How does the change of basis theorem work in linear algebra?

[bonfire09]

Let B={b₁,b₂} and C={c₁,c₂} be basis. Then the change of coordinate matrix P(C to B) involves the C-coordinate vectors of b₁ and b₂. Let
[b1]c=[x1] and [b2]c=[y1]
...[x2]...[y2].

Then by definition [c1 c2][x1]=b1 and [c1 c2][y1]=b₂. I don't get how you can
....... [x2].....[y2]
multiply the matrix with basis set C with the change of coordinate matrix P(C to B) to get back basis set B ?
Can anyone help me understand how the derive the fact that you can take the set C basis and matrix P to get basis b1 and b2? My textbook just says very little about it.

Here is an example of a problem relating to this idea.
There was a problem that stated find a basis {u₁,u₂,u₃} for R^3 such that P is the change of coordinates matrix from{u₁,u₂,u₃} to the basis {v₁,v₂,v₃}? P was given and v₁,v₂,v₃were given as well. I know how to do it but don't get the how it works?

 

[rdbateman]

Ok. Let's work in Two Dimensions because writing math text on this forum is slow for me.

Suppose there is a vector v = <x1,x2>. We are used to the regular cartesian basis vectors
${e}_1$ = <1,0> and ${e}_2$ = <0,1> such that

v = $<x1,x2>^{T}$ = x1${e}_1$ + x2${e}_2$

now let's say we want to express v in a new basis like ${f}_1$ = <1,1> and ${f}_2$ = <1,2> such that

v = $<y1,y2>^{T}$ = y1${f}_1$ + y2${f}_2$

since v is the same vector regardless of how it is represented, we can equate the two basis expressions

v = v

y1${f}_1$ + y2${f}_2$ = x1${e}_1$ + x2${e}_2$

<${f}_1$,${f}_2$>$<y1,y2>^{T}$ = <${e}_1$,${e}_2$>$<x1,x2>^{T}$

Let <${f}_1$,${f}_2$> = $F$ (2x2 Matrix with f basis as columns)
Let <${e}_1$,${e}_2$> = $E$ (2x2 Matrix with f basis as columns)

Then

$F$$<y1,y2>^{T}$ = $E$$<x1,x2>^{T}$

So

$<y1,y2>^{T}$ = $F^{T}$$E$$<x1,x2>^{T}$

Let $P$ = $F^{T}$$E$

Then

$<y1,y2>^{T}$ = $P$$<x1,x2>^{T}$

I hope this shows where the logic is coming from. If it is still not clear, let me know, and I will clarify. I will edit this better when I return home.

 

[bonfire09]

thanks this is what I was looking for. I get it.