Originally Posted by

**transgalactic** i will try to write in a more understandable way.

__i am given:__

v1..vn is a basis in $\displaystyle R^n$

$\displaystyle b\epsilon R^{n}$

$\displaystyle b\neq\{0\}$

**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 $\displaystyle \{v_1-v_n,\,v_2-v_n,\,...,\,v_{n-1}-v_n\} $ is an independent __set__ , then do as follows:

$\displaystyle a_1(v_1-v_n)+a_2(v_2-v_n)+...+a_{n-1}(v_{n-1}-v_n)=0\Longrightarrow a_1v_1+a_2v_2+...+a_{n-1}v_{n-1}+(a_1+...+a_{n-1})v_0$ . and now

use that [tex]\[v_1,...,v_n\}[\tex] is an indep. set and deduce directly that $\displaystyle a_1=a_2=...=a_{n-1}=0$ .

Tonio

Pd. What's your mother tongue?

__second question:__

if v1..vn are solutions to Ax=b system then

$\displaystyle \rho(A)$=1

??

__how i tried to solve the second question:__

$\displaystyle \rho(A)$ 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))+$\displaystyle \rho(A)$ we get n=n+$\displaystyle \rho(A)$

so i got

$\displaystyle \rho(A)$=0

but i am asked to prove that $\displaystyle \rho(A)$=1

where is my mistake

?