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I have the following problem:

Are the following vectors linearly independent in $\mathbb{R}^2$? \begin{bmatrix} -1 \\ 2 \end{bmatrix}\begin{bmatrix} 1 \\ -2 \end{bmatrix}\begin{bmatrix} 2 \\ -4 \end{bmatrix}

when I solve this using $c_1 v_1+c_2 v_2+ c_3 v_3=0$

I get an underdetermined system, can anyone help me to understand what this means for the linear independence?

Thanks in advance :)

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    $\begingroup$ $v_1 = -v_2 = -0.5v_3$. $\endgroup$ – Alex Silva Jan 3 '17 at 12:21
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    $\begingroup$ If the system is underdetermined, that means that there is more than one solution. In particular, there is a solution. $\endgroup$ – Carsten S Jan 3 '17 at 18:21
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Although other answers have pointed out the fast track (more than $n$ vectors cannot be linearly independent in an $n$-dimensional space), I just wanted to help you complete your own approach, which was to solve

$$c_1 \begin{bmatrix} -1 \\ 2 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ -2 \end{bmatrix} + c_3 \begin{bmatrix} 2 \\ -4 \end{bmatrix} = 0.$$

Correct answer

As you noticed, this gives you two equations in three unknowns

$$\begin{matrix} -c_1 &{} + c_2 &{} + 2c_3 &{} = {}& 0\\ 2 c_1 &{} - 2 c_2 &{} - 4 c_3 &{} = {}& 0 \end{matrix}$$

Since you have more variables than restrictions, if you can find any solution you will get infinitely many. For example, adding the first equation to the second one twice gives

$$0 = 0.$$

Though this is true, it gives you absolutely no information. It means the second equation is just a multiple of the first, so any solution to one is also a solution to the other. All we have then, is the first equation, which I will rewrite as

$$c_1 = c_2 + 2c_3$$

If we let $c_2 = a$, $c_3 = b$, the equation tells us that any combination of $(c_1, c_2, c_3)$ of the form $(a + 2b, a, b)$ ($a, b \in \mathbb R$) solves the system. If you set $a = b = 0$, you get the trivial solution, but also $(1, 1, 0)$ or $(13, 3, 5)$ are possibilities. So you have shown that $$c_1 v_1+c_2 v_2+ c_3 v_3=0$$ does not just have the trivial solution, in fact you have shown it has infinitely many of them because any linear combination of the form $(a + 2b) v_1 + a v_2 + b v_3 = 0$, hence the vectors are not independent.

Old answer

As you noticed, this gives you two equations in three unknowns

$$\begin{matrix} -c_1 &{} + c_2 &{} + c_3 &{} = {}& 0 & (!)\\ 2 c_1 &{} - 2 c_2 &{} - 4 c_3 &{} = {}& 0 \end{matrix}$$

(Note that the first equation marked with (!) is incorrect, due to a typo in the original post it has $c_3$ instead of $2c_3$, but I think that the rest of the post shows an important technique so I'm leaving this part in for reference).

Since you have more variables than restrictions, if you can find any solution you will get infinitely many. For example, adding the first equation to the second one twice gives

$$-2c_3 = 0 \implies c_3 = 0.$$

Then substituting back into the first one, you get

$$-c_1 + c_2 = 0$$ or $$c_1 = c_2,$$

which is all the information you can squeeze out of those two equations.

This means that any combination of $(c_1, c_2, c_3)$ of the form $(a, a, 0)$ ($a \in \mathbb R$) solves the system. If you set $a = 0$, you get the trivial solution, but also $(1, 1, 0)$ is a possibility. So you have shown that $$c_1 v_1+c_2 v_2+ c_3 v_3=0$$ does not just have the trivial solution, in fact you have shown it has infinitely many of them because any linear combination of the form $a v_1 + a v_2 = 0$, hence the vectors are not independent.

