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I am studying for a test. The textbook problem is:

The vectors $v_1, v_2, v_3, v_4$ are linearly independent. Determine if the following vectors are also independent.

$v_1-v_2, 2(v_2-v_3), 3(v_3-v_4)$

I have no idea how to determine this. I would greatly appreciate any help. Thank you!

2 Answers 2

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If you write down the linear independence equation for the new vectors, you have the following $$a(\mathbf{v_1} - \mathbf{v_2}) + 2b(\mathbf{v_2} - \mathbf{v_3}) + 3c(\mathbf{v_3} - \mathbf{v_4}) = \mathbf{0}$$ for scalars $a, b, c$. Now this can be rearranged to $$a\mathbf{v_1} + (2b - a)\mathbf{v_2} + (3c - 2b)\mathbf{v_3} -3c\mathbf{v_4} = \mathbf{0}$$ What can you say about the coefficients of this equation?

Edit: I feel I need to add a bit more to this.

First, recall the definition of linear independence. If a set of vectors $\{\mathbf{v_1}, \cdots, \mathbf{v_n}\}$ is linearly independent, then the only solution to the equation $$a_1\mathbf{v_1} + \cdots + a_n\mathbf{v_n}=0$$ is if all the scalars $a_1,\ \cdots,\ a_n$ are zero. Note we are not aiming to sum the vectors to zero, but rather we are interested in how they sum to zero. If there is a non-trivial solution, i.e. you can add the vectors such that a non-zero linear combination makes the zero vector, then the vectors are said to be linearly dependent. There is a dependence amongst the vectors in the sense that some of the vectors in the set can be written as a linear combination of the others.

In your case, we are interested in a set of four vectors. So we care about how these four vectors $\{\mathbf{v_1}, \mathbf{v_2}, \mathbf{v_3}, \mathbf{v_4}\}$ sum to zero. I could use $a_1, a_2, a_3$ as the scalar coefficients, or I can use $b_1, b_2, b_3$ but I chose $a, b, c$ for convenience (no subscripts). You should note that how we choose to represent the scalars (or the vectors) have no real effect on the question; it doesn't matter what we name the coefficients, they're just names.

Therefore, we are interested which $a$s, $b$s and $c$s can make this equation zero $$a(\mathbf{v_1} - \mathbf{v_2}) + 2b(\mathbf{v_2} - \mathbf{v_3}) + 3c(\mathbf{v_3} - \mathbf{v_4}) = \mathbf{0}$$ Now we know nothing about these vectors, but we know about the constituents. So we separate each of the vectors and we get $$a\mathbf{v_1} + (2b - a)\mathbf{v_2} + (3c - 2b)\mathbf{v_3} -3c\mathbf{v_4} = \mathbf{0}$$ This is an equation we recognize. We know that the set of vectors $\{\mathbf{v_1}, \mathbf{v_2}, \mathbf{v_3}, \mathbf{v_4}\}$ is linearly independent, so all the coefficients that solve the above equation must be 0. Namely $$a=0$$ $$2b - a = 0$$ $$3c - 2b = 0$$ $$-3c = 0$$ From this simple system, we can see that the only solutions for $a, b, c$ is $a=b=c=0$. That means the new vectors are indeed linearly independent. It would be impossible to find a non-zero solution to the linear independence equation.

