# Linear dependence for this particular example

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.

• What you have to find out is whether there is a couple of values (scalars) $\;a,\,b\;$ with at least one of them different from zero that will render $\;ac_1+bc_2=0\;$ . We know that if one of those is zero then the other one will be zero, too (as botyh given vectyors are not the zero vector...)...but we don't care. You already found a couple of non-zero scalars as above $\;\implies\;$ the set of two vectors is linearly dependent. – DonAntonio Apr 24 '18 at 9:26
• @DonAntonio You mixed coefficients and vectors. In OPs question $c_i$ are the coefficients and $v_i$ the vectors. – Christoph Apr 24 '18 at 9:50
• @Christoph It looks like it could be so, yet if you read after that the OP wrote $\;3c_1+6c_2=0\;$ and etc., making clear $\;c_i\;$ are the vectors...! This may happen when the OP is not careful enough as to give a name to her(his) vectors/scalars. – DonAntonio Apr 24 '18 at 10:03
• @DonAntonio "$3c_1+6c_2=0$" is just the second row of the equation $c_1 v_2+c_2v_2=0$, OP named all vectors and scalars... – Christoph Apr 24 '18 at 10:07
• $c_1$ and $c_2$ are scalers – Navdeep Apr 24 '18 at 11:52

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.

• Can you please justify it from the point of view of linear dependence taking into account the definition of linear dependence. – Navdeep Apr 24 '18 at 12:20
• If $u=\alpha v$ with $\alpha \ne 0$ then $1 \cdot u + (-\alpha) \cdot v = 0$, and therefore $u$ and $v$ are linearly dependent. – Lee Mosher Apr 24 '18 at 12:27
• I know that this holds true but I want to show that this set is linearly dependent taking into account infinite many solutions as shown in the question. Actually I am confused that, infinite many solutions still have one solution where $c_1=0$ and $c_2=0$, that makes it linearly independent. Please explain. – Navdeep Apr 24 '18 at 13:51
• You do not have to take those infinite solutions into account in any way to show that those two vectors are linearly dependent. You only need to take into account the fact that there is at least one solution that is not $c_1=c_2=0$. In linear algebra, dependency of a collection of vectors is all about whether or not a vector can be expressed as a linear combination of the remaining vectors of that collection. Here, there is no need to be interested in how many exist but rather if some exist. – Bill O'Haran Apr 24 '18 at 13:56
• Thanks a lot!!! – Navdeep Apr 24 '18 at 15:35

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 $c_1 \ne 0$.