Base change Matrix from 2 given Matrices that perform the same function in 2 Bases I have 2 Matrices. Both of them perform the nilpotent function f, but they are in different Bases.
$D_{BB}(U)$ and $D_{CC}(U)$.
I need $D_{CB}(U)$.
I also have the bases given.
How would I go about that theoretically?
Thanks!
 A: Apparently you consider a linear transformation $f: V \rightarrow V$ where $[f]_{BB}=D_{BB}(U)$ and $[f]_{CC}=D_{CC}(U)$ (this notation is strange in my view, what is $U$, should those $U$'s be $f$?). My notation is this if $B = \{ v_1, \dots, v_n \}$ is a basis for the vector space $V$ and $e_1, \dots e_n$ is the standard basis of $\mathbb{R}^n$ then
$$ \Phi_{B}(x_1v_1+ \cdots + x_nv_n) = x_1e_1+ \cdots + x_ne_n$$
that is $\Phi_{B}(v_j)=e_j$ for $j=1, \dots , n$ and extend linearly. In short, the coordinate map $\Phi_{B}$ converts the abstract basis $B$ to the standard basis. Likewise, if $C = \{ w_1, \dots , w_n \}$ is a basis of $V$ then $\Phi_{C}: V \rightarrow \mathbb{R}^n$ is defined by $\Phi_{C}(w_j)=e_j$ extended linearly. The coordinate maps give linear isomorphisms of $V$ and $\mathbb{R}^n$. I use notation $L_A$ for $L_A(v)=Av$, this is the linear transformation on $\mathbb{R}^n$ which is induced by the matrix $A$. We can trade the abstract linear transformation $f$ for a matrix transformation by:
$$ f = \Phi_B^{-1} \circ L_{[f]_{BB}} \circ \Phi_B$$
or
$$ f = \Phi_C^{-1} \circ L_{[f]_{CC}} \circ \Phi_C$$
or if we use basis $C$ in the domain and basis $B$ in the range
$$ f = \Phi_B^{-1} \circ L_{[f]_{CB}} \circ \Phi_C$$
If we use $[L_A]=A$ to denote the standard matrix map we may also express the above in terms of the matrix relations:
$$ [\Phi_B \circ f \circ \Phi_B^{-1} ]= [L_{[f]_{BB}}] =[f]_{BB}$$
$$ [\Phi_C \circ f \circ \Phi_C^{-1} ]= [L_{[f]_{CC}}] =[f]_{CC}$$
$$ [\Phi_B \circ f \circ \Phi_C^{-1} ]= [L_{[f]_{CB}}] =[f]_{CB}$$
So, multiply by $[\Phi_C \circ \Phi_B^{-1}]$ the last equation to obtain:
$$ [\Phi_C \circ \Phi_B^{-1}][\Phi_B \circ f \circ \Phi_C^{-1} ]= [\Phi_C \circ \Phi_B^{-1}][f]_{CB}$$
hence,
$$ [\Phi_C^{-1} \circ f \circ \Phi_C ]= [\Phi_C \circ \Phi_B^{-1}][f]_{CB}$$
which is to say $[f]_{CC}=[\Phi_C \circ \Phi_B^{-1}][f]_{CB}$
but, probably we want to solve for $[f]_{CB}$,
$$ [f]_{CB} = [\Phi_B \circ \Phi_C^{-1}][f]_{CC} $$
or, in your notation,
$$ D_{CB}(U) = [\Phi_B \circ \Phi_C^{-1}]D_{CC}(U)$$
The change of basis matrix above is calculated as follows:
$$ col_j([\Phi_B \circ \Phi_C^{-1}]) = \Phi_B(w_j)$$
that is; the $j$-th column of the change of basis matrix is the coordinate vector of the $j$-th domain basis element in the range-coordinates. Certainly I have given you more information than you require here, perhaps you can cipher a minimal path for the problem you face.
