Lenear algebra theoretical question

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The discussion centers on the linear independence of a series of vectors within the vector space V = Q^4 over the rational numbers Q. It is established that a vector space of dimension 4 cannot accommodate more than 4 linearly independent vectors. The confusion arises from the interpretation of "a vector series v with length 6," which is clarified to mean 6 vectors, not vectors with 6 components. Consequently, the set of 6 vectors cannot be linearly independent in Q^4, as it exceeds the dimensionality of the space.

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there is a space V=Q^4 over Q (i thing Q means rational)


there is a vector series v with length 6 on V
does v linearly independent?
if it is explain is not ,give a counter example ??



Q^4 means we have 4 coordinates in each vector

but v has vectors with 6 coordinates

so i can't put them together

??
 
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transgalactic said:
there is a space V=Q^4 over Q (i thing Q means rational)


there is a vector series v with length 6 on V
What exactly do you mean by "a vector series v with length 6"? Do you mean 6 vectors in V or one or more vectors with 6 components? vectors with other than 4 components are not even in Q^4.

does v linearly independent?
if it is explain is not ,give a counter example ??
You appear to mean 6 vectors in V. Do you mean to ask if a set of 6 vectors can be linearly independent? You know, I hope, that Q^4 has dimension 4 as a vector space over Q. Look at av1+ bv2+ cv3+ dv4+ ev4+ fv5+ gv6= 0. Rewriting v1, v2, v3, v4, v5, and v6 in terms of a basis for V and combine the same basis vectors. That will give you 4 equations for the 6 numbers a, b, c, d, e, f, g.



Q^4 means we have 4 coordinates in each vector

but v has vectors with 6 coordinates
If that is true then they are not even vectors in V. I feel sure there are supposed to be 6 vectors, not vectors with 6 coordinates.

so i can't put them together

??
 

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