Linear Dependence/Independence

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In n-dimensional Euclidean space, a minimum of n vectors is required for linear independence, while fewer than n can still be independent. The discussion clarifies that n-dimensional space can contain (n-1)-dimensional subspaces, which can have families of n-1 linearly independent vectors. For instance, in three-dimensional space, single and pairs of vectors can be independent, but any set of four vectors will be dependent. The confusion arose from misinterpreting the phrase "at least" in the context of vector independence. Ultimately, the key takeaway is that while n vectors are necessary for independence in Rn, fewer can also be independent.
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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?
 
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Gear300 said:
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?
 
I get what you're saying. I mixed stuff up (thought too hard about the words "at least"). Thanks.
 
More than n vectors cannot be independent in Rn. Less can be.
 

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