If S is a linearly dependent set, then each vector in S is a linear combination of other vectors in S

[Syringer]

I thought it was true, but it's listed as false. Why? I thought this was essentially the definition of linear dependence.

 

[Shooters]

Not quite the definition of linear dependence.

In order for the statement to be true it would have to read:

If S is a linearly dependent set, then at least one vector in S is a linear combination of other vectors in S.

 

[upsciLLion]

Originally posted by: Shooters
Not quite the definition of linear dependence.

In order for the statement to be true it would have to read:

If S is a linearly dependent set, then at least one vector in S is a linear combination of other vectors in S.

Yup. The word, "each," is what makes the statement false.

 

[Syringer]

But if linear dependence says that you can write a bunch of a lin dep vectors as..

a1v1+a2v2+...+aivi+...+anvn = 0

Couldn't the arbitary vector vi be written as..

(1/ai)(-a1v1-a2v2-anvn) ?

 

[Shooters]

Originally posted by: Syringer
But if linear dependence says that you can write a bunch of a lin dep vectors as..

a1v1+a2v2+...+aivi+...+anvn = 0

Couldn't the arbitary vector vi be written as..

(1/ai)(-a1v1-a2v2-anvn) ?

not if ai = 0.

 

[Syringer]

Ah, touche.

 

[Orsorum]

Heh, I vaguely remember this. Stupid linear algebra.

 

[helpme]

Linear Algrbra