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Gear300
Jul1-09, 08:29 PM
It is stated that for n-dimensional Euclidean space, n vectors are needed at least for linear independence. But if an n-dimensional Euclidean space also includes (n-1)-dimensional Euclidean space, then why can't it also include a family of n-1 linearly independent vectors?

Petek
Jul1-09, 08:52 PM
It is stated that for n-dimensional Euclidean space, n vectors are needed at least for linear independence. But if an n-dimensional Euclidean space also includes (n-1)-dimensional Euclidean space, then why can't it also include a family of n-1 linearly independent vectors?

It can. For example, in three dimensional Euclidean space the single vector (1,0,0) is linearly independent, as are the two vectors {(1,0,0), (0,1,0)} and the three vectors {(1,0,0), (0,1,0), (0,0,1)}. However, any four vectors in this space will necessarily be linearly dependent. Does that help?

Gear300
Jul1-09, 09:29 PM
I get what you're saying. I mixed stuff up (thought too hard about the words "at least"). Thanks.

HallsofIvy
Jul2-09, 06:49 AM
More than n vectors cannot be independent in Rn. Less can be.