The proof of the theorem about norms

Theorem. Let V be a Euclidean vector space then the norm has the following properties:

||A||> or equals 0, ||A||=0 if and only if A=0.
||kA||=|k|||A||.
|<A,B>|< ||A|| ||B|| (the Cauchy-Schwartz inequality).
||A+B||< or equal ||A||+||B|| (the triangle inequality).

Proof. 1. This follows directly from the definition of the norm:

||A||=sqrt(<A,A>).

2. Indeed,

||kA||=sqrt(<kA,kA>)=(by the properties of dot products)=sqrt(k²)A*A))=|k|sqrt(A*A)=|k|||A||.

3. Notice that if B=0 then <A,B>=0 and ||A||||B||=0, so the statement is true. So suppose that B is not zero. Take an arbitrary number x and consider the vector A+xB. By the fourth property of dot products we have:

<(A+xB),(A+xB)> > = 0.

Using other properties of dot products we can transform this inequality:

(B*B)x² + 2(<A,B>)x+(<A,A>)> =0

Notice that <B,B>, <A,B> and <A,A> are real numbers. Since B is not 0, <B,B> is not zero (by the properties of dot products. So we have a quadratic polynomial which is non-negative for all values of the variable x. From high school mathematics we know that this can happen only if the discriminant of this polynomial is non-positive. Thus we have

(<A,B>)²-(<A,A>)(<B,B>)< =0

This implies

|<A,B>|< = sqrt(<A,A>)sqrt(<B,B>) = ||A|| ||B||

The proof is complete.

Property 4 is left as an exercise.

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