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If I would like to prove, that two vectors $a = \binom{1}{2}$ and $b = \binom{0}{1}$ are linearly dependent, I solve the equation $x_1 a + x_2 b = 0$. If $x_1 = x_2 = 0$ the vectors are linear independent. To solve $x_1$ and $x_2$ in more complex cases one can put the vectors in matrices and solve it.
And regarding this I have a question: Does it matter if I put the vectors in columns or in rows?

For the given vectors, does it matter if I solve $$ \left[ \begin{array}{cc|c} 1&2&0\\ 0&1&0 \end{array} \right] $$ or

$$ \left[ \begin{array}{cc|c} 1&0&0\\ 2&1&0 \end{array} \right] $$

Maybe in this case, it works for both variants. But how is this in general? And why does it (not) matter?

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  • $\begingroup$ It is the same in general since the rank of a matrix is invariant under transposition. $\endgroup$ – Mundron Schmidt Sep 14 '17 at 14:12
  • $\begingroup$ But $a = \binom{1}{2}$ and $b = \binom{0}{1}$ are not linearly dependent. We just need that the rank of the matrix $A$ is $2$ for linear independence. But $rk(A)=rk(A^ T)$, so both variants always work. $\endgroup$ – Dietrich Burde Sep 14 '17 at 14:12
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One obvious difficulty here is that your two given vectors, $a= \begin{pmatrix}2 \\ 1 \end{pmatrix}$ and $b= \begin{pmatrix}0 \\ 1 \end{pmatrix}$ are not dependent! The equation $ax_1+ bx_2= 0$ easily gives $x_2= \frac{-a}{b}x_1$. That is, two vectors are "dependent" if and only if one is a multiple of the other. For more vectors (in a vector space of dimension greater than 2) it is more complicated.

Yes, you can set the vectors in matrix, either as rows or columns, it really doesn't matter which. And then the set of vectors is "dependent" if and only if that matrix is invertible- which is true if and only if the determinant is 0. But neither $\begin{pmatrix}2 & 2 \\ 1 & 0\end{pmatrix}$ nor $\begin{pmatrix}2 & 1 \\ 0 & 1\end{pmatrix}$ has determinant 0- the first has determinant -2 and the second has determinant 2.

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  • $\begingroup$ The consideration of the determinant fails if you have $m$ vectors in $\mathbb{R}^n$, with $m<n$. $\endgroup$ – egreg Sep 14 '17 at 15:14
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There is some confusion here. You wrote that “If $x_1=x_2=0$ the vectors are linear independent”. It shoud be: If the only solution is $x_1=x_2=0$, the vectors are linear independent.

On the other hand, asserting that the vectors $\begin{pmatrix}a\\b\end{pmatrix}$ and $\begin{pmatrix}c\\d\end{pmatrix}$ are linearly independent is equivalent to asserting that the vectors $(a,b)$ and $(c,d)$ are linearly independent because, if $x_1,x_2\in\mathbb R$, then$$x\begin{pmatrix}a\\b\end{pmatrix}+y\begin{pmatrix}c\\d\end{pmatrix}=\begin{pmatrix}ax+cy\\bx+dy\end{pmatrix}\text{ and }x(a,b)+y(c,d)=(ax+cy,bx+dy).$$So, it is the same system that you have to solve.

It turns out that $(a,b)$ and $(c,d)$ are linearly independent if and only if $(a,c)$ and $(b,d)$ are linearly independent, because both assertions are equivalent to $ad\neq bc$.

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