The linear independence of vectors $(u_1+iv_1)$ and $(u_2+iv_2)$ in complex vector space $V_C$

Question

Suppose $V_C$ is a complex vector space and hence the vectors in it should be in the form of $(u+iv)$ where $u,v \in V$ and $V$ is a real vector space

Supposed there are two vectors $(u_1+iv_1)$ and $(u_2+iv_2)$ are linearly independent in $V_C$. Is there any conclusion about the linear independence or dependence that we can make about the vectors $u_1,v_1,u_2,v_2\in V$?

Since $(u_1+iv_1)$ and $(u_2+iv_2)$ are linearly independent in $V_C$, it's clear to state that $$(a_1+ib_1)(u_1+iv_1)+(a_2+ib_2)(u_2+iv_2)=0\implies a_1=b_1=a_2=b_2=0$$ where $a_1,b_1,a_2,b_2\in \mathbb R$

This is equivalent to say(that is expand the above equation): $$\begin{cases}a_1u_1-b_1v_1+a_2u_2-b_2v_2=0 & \text{real part} \\ a_1v_1+b_1u_1+a_2v_2+b_2u_2=0 &\text{imaginary part} \end{cases} \iff a_1=b_1=a_2=b_2=0$$

Then I don't know whether this implies that $(u_1,v_1,u_2,v_2)$ are linearly independent or other relationships about linear independence and dependence within these four vectors.

Thus any help? Thanks!

Answer 1

No conclusion can be reached about the dependence of $u_1,u_2,v_1,v_2$. As a simple example, note that the vectors $$ (1+i,1+i), \quad (1+i,1-i) $$ are linearly independent in $\Bbb C^2$, but the real vectors $(1,1),(1,1),(1,1),(1,-1)$ are clearly linearly dependent over $\Bbb R^2$.