i will try to write in a more understandable way.

__i am given:__
v1..vn is a basis in

**first question:**
proof that {v1-vn,v2-vn,...,vn-1 - vn} is an independant group

**how i tried to solve the first question:**
by definition:

in order to prove that a group is independant,

we need to show the the only way for the bottom equation to exist

a1(v1-vn)+a2(v2-vn) ..+an(vn-1 -vn)=0

is if a1=..=an=0 all the coefficient has to be zero.

so it rang a bell "i need to get a trivial solution "

but trivilal solution is in Ax=0 system could be should if |A| differs zero.

but A is a square matrices by definition.

how to construct from this single equation a square matrices?

??

Several things:

1) Stop calling "group" to sets of vectors. This is incorrect and confusing since groups are very different things.

2) If all you have to do is to prove that is an independent __set__ , then do as follows:

. and now

use that [tex]\[v_1,...,v_n\}[\tex] is an indep. set and deduce directly that .

Tonio

Pd. What's your mother tongue? __second question:__
if v1..vn are solutions to Ax=b system then

=1

??

__how i tried to solve the second question:__ is the dimention of row or column space

if v1..vn are solving this system

then the dimention of the solution space is n, dim(P(A))=n

and from the formula where n=dim(P(A))+

we get n=n+

so i got

=0

but i am asked to prove that

=1

where is my mistake

?