Proof of Linearly independence

pyroknife · Sep 21, 2012

The problem is attached. I just wanted to see if the way I proved my statement is correct.

My answer: No, because there exists more columns than rows, thus at least one free variable always exists, thus these vectors are linearly dependent.

HallsofIvy · Sep 21, 2012

You are right but, personally, I wouldn't phrase it in terms of "columns" and "rows" since nothing is said of matrices in the problem. Rather, I would not that R^m has dimension m so there is a basis containing m vectors. Every one of the n vectors in the set can be written in terms of the the m vectors in the basis so they cannot be independent.

(A basis for an m-dimensional vector space has three properties:
1) they span the space
2) they are independent
3) there are m vectors in the set
Any set containing fewer than m vectors cannot span the space, any set with more than m vectors cannot be independent.)

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Proof of Linearly independence

pyroknife

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HallsofIvy

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