Change of Basis of Linear Transformation

Hi all,

Let B be a basis for R^2. Let L be the linear transformation represented in standard coordinates by the matrix R. Then, to represent L with respect to the standard coordinates, we take A^(-1)RA where A is a column matrix of the B vectors.

Now, what if R was expressed in terms of the basis B and we want to express L with respect to the standard coordinates? How would we go this way?

Thanks

Re: Change of Basis of Linear Transformation

Hey sfspitfire23.

The easiest way to look at this is to go from first principles.

Now in one basis we have some vector v in Basis B, and in another basis we have v' in Basis B'. Let x be the true vector that is invariant to both bases: then this implies

Bx = v and

B'x = v'. If B and B' are both basis then B and B' are invertible giving

x = B^(-1)v = B'^(-1)v'. Writing these equations to get v in terms of v' (and vice versa) gives:

B*B'^(-1)v' = v and

B'*B^(-1)v = v'

Now a linear transformation acting on R^2 is simply Ax = b taking x and mapping it to b, and the above can relate Av to Av' in a straight-forward manner.

Re: Change of Basis of Linear Transformation

suppose B and B' are two bases for a vector space V, and L:V-->V is a linear transformation.

if [L]_{B} = R, and the change-of-basis matrix from B' to B is A (whose columns consist of the basis vectors of B' expressed in B-coordinates), then:

[L]_{B'} = A^{-1}RA, as you correctly deduced.

it therefore stands to reason that:

R = A[L]_{B'}A^{-1} (use matrix multiplication).

here is how it works:

A^{-1}([v]_{B}) = [v]_{B'} (since A changes B'-vectors to B-vectors, A^{-1} "changes them back" to B'-vectors).

[L]_{B'}([v]_{B'}) = [Lv]_{B'} (the matrix for L in the basis B' takes v in B'-coordinates to Lv in B'-coordinates)

A([Lv]_{B'}) = [Lv]_{B} (A turns Lv in B'-coordinates to Lv in B-coordinates).

so A[L]_{B'}A^{-1}([v]_{B'}) = A([L]_{B'}(A^{-1}([v]_{B}))) = A([L]_{B'}([v]_{B'})) = A([Lv]_{B'}) = [Lv]_{B}, that is:

A[L]_{B'}A^{-1} is a linear transformation which takes [v]_{B} to [Lv]_{B} for every v in V, and so is the matrix for L in the basis B, that is: R.

the algebra "looks prettier" if we call the matrix for L in the basis B', S.

then S = A^{-1}RA so:

AS = AA^{-1}RA = IRA = RA

ASA^{-1} = RAA^{-1} = RI = R