Change of Basis of Linear Transformation

Oct 2009
273
2
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
 
Last edited:

chiro

MHF Helper
Sep 2012
6,608
1,263
Australia
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.
 

Deveno

MHF Hall of Honor
Mar 2011
3,546
1,566
Tejas
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-1RA, 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-1RA so:

AS = AA-1RA = IRA = RA

ASA-1 = RAA-1 = RI = R