Linearly dependent Watch

redrose_ftw
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Hoping that someone can clear this up for me.

I know that when a set of vectors such as {U1,U2,0}, includes a zero vector, then the set of vectors is linearly dependent. However I don't really understand why that is so?! Could someone explain?

Thanks
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DarthVador
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(Original post by redrose_ftw)
Hoping that someone can clear this up for me.

I know that when a set of vectors such as {U1,U2,0}, includes a zero vector, then the set of vectors is linearly dependent. However I don't really understand why that is so?! Could someone explain?

Thanks
Just put the vectors into a square matrix. Calculate the determinant, if it's zero, then the two vectors are linearly dependent.
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DFranklin
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(Original post by redrose_ftw)
Hoping that someone can clear this up for me.

I know that when a set of vectors such as {U1,U2,0}, includes a zero vector, then the set of vectors is linearly dependent. However I don't really understand why that is so?! Could someone explain?

Thanks
A set of vectors v_1, ...v_k are linearly dependent if you can find scalars \lambda_1, ..., \lambda_k not all zero such that \sum \lambda_i v_i = 0.

If one of the vectors is 0, v_j = 0, say, then you can take \lambda_i = 0 for i \neq j, \lambda_j = 1, and then \sum \lambda_i v_i = 0.
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redrose_ftw
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(Original post by DarthVador)
Just put the vectors into a square matrix. Calculate the determinant, if it's zero, then the two vectors are linearly dependent.
I did read on that, I understand how that would work for a square matrix, but what about when the set of vectors are in R5? or any where Rn, n being greater than the number of vectors in the vector space.
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DFranklin
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(Original post by redrose_ftw)
I did read on that, I understand how that would work for a square matrix, but what about when the set of vectors are in R5? or any where Rn, n being greater than the number of vectors in the vector space.
In this case, I'd rather be clear than polite; I don't think DarthVador's suggestion is meaningful here.
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placenta medicae talpae
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To be linearly independent, you'd need the only constants satisfying \alpha_1{\bf u}_1+\alpha_2{\bf u}_2+\alpha_3{\bf 0}={\bf 0}$ to be $(\alpha_1, \alpha_2, \alpha_3)={\bf 0}.

However, we can have for example (\alpha_1,\alpha_2,\alpha_3)=(0,  0,4).

So these vectors are not linearly independent
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redrose_ftw
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Thanks guys, big help
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