Linear dependence for this particular example

Question

We have two vectors \begin{bmatrix} 2 \\ 3 \\ \end{bmatrix} and \begin{bmatrix} 4 \\ 6 \\ \end{bmatrix}

If we try to determine whether this set of given vectors is linearly dependent or not, we have to use the equation $c_1v_1+c_2v_2=0$ (where atleast one of the scaler quantity $c_1$ or $c_2$ must be non-zero) to conclude that the set is linearly dependent. The equations that we get are: $2c_1+4c_2=0$ and $3c_1+6c_2=0$. If we solve these equations, we get infinite many solutions for $c_1$ and $c_2$ ranging from $0,1,2,.....$ but if either of $c_1$ or $c_2$ is $0$ the other will definitely be $0$ thus concluding that this set is linearly independent but we know that it is linearly dependent. I am confused as to where I am wrong.May be I am doing wrong math. Please guide me.

Answer 1

This set is linearly dependent. In the case of two vectors $u,v$, all you have to do is check that they are colinear (that is $\exists \alpha\neq 0, u=\alpha v$).

In the equation you wrote, $c_1=c_2=0$ will always be a solution whatever the vectors. What you are interested in is whether or not it is the only solution (in that case the set is linearly independent). But here, as you noticed, there are infinitely many solutions, hence the set is linearly dependent.

Answer 2

What you need to prove is $ax+by=0 \implies a=b=0$. This is clearly not true since $2v_1-1v_2=0$.

$2c_1+4c_2=0 \implies c_1= -2c_2$ and $3c_1+6c_2=0 \implies c_1=-2c_2$. Yes, if one equals $0$, then the other must equal $0$. But that is not the definition of linear independence. What you should be more interested in is,if $c_2 \ne 0$, then $c_1 \ne 0$.