## linearly independent

I dont know if this is a rule, but can a set of vectors be linearly independent if their determinant is not equal to zero?

say 4 vectors are given in R^4, if I took the determinant of the 4 vectors such that det{v1 , v2, v3, v4} is not equal to zero, could i say that these vectors are linearly independent?

if they are linearly independent, then does it mean that these 4 vectors span R^4?

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 Quote by UrbanXrisis I dont know if this is a rule, but can a set of vectors be linearly independent if their determinant is not equal to zero? say 4 vectors are given in R^4, if I took the determinant of the 4 vectors such that det{v1 , v2, v3, v4} is not equal to zero, could i say that these vectors are linearly independent? if they are linearly independent, then does it mean that these 4 vectors span R^4?
If the determinate is 0, then the matrix is not invertible. If a matrix is not invertible, then the only solution of the homogenous system is the trivial solution. And it is also true that if you have n linearly independent vectors, then they span a vector space of dim=n.

I'm taking linear algebra too, so my knowledge isn't that great, but I'm pretty sure everything I stated is correct.

 Recognitions: Gold Member If I'm understanding you correctly, you are saying that if you took four vectors, say collumn vectors, and formed them as the collumns of a matrix, and took the determinant of that matrix and found the determinant of that matrix and it was not zero, then could you say that the vectors are linearly independant? If thats what you mean then yes, because you know from the properties of the determinant that if one of the collumns was a linear combination of the others, then the determinant would be zero. Since it is not zero then no collumn can be a linear combination of the others. Therefore the vecors are linearly independant.

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## linearly independent

 Quote by Geekster If the determinate is 0, then the matrix is not invertible. If a matrix is not invertible, then the only solution of the homogenous system is the trivial solution. And it is also true that if you have n linearly independent vectors, then they span a vector space of dim=n. I'm taking linear algebra too, so my knowledge isn't that great, but I'm pretty sure everything I stated is correct.

you have it exactly backwards. Determinant *not* zero means that there is a unique solution. (consider the zero matrix if need be to see where you've gone wrong)

 if a set of vector spans a matrix, doesnt the vectors have to be linearly independent? So what is the need for a basis? a basis is a set of vectors that span a vector space and is LI, but isnt a vector that spans V already LI?

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 Quote by UrbanXrisis if a vector spans a matrix, doesnt it have to be linearly independent? So what is the need for a basis? a basis is a set of vectors that span a vector space and is LI, but isnt a vector that spans V already LI?
Huh? What do you mean "if a vector spans a matrix"? And as far as a single vector being linearly independant, that is true as long as it is not zero. And why are you asking what the need is for a basis? And what do you mean by "a vector that spans V"? If a single vector spans V then it is one dimensional at most. Also, if V consisted only of zero, then the zero vector could span it, but not be linearly independant.

 yeah, i caught that and changed the post