# Linearly dependentWatch

#1
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
0
5 years ago
#2
(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.
0
5 years ago
#3
(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 are linearly dependent if you can find scalars not all zero such that .

If one of the vectors is 0, , say, then you can take for , , and then .
0
#4
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.
0
5 years ago
#5
(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.
0
5 years ago
#6
To be linearly independent, you'd need the only constants satisfying .

However, we can have for example .

So these vectors are not linearly independent
0
#7
Thanks guys, big help
0
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