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Let set of vectors $\{x,y,z\}$ be linearly independent. Then would $\{x,y,z,x\}=\{x,y,z\}$ be linearly dependent, also?

If so, that seems like a problem (since $\alpha x+\beta y+\gamma z+(-\alpha)x=0$ would allow for a non-zero $\alpha$) unless it is understood that each vector in a set of vectors is used only once in a linear combination of that set.

I've seen one author use tuples to get around this. Is that the way to go?

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up vote 2 down vote accepted

Yes, to properly deal with linear dependence it helps to have a collection concept that permits multiple occurrences of members. Multisets are a natural choice. Tuples work too, but one doesn't need the extra order structure.

Beware that many linear algebra textbooks fail to distinguish between sets and multisets, so their presentation of linear dependence is not rigorous.

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Thanks, I hadn't heard of multisets. – ohmygoodness Sep 24 '12 at 0:11
To be clear, are the following statements correct? If a $X$ is a linearly independent multiset of vectors, then the mutiplicity of each vector in $X$ is $1$. If $Y$ is a multiset of vectors that contains a vector of multiplicity greater than one, then $Y$ is linearly dependent. – ohmygoodness Sep 24 '12 at 0:27
@Grad Yes. The two statement are equivalent (contrapositives). – Bill Dubuque Sep 24 '12 at 0:41

Yes. Tuples.

Or, using the implicit assumption that you wrote: ' it is understood that each vector in a set of vectors is used only once in a linear combination of that set'.

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How many vectors are in {x, y, z, x}? Only three. They are linearly independent, per the assumption.

Remember what you wrote: {x, y, z, x} = {x, y, z}. Denote them instead a, b, respectively. Since a = b, and a is linearly independent, then b must be linearly independent also -- they're the same set.

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What you are talking about is really a list. A list can contain duplicate elements. I think it's more precise here.

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