Linear Algebra - Linear Transformations, Change of Basis

[sassie]

Homework Statement

I need to prove this formula, but I'm not sure how to prove it.[T]_C = P_(C<-B).[T]_B.P_(C<-B)^-1

whereby B and C are bases in finite dimensional vector space V, and T is a linear transformation. Your help is greatly appreciated!

Homework Equations

T(x)=Ax
[x]_C=P_(C<-B)[x]_B
Similar matrices.

The Attempt at a Solution

T(x)=Ax
[T]_C=[[T(c1)]_C...[T(cn)]_C

...

 

[HallsofIvy]

Suppose [x]_B is some vector x represented in basis B and that Tx= y (independent of the basis).

Then, by definition of "P_B<-C", P_C<-B is P^-1_B<-C is so P^{-1}_B<-C[x]_C= [x]_B.

Then T_BP^{-1}_B<-C[x]_Cx_{C= TB[x]B= [y]B and, finally, PB<-CTBP^{-1}B<-C[x]C= PB<-C[y]C= [y]_B.}

 

[sassie]

Suppose [x]_B is some vector x represented in basis B and that Tx= y (independent of the basis).

Okay, thanks! I did manage to get something like that before, but I got stuck because I didn't do the tx=y thing.

But may I ask, why does Tx= y need to be independent of the basis?

 

[HallsofIvy]

Okay, thanks! I did manage to get something like that before, but I got stuck because I didn't do the tx=y thing.

But may I ask, why does Tx= y need to be independent of the basis?

A linear transformation, T, is a function from vector space U to vector space V such that T(au+bv)= aT(u)+ bT(v). There are no "bases" required for its definition. Bases are only necessary to write the linear transformation as a matrix.