null vector in complex space let is vector scalar product of which to itself is zero, for example let us take vector scalar product to itself
$(1,i)*(1,i)=1-1=0$
let us consider all null vector and show that they represent subspace, vector space is called subspace if it is belong to main space and two operation : inner operation called as sum (+) and outer operation (*) is satisfied, for instance
$(1,i)+(1+i)=(2,2i)$
its scalar product to itself will be
$(2,2i)*(2,2i)=4+4*(-1)=0$
$\alpha(1,i)=(\alpha,\alpha*i)$
and scalar product is $(\alpha,\alpha*i)*(\alpha,\alpha*i)=\alpha^2-\alpha^2=0$
short question will be: we can easily generate this statements for all type of null vector right?because for more complex vectors, writing is difficult and therefore i decided to show by simple example
EDITED : also if given vector is null vector, then its corresponding conjugate is also null vector
for instance $(1,i)$ its conjugate is $(1,-i)$ , scalar product will be
$(1,-i)*(1,-i)=1+(-i*-i)=1+(-1)=0$
It's not a subspace. Here's a counterexample to the sum operation: take $v = (a, ia)$ and $u = (x, -ix)$, where $a, x\neq0$. $$\begin{align} v*v &= a^2-a^2 = 0\\ u*u &= x^2-x^2 = 0\\ u+v &= (a+x, i(a-x))\\ (u+v)*(u+v) &= (a+x)^2 - (a-x)^2\\ &= a^2+2ax+x^2-a^2+2ax-x^2\\ &= 4ax \neq 0 \end{align}$$