The short way

In the two variations above, we have basically explored why a system with fewer equations than unknowns is underdetermined. Either two or more equations are multiples of one another, in which case you can remove the "duplicates". The $k$ independent equations in the $n$ unknowns that you have left fix $k$ of the variables, but you get an $n - k$ dimensional solution space. In the correct answer, $n = 3, k = 1$ while in the first version I accidentally had $k = 2$ but still needed one constant, any arbitrary value of which gives a solution.

Of course, you might have seen this right away, since $v_1 = -v_2$, and $v_3 = 2v_2$, and any of these is sufficient.

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    $\begingroup$ Your first equation should be $-c_1 + c_2 + 2c_3 = 0$, which is why $c_3$ is not determined by this system either. All three vectors are parallel. $\endgroup$ – Paul Sinclair Jan 3 '17 at 17:12
  • $\begingroup$ thanks for putting so much effort in helping me :) $\endgroup$ – Amaluena Jan 3 '17 at 17:18
  • $\begingroup$ Downvoted, for the simple reason that you have a calculation error in solving the system which makes most of what you've written incorrect. $\endgroup$ – Meni Rosenfeld Jan 3 '17 at 21:23
  • $\begingroup$ @MeniRosenfeld sorry about that, should have seen it when I wrote the final line. Fixed now (I hope there are no more algebra errors, just got up). $\endgroup$ – CompuChip Jan 4 '17 at 8:25
  • $\begingroup$ @CompuChip: Sure, no problem. Thanks for the fix, I've removed the downvote. $\endgroup$ – Meni Rosenfeld Jan 4 '17 at 11:48
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$\mathbb R^2$ has dimension $2$, so a set of $3$ vectors from $\mathbb R^2$ can never be linearly independent.

(In your case, the three vectors are even more dependent than they have to be, since they are all parallel).

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$$0v_1+2v_2-v_3=0$$

In general, you can never have more than $k$ linearly independent vectors in a $k$-dimensional vector space

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  • $\begingroup$ There's no "in general" about it, it's an absolute. No need to qualify yourself :) $\endgroup$ – TheEnvironmentalist Jan 3 '17 at 22:53
  • $\begingroup$ @TheEnvironmentalist that's what in general means in mathematics. "In the general case" $\endgroup$ – Stella Biderman Jan 3 '17 at 22:54
  • $\begingroup$ In common parlance, in general implies exceptions. Thesaurus.com lists "usually" as a synonym $\endgroup$ – TheEnvironmentalist Jan 3 '17 at 22:58
  • $\begingroup$ @TheEnvironmentalist Yes, but it doesn't mean that in mathematics. $\endgroup$ – Stella Biderman Jan 3 '17 at 22:58
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    $\begingroup$ @TheEnvironmentalist I don't think the OP was confused. Sometimes people get confused by mathematics terminology, and that turns into a good teaching moment. I don't think avoiding universally standard mathematics terminology is the right decision. Besides, it seems condescending to presume that people don't know universally standard terminology. $\endgroup$ – Stella Biderman Jan 3 '17 at 23:03
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The line $2x+y=0$ joining any two of the given points also passes through the origin, so they are not linearly independent.

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Actually all three vectors are multiples of each other, as is seen here without any calculation. This is enough to say that the vectors are linearly dependent. Even any two of your vectors are not linearly independent, but linearly dependent.

I'm not fluent in writing math symbols (and not yet allowed to add comment), therefore I write somehow clumsy:

  vector2 (1,-2) = (-1)*vector1 (-1,2)

  vector3 (2,-4) =   2*vector2 (1,-2) = (-2)*vector1 (-1,2)

What concerns your equation, which was your question: because your equation is underdetermined, this means the vectors are linearly dependent. This is the normal behavour of such an equation for linearly dependent vectors.

In your case, even when you would set c1=1, the equation would remain undetermined, because v2 und v3 are linearly dependent. This would not happen if two of the vectors, say v2 and v3, were linearly independent, or they were not simply parallel as your vectors are.

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