$V$ Is a vector space, $S:V\to V\;$, $T:V\to V\;$ are linear transformations, $B,C$ are ordered bases for $V$

Question

Also, $\left [ S \right ]_{C}^{C} = \left [T \right ]_{B}^{B}$

I need to show that there is a one-to-one and onto linear trasformation $R:V\to V\;$ such that $T = R^{-1}\circ S \circ R$

I understand that this question is related to change of basis matrices, but I didn't figure out how to do this proof.

Answer 1

So, the question is, how can you use $S$ to compute $T$?

Take a vector $v \in V$. One way to compute $Tv$ is to first take $[v]_B$, the coordinate column vector of $v$ with respect to $B$, then multiply it to $[T]_B^B$. The result is $[T]_B^B[v]_B = [Tv]_B$, so the resulting column vector can be expanded by way of a linear combination with respect to vectors of $B$, to produce the vector $Tv$.

Of course, we also happen to know that $[T]_B^B = [S]_C^C$. Essentially, we need to turn $[v]_B$ into a column vector $[w]_C$ for some $w$. Then, the exact same multiplication $[T]_B^B[v]_B = [Tv]_B$ can now be interpreted differently as $[S]_C^C[w]_C = [Sw]_C$. The same process that translates finds $w \in V$, given $v \in V$, such that $[w]_C = [v]_B$, can be reversed to turn $[Sw]_C$ into some coordinate column vector $[x]_B$. But, this vector $x$ must be $Tv$, as we produced it in the following fashion: $$[x]_B = [Sw]_C = [S]_C^C[w]_C = [T]_B^B[v]_B = [Tv]_B.$$

So, in total, this is what we're planning: $$v \mapsto [v]_B = [w]_C \mapsto [S]_C^C[w]_C = [Sw]_C = [x]_B \mapsto x = Tv,$$ for some $w, x \in V$, where $w$ depends on $v$, and $x$ depends on $Sw$.

Let's give some of these maps names. We have $\phi_B, \phi_C : V \to \Bbb{R}^{n \times 1}$, the coordinate maps mapping $v$ to $[v]_B$ and $[v]_C$ respectively. Note that these are isomorphisms, and invertible.

The map that takes $v$ to $w$ is the composition $\phi_C^{-1} \circ \phi_B : V \to V$; $\phi_B$ takes $v$ to the coordinate column vector, and $\phi_C^{-1}$ pretends that this column vector was in terms of $C$, and expresses the vector in $V$ that would produce such a column vector. So, simplifying the above picture a little, $$v \underset{\phi_C^{-1} \circ \phi_B}{\longrightarrow} w \underset{S}{\longrightarrow} Sw \underset{\phi_B^{-1} \circ \phi_C}{\longrightarrow} x = Tv.$$ This means our $R$ is $\phi^{-1}_C \circ \phi_B$. This is a composition of isomorphisms, and hence is an isomorphism itself, making it linear, one-to-one, and onto.

Answer 2

Hint: What about R mapping the $n$-th vector of B into the $n$-th vector of C?

Answer 3

Another way to see the problem is to realize that $R$ is simply the matrix representation of the identity transformation where "domain" vectors are expressed in the $B$ basis and "image" vectors are expressed in the $C$ basis.