The way to prove that a set of vectors is linearly independent, is to show that they cannot "make" the zero vector in a non-trivial way. What I mean is the following. Say we work in $\mathbb{R}^3$. Then the equation $$a_1\begin{pmatrix}1 \\ 0 \\ 0\end{pmatrix} + a_2\begin{pmatrix}0 \\ 1 \\ 0\end{pmatrix} + a_3\begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix} = \begin{pmatrix}0 \\ 0 \\ 0\end{pmatrix}$$ is the equation which determines if the three vectors above are linearly independent. Now we don't know what $a_1$ or $a_2$ or $a_3$ is, but we'd like to find out. Certainly there does exist a solution, namely $a_1 = a_2 = a_3 = 0$ will solve the equation. But this solution is too "obvious" and rather uninteresting, so we say this is the trivial solution. Now we can ask, are there other solutions to the equation? Non-trivial solutions where the coefficients are not all zero? The answer for this particular set of vectors is no, and it's rather obvious to see. $a_1$ must be zero or the first component will be non-zero. Likewise $a_2$ must be zero or the second component will be non-zero, same with $a_3$. So we've shown that the above equation has only one solution, the trivial solution. Linearly independent sets are by definition the sets in which there exists only the trivial solution. In that sense, we have proven that the above set is linearly independent. If there exists non-trivial solutions, then the set is call linearly dependent. One example is the following $$a_1\begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix} + a_2\begin{pmatrix}1 \\ 2 \\ 1\end{pmatrix} + a_3\begin{pmatrix}1 \\ 0 \\ 0\end{pmatrix} + a_4\begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix} = \begin{pmatrix}0 \\ 0 \\ 0\end{pmatrix}$$ Again, we want to find out which values of $a_1, \cdots, a_4$ solve the equation. We still have the trivial solution. It's always there. But more importantly, you can verify that $$a_1 = -2, a_2 = 1, a_3 = 1, a_4 = -1$$ also solves the equation. In fact, there exists infinitely many values of the coefficients which solves the above equation. Since there exists non-trivial solutions to the above equation, we have proven that the above set is linearly dependent.

This is how the method works in general, perhaps not as easily and clearly as the above examples, but the same principles carry over. To prove that a set is linearly independent, you must prove that the set cannot add to the zero vector in a non-trivial way. If you have anymore confusion after this, please ask me.

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    @Khromonkey Why not?2012-09-11
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    Looks correct to me...2012-09-11
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    To downvoter, would you please explain why?2012-09-11
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    Can you link me to information about this equation? I have not heard of this before. I'm a bit lost as to where the incrementing constants are coming from.2012-09-11
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    What do you mean? A set of vectors is linearly independent if and only if the only solution to the equation above is $a=b=c=0$. This is the basic definition of linear independence.2012-09-11
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    That equation and method is totally fine, now what you need to do is rearrange the terms to get or not a contradiction, now you show that this set is independent by showing that you can´t write one in terms of the rest, please try something and if you can´t i help you, but all you need to use is the definition and a bit of reasoning.2012-09-11
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    If a = b = c = 0, wouldn't the coefficients just be 0? Did you mean the sum of the vectors in a family with different constants = 0?2012-09-11
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    no, by definition a1,a2,a3,a4 must be different than 0 if the set is independent.2012-09-11
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    I'm a bit confused as there are no a1, a2, a3 or a4.2012-09-11
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    With more advanced approach but may be easiest to understand, you can use steinitz lemma.2012-09-11
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    @EuYu, The coefficients of v2 and v3 are linear combinations of one another. Thus, the family is not independent?2012-09-11
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    We don't care if the coefficients of the vectors are linearly independent (and it doesn't really even make sense to consider linear combinations of scalars). We want to show that the vectors can't be written as linear combinations of each other.2012-09-11
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    @EuYu I feel like I'm missing the obvious. I don't see how to prove that they are NOT linear combinations. As far as finding a combination goes, I don't know how to prove dependence with just variables. You've pretty much narrowed it down for me, but I still don't get it.2012-09-12
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    Hmm, perhaps a more concrete example will help. Let me elaborate a bit more.2012-09-12
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    by "the answer is no right?" I meant the answer to whether they are dependent or not is: They are not dependent. Sorry for controversy2012-09-12
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    I am convinced that they are independent so sorry for confusing you.2012-09-12
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    You are right, they are **not** dependent. They're independent.2012-09-12
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    Oh I see now. That makes much more sense, thank you!2012-09-12
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Since ${\mathbf{v_1},\mathbf{v_2},\mathbf{v_3},\mathbf{v_4}}$ is a set of linearly dependent vectors in $\mathbb{R}^n$, that means that there exist a set of numbers of cardinality 4 different that zero, such that $a_1\mathbf{v_1}+a_2\mathbf{v_2}+a_3\mathbf{v_3}+a_4\mathbf{v_4}=0$. Thats the key, now try to rearrange the terms of the equality that you need to prove if the last set were independent.

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    The first set given is linearly independent, not dependent--is that what you meant?2012-09-11
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    by definition the firs set is independent, so the first equality holds, now you set the constants as you need with the rearrangement that EuYu did.2012-09-11
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    I think you are making typos that confused me, sorry for the confusion.2012-09-11