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Math Help - Necessary and sufficient condition for linear dependence

  1. #1
    Member Mollier's Avatar
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    Necessary and sufficient condition for linear dependence

    Hi,

    problem:

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

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

    attempt:

    (a)
    <br />
\begin{aligned}<br />
a(\xi_1,\xi_2,\xi_3)+b(\eta_1,\eta_2,\eta_3)=&\;(0  ,0,0)\\<br />
(a\xi_1+b\eta_1,a\xi_2+b\eta_2,a\xi_3+b\eta_3)=&\;  (0,0,0)\\<br />
\end{aligned}<br />

    <br />
\begin{aligned}<br />
1.a\xi_1+b\eta_1=&\;0\\<br />
2.a\xi_2+b\eta_2=&\;0\\<br />
3.a\xi_3+b\eta_3=&\;0<br />
\end{aligned}<br />

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

    (b)
    <br />
\begin{aligned}<br />
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)\\<br />
(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)<br />
\end{aligned}<br />

    <br />
\begin{aligned}<br />
1.a\xi_1+b\eta_1+c\zeta_1=&\;0\\<br />
2.a\xi_2+b\eta_2+c\zeta_2=&\;0\\<br />
3.a\xi_3+b\eta_3+c\zeta_3=&\;0<br />
\end{aligned}<br />

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

    det = 0:

    <br />
\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<br />

    Not sure how to proceed.

    Thanks people !
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  2. #2
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    Your work is essentially correct, although a little too detailed.

    The linear dependence of 2 vectors u and v is that there exists a and b such that au+bv=0. This is what you've done, although in component form. Your result is basically that one is a scaled version of the other.

    The linear dependence of 3 vectors is the same, there exist 3 constants such that au+bv+cw=0. If you want to interpret these vectors as vectors in 3 dimensional space, then yes the determinant will be zero since the 3 vectors won't fill out a solid volume. Instead they will only span a plane or line.
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  3. #3
    Member Mollier's Avatar
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    Quote Originally Posted by Mollier View Post
    <br />
\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<br />
    Can I say that the vectors are dependent if \eta_2\zeta_3=\eta_3\zeta_2 and \xi_2\zeta_3=\xi_3\zeta_2 and \xi_2\eta_3=\xi_3\eta_2.
    They are also dependent when \xi_1=\eta_1=\zeta_1=0..
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  4. #4
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    There are lots of ways to say they're dependent. Visually it means they point in the same direction. You can't use the zero vector because it doesn't point in any direction.
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