Real vector on complex planes?

Question

Consider:

• the vector $\mathbf v=[a,0,b,0]$, where $a,b \in \mathbb R$
• the plane $L_1$ spanned by the vectors $\frac{1}{2}[1-i,0,1+i,0]$ and $\frac{1}{2}[0,1+i,0,1-i]$
• the plane $L_2$ spanned by the vectors $\frac{1}{2}[1+i,0,1-i,0]$ and $\frac{1}{2}[0,1-i,0,1+i]$

Could $\mathbf v$ be a vector of $L_1$ or $L_2$?

Answer 1

The answer is: No.

As a proof you could use the vector $v=[1,0,0,0]$.

Is $v$ in $L_1$? Then it should be a linear combination of the basis vectors $$ v=[1,0,0,0] \overset{!}{=} c_1 [1-i,0,1+i,0] + c_2[0,1+i,0,1-i] \,. $$ By comparing the 1st and 3rd elements one deduces the contradiction $$ 1= c_1(1-i)\,, \quad 0 = c_1(1+i)\,. $$ Thus, $v \notin L_1$.

Similarly, one can show $v \notin L_2$.