares of real numbers can be zero if and only if e|AC|h of those numbers is zero. Suppose pp| = 0 if and only if |p|2 = 0 if and only if x2 + y2 = 0 if and only if x=y=0 if and only if (x,y) = (0,0) if and only if p = 0.
A similar proof exists if p = (x,y,z).
It can be shown that
|(x1,y1) + (x2,y2)| ≤
|(x1,y1)| + |(x2,y2)|
by squaring both sides and proving that inequality using much algebraic computation.
Similarly it can be shown that
|(x1,y1,z1) + (x2,y2,z2)| ≤ |(x1,y1,z1)| + |(x2,y2,z2)|
by squaring both sides and proving that inequality using much algebraic computation. However, in the process of either computation an expression equivalent to the Schwarz inequality is involved. But this inequality is discussed in section 2 of the main text, and proven in the next section below. Below is a more simple proof of the triangle inequality. Since it does not depend upon arrays the proof of the triangle inequality is true for all vectors in sp|AC|es that have an inner product.
Start with the left side of the inequality |u + v| ≤ |u| + |v|, or more ex|AC|tly, start with square of both sides of it:
(#) |u + v|2 = (u + v)2 = (u + v)⋅(u + v) = u⋅u + 2u⋅v + v⋅v ≤ |u|2 + 2|u| |v| + |v|2 = ( |u| + |v|)2.
Therefore, |u + v|2 ≤ ( |u| + |v|)2.
Take square roots of both sides to get: |u + v| ≤ |u| + |v|.
gain it is easier to work with the squares of both sides of the equality. If p = (x,y) then
|λ(x,y)|2 = |(λx,λy)|2 = (λx)2 + (λy)2 = λ2x2 + λ2y2 = λ2(x2 + y2) =
|λ|2 |(x,y)|2.
A similar chain of equalities exists for p = (x,y,z).