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I've come across a question in Linear Algebra that I can't quite figure out. I've tried a multitude of things that either don't work or aren't sufficient enough to convince me I understand linear independence well enough.

I know a set of vectors, S, in vector space V are linearly independent if their linear combination, that is,

$\lambda_1 \mathbf{v}_1 + ... + \lambda_n \mathbf{v}_n = \mathbf{0}$

means all scalars are equal to each other and 0,

$\lambda_1 = ... = \lambda_n = 0.$

I can also show a set of vectors S is linearly independent if I'm given a set of vectors with numerical values - by creating a matrix and reducing it to row echelon form. However, my understanding isn't great enough that I can expand on this and answer questions such as the following:

Assume the vectors u, v and w are linearly independent elements of a vector space V. For each of the following sets decide whether it is linearly independent.

A. {u + v + w, v - 2w, 2u + 3w}

B. {u + 2w, v + 2w, 2w}

C. {x, y, z}

where,

x = u + 2v - w,

y = 2x + u + 2v - w,

z = 3x - 2y.

If anyone can explain to me the connection between this type of question and the definition of linear independence by answering A or providing a guideline of how to answer A then hopefully I can tackle B and C and any related questions. Thanks.

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You basically need to write the definition of linear independence. Suppose that $$ \alpha(u+v+w)+\beta(v-2w)+\gamma(2u+3w)=0. $$ We can rewrite this as $$ (\alpha+2\gamma)\,u+(\alpha+\beta)\,v+(\alpha-2\beta+3\gamma)\,w=0. $$ Since $u,v,w$ are linearly independent, we get the equalities $$ \alpha+2\gamma=0,\ \ \alpha+\beta=0,\ \ \alpha-2\beta+3\gamma=0. $$ Now you can analyze this system. If the only solution is $\alpha=\beta=\gamma=0$, you'll know that the three vectors in A are linearly independent. If you produce a nonzero solution, you'll know that they are linearly dependent.

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  • $\begingroup$ Thank you. So can I put these equations into a matrix and row reduce? And if it reduces to the identity matrix does that mean it is linearly independent as the solution is trivial? $\endgroup$ – user431718 Jul 21 '17 at 18:23
  • $\begingroup$ Exactly. $ \ \ $ $\endgroup$ – Martin Argerami Jul 21 '17 at 18:44
  • $\begingroup$ Brilliant thanks very much $\endgroup$ – user431718 Jul 21 '17 at 19:06
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Hint: In both cases, write the coordinates of the vectors in the set with respect to the original set as rows in a matrix. Row reduce this matrix.

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    $\begingroup$ While this is clever, I doubt if OP would be able to see why this method works $\endgroup$ – user160738 Jul 21 '17 at 17:05
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for A: we have $$\alpha(u+v+w)+\beta(v-2w)+\gamma(2u+3w)=0$$ this is equivalant to $$u(\alpha+2\gamma)+v(\alpha+\beta)+w(\alpha-2\beta+3\gamma)=0$$ can you finish? for B: we get $$u\alpha=0,v\beta=0,w(2\alpha+2\beta+2\gamma)=0$$

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You just need to use the definition. For the first case you can write:

$$\lambda_1(u+v+w)+\lambda_2(v-2w)+\lambda_3(2u+3w)=0$$

now rewrite like:

$$(\lambda_1+2\lambda_3)u+(\lambda_1+\lambda_2)v+(\lambda_1-2\lambda_2+3\lambda_3)w=0$$

Now use that $u,v,w$ are independent, solve the system and find $\lambda_1,\lambda_2,\lambda_3$.

Can you finish?

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