Linear Independance: Show {v1,v2,...,vn} is Basis of V

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The discussion confirms that if {v1, v2, ..., vn} is a minimal spanning set for a vector space V, then it is indeed a basis for V. This is established by demonstrating that if any vector in the set were linearly dependent, the number of vectors required to span V would be less than n, contradicting the definition of minimality. Therefore, the set must be linearly independent, fulfilling the criteria for being a basis.

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Suppose that {v1,v2,...,vn} is a minimal spanning set for a vector space V. That is V = span {v1,v2,...,vn} and V cannot be spanned by fewer than n vectors. Show that {v1,v2,...,vn} is a basis of V

to be a basis for V then V = span (v1,v2,...,vn}
we already have that
suppose one of thise vectors was not linearly dependent then the number of vectors in the span is less than n. But V = span of n vectors
so the set of vectors must be lienarly dependent
 
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Surely the definition of a basis is that it is a minimal spanning set?

Or are you starting from linearly independent spanning set and showing any such thing is minimal, and vice versa?
 
stunner5000pt said:
Suppose that {v1,v2,...,vn} is a minimal spanning set for a vector space V. That is V = span {v1,v2,...,vn} and V cannot be spanned by fewer than n vectors. Show that {v1,v2,...,vn} is a basis of V

to be a basis for V then V = span (v1,v2,...,vn}
we already have that
suppose one of thise vectors was not linearly dependent then the number of vectors in the span is less than n. But V = span of n vectors
so the set of vectors must be lienarly dependent
Yes, to be minimal, they have to be independent. Suppose not, then you could do away with at least one (how?) and still span V (you should work out an example here), which contradicts "minimal."

See http://en.wikipedia.org/wiki/Vector_basis
 
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