About a norm : $p(uv)=p(u)p(v)$ all the time?

Question

Say, $p$ is a norm on a vector space.

Then can we say that

$$p(uv)=p(u)p(v)$$ all the time?

Thanks.

Answer 1

Besides the obvious difficulty to make sense of vector-times-vector multiplication, there are vector spaces, where the inequality is true: $V=\mathbb R$ or $V=\mathbb C$.

It fails however for matrices. Take for instance $$ A=\pmatrix{0 & 1\\ 0& 0 }. $$ Take any matrix norm. Then $\|A\|\ne0$, as $A\ne0$. However it holds $A^2=0$, hence $$ 0 = \|A\cdot A\| \ne \|A\|\cdot\|A\| = \|A\|^2 . $$

Answer 2

Norm p is a map from a vector space V to the reals R. Assuming uv is the inner product (which maps VxV to some set of scalars S) then p(uv) is only defined when V = S. In that case you would have something like p(uv)=|uv| and p(u)p(v)=|u||v|. Otherwise p(uv) is not really defined for more general vector spaces.