1. ## Linear Algebra- Basis

Let V be a finite n-dim vector space

v1,v2,....,vn
w1,w2,...,wn

be two set of basis

Let v be a vector in V be represented in this basis as follows:
v = a1.v1 + a2.v2 + .... + an.vn
also
v = a1.w1 + a2.w2 + ... + an.wn
(note the coefficients a1,a2,...,an are same for both basis)

Can we conclude that v1=w1; v2 = w2' ....vn = wn from this?

Is yes - why?
(obviously I assume v is not equal to 0)

2. Originally Posted by aman_cc
Let V be a finite n-dim vector space

v1,v2,....,vn
w1,w2,...,wn

be two set of basis

Let v be a vector in V be represented in this basis as follows:
v = a1.v1 + a2.v2 + .... + an.vn
also
v = a1.w1 + a2.w2 + ... + an.wn
(note the coefficients a1,a2,...,an are same for both basis)

Can we conclude that v1=w1; v2 = w2' ....vn = wn from this?

Is yes - why?
(obviously I assume v is not equal to 0)
no! for example in $\displaystyle V=\mathbb{R}^2$ let $\displaystyle \{e_1,e_2 \}$ be the standard basis. then $\displaystyle \{e_1-e_2, 2e_2 \}$ is also a basis for $\displaystyle V$ and $\displaystyle e_1+e_2=e_1-e_2 + 2e_2.$

3. Thanks - a very neat counter example. But does my equation put some kind of constraint on the basis - I mean is there a special relation between v1,v2...vn and w1,w2,...,wn for such a thing to happen? I am kind of confused here

4. Originally Posted by aman_cc
Thanks - a very neat counter example. But does my equation put some kind of constraint on the basis - I mean is there a special relation between v1,v2...vn and w1,w2,...,wn for such a thing to happen? I am kind of confused here
what i can say now is that we'll have this situation whenever we have an $\displaystyle n \times n$ matrix $\displaystyle C=[c_{ij}]$ with $\displaystyle c_{ij} \in F$ (the base field) with these properties that $\displaystyle \det C \neq 0$ and $\displaystyle \det (C - I) = 0.$

to see that let $\displaystyle \{v_1, \cdots , v_n \}$ be any basis for $\displaystyle V$ and define $\displaystyle w_j=c_{1j}v_1 + \cdots + c_{nj}v_n, \ 1 \leq j \leq n.$ since $\displaystyle \det C \neq 0,$ the set $\displaystyle \{w_1, \cdots , w_n \}$ is a basis for $\displaystyle V.$ also since $\displaystyle \det(C-I)=0,$ there

exists a vector $\displaystyle \bold{0} \neq \bold{x}=\begin{bmatrix}a_1 & . & . & . & a_n \end{bmatrix}^T \in F^n$ such that $\displaystyle C \bold{x}=\bold{x},$ which is equivalent to $\displaystyle a_1v_1 + \cdots + a_nv_n = a_1w_1 + \cdots + a_nw_n.$