Show that two vectors $\vec a_{1} = (a_{11},a_{12}), \vec a_{2}=(a_{21},a_{22}) $ in plane are linearly independent if and only if $ a_{11}a_{22}-a_{12}a_{21} \neq 0. $

So here is my textbook's description on linear dependence and linear independence:

In general, given k vectors $ \vec a_{1}, \dots , \vec a_{k} $, if any one of $ \vec a_{1}, \dots , \vec a_{k} $ is a linear combination of the other vectors, $ \vec a_{1}, \dots \vec a_{k} $ are called linearly dependent. If vectors $ \vec a_{1}, \dots , \vec a_{k} $ are not linearly dependent, then they are called linearly independent.

Using this description, I tried solving the problem but I just keep getting stuck on how the equation $ a_{11}a_{22}-a_{12}a_{21} \neq 0 $ could be related to the linear independence of the two vectors in the question, $ \vec a_{1} $ and $ \vec a_{2} $.

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    $\begingroup$ For them to be linearly dependent, then one must be a multiple of the other. $\endgroup$ – Lord Shark the Unknown Jul 13 at 7:57
  • $\begingroup$ Oh I see. Guess I was stuck with this question because I didn't come up with thinking of it in terms of linear dependence instead of linear independence. Thanks $\endgroup$ – linearAlg Jul 13 at 8:02

If $\vec a_1=\lambda\vec a_2$ or equivalently $a_{11}=\lambda a_{21}$ and $a_{12}=\lambda a_{22}$ then it is easy to verify that $a_{11}a_{22}-a_{12}a_{21}=0$.

Same story if $\vec a_2=\lambda\vec a_1$.

If conversely $a_{11}a_{22}-a_{12}a_{21}=0$ then $a_{22}\vec a_1=a_{12}\vec a_2$.

This gives three options:

  • $a_{22}=a_{21}=0$ or equivalently $\vec a_2=0$ so that $\vec a_2=0\vec a_1$
  • $a_{22}\neq 0$ so that $\vec a_1=\frac{a_{12}}{a_{22}}\vec a_2$
  • $a_{12}\neq 0$ so that $\vec a_2=\frac{a_{22}}{a_{12}}\vec a_1$

So in all cases $\vec a_1$ and $\vec a_2$ appear to be linearly dependent.


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