# elements in base does not depend on the basis

[CubicFlunky77]

Essentially, I have to show that where {$e_1,...,e_n$} forms the basis of $L$, no family of vectors {$e'_1 ,..., e'_m$} with $m$>$n$ can serve as the basis of $L$. The book shows this by saying there exists a $0$ vector such that $0 = \sum_{i=1}^{m}x_ie'_i$, where not all $x_i$ vanish. I wanted to show it by distinguishing the sets to which $e$ and $e'$ belong and showing that if one belongs to a set in n-dimension and the other belongs to a different set in g-dimension where g>n, the n-dimension cannot encompass the g-dimension. I'm not sure if this is doable or too longwinded. Also, I do not understand their reasoning or where the $0$ vector comes from. Help?

 

[mfb]

If n vectors form a basis of L, then the dimension of L is n, and not g. Where does that g come from, and what is it supposed to be?
It is possible to assume that there is a different basis with m>n vectors, and show that the first n vectors cannot form a basis then (proof by contradiction).

What do you mean with "where the 0 vector comes from"? The 0 vector is part of L.

 

[Stephen Tashi]

Essentially, I have to show that where {$e_1,...,e_n$} forms the basis of $L$, no family of vectors {$e'_1 ,..., e'_m$} with $m$>$n$ can serve as the basis of $L$. The book shows this by saying there exists a $0$ vector such that $0 = \sum_{i=1}^{m}x_ie'_i$, where not all $x_i$ vanish.

You should should state exactly what book's proof says.

I wanted to show it by distinguishing the sets to which $e$ and $e'$ belong and showing that if one belongs to a set in n-dimension and the other belongs to a different set in g-dimension where g>n, the n-dimension cannot encompass the g-dimension. I'm not sure if this is doable or too longwinded.

That's doesn't describe a specific method, so I'm not sure either.

Also, I do not understand their reasoning or where the $0$ vector comes from. Help?

You'll have to reveal the book's proof if you want someone to evaluated it.