Linearly dependent

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 :smile:

Reply 1

DarthVador

8

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 :smile:

Just put the vectors into a square matrix. Calculate the determinant, if it's zero, then the two vectors are linearly dependent.

Reply 2

DFranklin

18

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 :smile:

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

.

Reply 3

redrose_ftw

OP

7

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.

Reply 4

DFranklin

18

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.