Hi,

problem:

(a) Find a necessary and sufficient condition for the linear dependence of two vectors in $\displaystyle \mathbb{C}^3$.

(b) Do the same for three vectors in $\displaystyle \mathbb{C}^3$.

attempt:

(a)

$\displaystyle

\begin{aligned}

a(\xi_1,\xi_2,\xi_3)+b(\eta_1,\eta_2,\eta_3)=&\;(0 ,0,0)\\

(a\xi_1+b\eta_1,a\xi_2+b\eta_2,a\xi_3+b\eta_3)=&\; (0,0,0)\\

\end{aligned}

$

$\displaystyle

\begin{aligned}

1.a\xi_1+b\eta_1=&\;0\\

2.a\xi_2+b\eta_2=&\;0\\

3.a\xi_3+b\eta_3=&\;0

\end{aligned}

$

$\displaystyle \mathrm{dependent\;if:}\;a\xi_1=b\eta_1,a\xi_2=b\e ta_2\;and\;a\xi_3=b\eta_3 $

(b)

$\displaystyle

\begin{aligned}

a(\xi_1,\xi_2,\xi_3)+b(\eta_1,\eta_2,\eta_3)+c(\ze ta_1,\zeta_2,\zeta_3)=&\;(0,0,0)\\

(a\xi_1+b\eta_1+c\zeta_1,a\xi_2+b\eta_2+c\zeta_2,a \xi_3+b\eta_3+c\zeta_3)=&\;(0,0,0)

\end{aligned}

$

$\displaystyle

\begin{aligned}

1.a\xi_1+b\eta_1+c\zeta_1=&\;0\\

2.a\xi_2+b\eta_2+c\zeta_2=&\;0\\

3.a\xi_3+b\eta_3+c\zeta_3=&\;0

\end{aligned}

$

Perhaps I could find the deteminant and set it to zero to check when the system is singular?

det = 0:

$\displaystyle

\xi_1(\eta_2\zeta_3-\eta_3\zeta_2)-\eta_1(\xi_2\zeta_3-\xi_3\zeta_2)+\zeta_1(\xi_2\eta_3-\xi_3\eta_2)=0

$

Not sure how to proceed.

Thanks people !