# change of basis help?

• Dec 14th 2012, 12:24 PM
bonfire09
change of basis help?
Lets say we have two basis. Let B={b1,b2} and C={c1,c2} 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 dont get how you can multiply the matrix with basis set C with the change of coordinate matrix P(C to B) to get back basis set B ?
/////////////////////////////[x2]/////////////////// [y2]

Can anyone elaborate on that thanks.
• Dec 14th 2012, 01:05 PM
chiro
Re: change of basis help?
Hey bonfire09.

Let x be the vector in the standard basis. Then given x_a in basis a and x_b in basis b we have the property:

x_a = A*x
x_b = B*x

where A and B are the basis operators for each space.

So lets look at going between bases: we know that A^(-1)*x_a = B^(-1)*x_b which means that:

x_a = A*B^(-1)*x_b and
x_b = B*A^(-1)*x_a
• Dec 14th 2012, 02:41 PM
bonfire09
Re: change of basis help?
im still stuck. I just don't get how it works in reverse. meaning you can just multiply a matrix with set C basis and the change of coordinates matrix P to get back to the original basis set B.
• Dec 14th 2012, 03:45 PM
chiro
Re: change of basis help?
I've basically derived the change of basis formula for you above.

What I've done above is I've made x be in the standard basis which is just the identity matrix and linked x_a and x_b to x or in other words, linked both bases to the standard basis which is just the identity matrix.

The rest is just algebra of matrices which assume that all matrices are invertible (and a basis matrix must be invertible to be a basis).
• Dec 14th 2012, 09:34 PM
bonfire09
Re: change of basis help?
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,v3 were given as well. I know how to do it but don't get the how it works?
P was given as [1,2,-1][-3,-5,0][4,6,1]. v1=[-2,2,3], v2=[=8,5,2] v3=[-7,2,6]. Then [u1 u2 u3]=[v1 v2 v3][ (u1)v (u2)v (u3)v]=[v1 v2 v3]P which was the way I went about it after messing around. I don't know why it works. Hopefully im more clear where im trying to come from.
• Dec 14th 2012, 09:36 PM
chiro
Re: change of basis help?
Take a look at my post again: it derives the formula from first principles.

If you have specific questions about the derivation I'll do my best to answer them.
• Dec 14th 2012, 09:45 PM
bonfire09
Re: change of basis help?
Oh here i just gave you an example. Look at my previous post.
• Dec 14th 2012, 09:56 PM
Deveno
Re: change of basis help?
here is an example:

suppose our vector space is R2, and we have:

B = {(1,0),(1,1)} = {b1, b2}

C = {(2,1),(-1,1)} = {c1, b2}.

now when i write [a,b]B, what i mean is: a(1,0) + b(1,1), and the same for C-coordinates: [a,b]C = a(2,1) + b(-1,1).

what is the matrix P that takes C-coordinates to B-coordinates?

well, such a matrix has columns P([1,0]C), and P([0,1]C). let's figure out what those columns are.

first of all, what is c1 = (2,1) in the B-basis? (2,1) = (1,0) + (1,1), that is [1,0]C = [1,1]B<---this is our first column.

next, what is c2 = (-1,1) in the B-basis? (-1,1) = -2(1,0) + (1,1), that is [0,1]C = [-2,1]B<---this is our second column.

so:

$P = \begin{bmatrix}1&-2\\1&1 \end{bmatrix}$

let's verify that P actually does what i say it does.

we'll take a random vector in C-coordinates, say [3,4]C. applying P to it, we get [3,4]C = [-5,7]B. is this true?

well 3c1+4c2 = 3(2,1) + 4(-1,1) = (6,3) + (-4,4) = (2,7).

and -5b1 + 7b2 = -5(1,0) + 7(1,1) = (-5,0) + (7,7) = (2,7).

note that the columns of P are the C-basis vectors IN THE BASIS B (the last part is IMPORTANT). unless B is the "standard" basis, you can't just write P down, you have to calculate what the C-basis vectors would be in the B-basis.
• Dec 14th 2012, 09:58 PM
chiro
Re: change of basis help?
Basically what your matrix is doing is if your basis is A = [u v w] where u,v,w are column vectors then multiplying A by a vector x gives:

Ax = u*x0 + v*x1 + w*x2 where x = <x0,x1,x2>

Recall that the standard basis is just Ix = i*x0 + j*x1 + k*x2 where <i,j,k> are the standard basis <1,0,0>, <0,1,0>, <0,0,1>.

So what you are trying to do is go from <u,v,w> to <a,b,c> in terms of one basis to another.

To do this you get both basis in terms of the identity and this is what I did above. We know that if we want to go from identity to other basis then we just multiply the vector in identity space by the basis matrix as a premultiplication.

So this means we can relate the identity basis vector to two arbitrary basis vectors and from this I derive the formula for a change of basis which I posted above.