# what is the intuition behind hermitian product

There is a claim in my book Linear Algebra - Serge Lang that

Since the dot product of vectors with complex coordinates maybe equal to zero without vectors being equal to $0$, we must change something in the definition.

I don't understand it. Can anyone explain it with an example?

-

The length of a vector $v$, using the standard dot product, is supposed to be $\sqrt{v\cdot v}$. However, if we take $v=(1,i)$, then $v\cdot v=1^2+i^2=1-1=0$, so the length would be $0$. However, this is a non-zero vector, so it should have non-zero length. So we change the dot product so that:
$$v\cdot w=\sum_{i=1}^nv_i\overline{w_i}$$
Then $v\cdot v=1\times 1+i\times(-i)=2$, and the length is $\sqrt{2}$ as we expect.
What would we get if we allowed $0$ length? – Berci Feb 5 '13 at 17:54
Nothing too horrible I suppose, but it shouldn't really be called length. Things would also be worse, in the sense that there would be vectors with $v\cdot v$ imaginary, and then it makes no sense to take a positive square root at all. Without knowing exactly what point Lang wants to make (the quoted sentence is extremely vague), it's hard to work out exactly what pathology he wants to avoid. – Matthew Pressland Feb 5 '13 at 18:06