Proving b' is a Basis of V: Let B be an Invertible Matrix

[ak416]

Let b = {x_1,...,x_n} be a basis for a vector space V over a field F.
let x'_j = Sum i=1 to n[Bij x_j] where Bij are the entries of any matrix (n x n). Prove that b'={x'_1,...,x'_n} is a basis for V and therefore B is the change of coordinate matrix.

Ok, the final connection between the proof and the coordinate matrix is obvious (that's how its defined). But I am not sure how to prove that b' is a basis. I know that
Sum j=1 to n[a_j x'_j]
= Sum j=1 to n[a_j Sum i=1 to n[Bij x_i]]
= Sum j=1 to n[Sum i=1 to n[a_j Bij] x_i]

So you can consider the Sum i=1 to n[a_j Bij] as the coefficient, call it c_i such that each element of the vector space v = Sum i=1 to n[c_i x_i] for unique scalars c_1,...,c_n.
But this would mean b' is a basis only if for each combination c_1,...,c_n there exists a a_1,...,a_n that generates this c_1,...,c_n. I am having trouble showing this.
I know there's another way to look at it. For example, what i just said is equivalent to having:
c_1 = Sum j=1 to n[a_j B1j]
c_2 = Sum j=1 to n[a_j B2j]
.
.
.
c_n = Sum j=1 to n[a_j Bnj]

so if i can prove that given these n equations and n unknows, there exists for every combination, c_1,...c_n a solution for j_1,...,j_n then i would be done.
But the book I am reading still didnt get to the chapter about solving equations so there must be a more elementary way of showing this.
So ya, if anyone could help that would be cool.

edit: the matrix is invertible

 

[ak416]

maybe this is more suited as a homework question.

 

[AKG]

Let b = {x_1,...,x_n} be a basis for a vector space V over a field F.
let x'_j = Sum i=1 to n[Bij x_j] where Bij are the entries of any matrix (n x n). Prove that b'={x'_1,...,x'_n} is a basis for V and therefore B is the change of coordinate matrix.

Any matrix (n x n)? Then this is wrong. Take B = 0. Then x'_j = 0 for all j, and b' = {0} is not a basis for V. You can prove that:

b' is a basis if and only if B is a change of coordinate matrix

In other words, you can prove that:

1) if b' is a basis, then B is a change of coordinate matrix, and 2) if B is a change of coordinate matrix, then b' is a basis

Do you want to prove, 1), 2), or both? Also, how does your book define "change of coordinate matrix"? Finally, do you know that since {x₁, ..., x_n} is a basis of V, then

a) any set of n vectors which span V are a basis of V
b) any set of n linearly independent vectors are a basis of V

 

[HallsofIvy]

More generally, if B is an invertible matrix, then B is a "change of coordinate matrix".

 

[ak416]

Ya i wrote under "edit" that B is invertible. But how do I prove, given that definition for x'_j that {x'_1,...,x'_n} is a basis for V. Once I know that then the change of coordinate matrix statement immediately follows.

 

[ak416]

Actually, i think i have the answer. Tell me if this is right.
Remember from my first post I said that if i can choose for every combination (c_1,...,c_n) a combination (a_1,...,a_n) that would generate that (c_1,...,c_n) then I would be done. Well i can. Knowing that B is invertible, this means there exists a B^-1 s.t B B^-1 = I. Then this means, for each i there exists a (B B^-1)ii = Sum j=1 to n[Bij B^-1ji] = 1. So for each c_i, choose a_j = B^-1ji * c_i and were done. Right?