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Theorem: Let $T$ be a linear operator over a finite dimensional vectorspace $V$, and let $\beta$ and $\gamma$ be two ordered bases for $V$. If $P$ is the change of basis matrix from $\beta$ to $\gamma$, then we have \begin{align*} [T]_{\beta} = P^{-1} [T]_{\gamma} P. \end{align*}

Proof: Let $Id: V \rightarrow V$ be the identical transformation. Then we have $T = Id \circ T = T \circ Id$. Now, since $P = [Id]_{\beta}^{\gamma}$, we also have that \begin{align*} P[T]_{\beta} = [Id]_{\beta}^{\gamma} [T]_{\beta}^{\beta} = [Id \circ T]_{\beta}^{\gamma} = [T \circ Id]_{\beta}^{\gamma} = [T]_{\gamma}^{\gamma} [Id]_{\beta}^{\gamma} = [T]_{\gamma} P, \end{align*} which implies that $[T]_{\beta} = P^{-1} [T]_{\gamma} P$.

Q.E.D.

Now, for the more general case, let $T : V \rightarrow W$ be a linear map, and let $\beta, \beta'$ be two bases for $V$, and let $\gamma, \gamma'$ be two bases for $W$. Also, let $A = [T]_{\beta}^{\gamma}$ and $B = [T]_{\beta'}^{\gamma'}$ be two matrixrepresentations of $T$. We search the relationship between $A$ and $B$. Let $P$ be the change of basis matrix from $\beta$ to$\beta'$, and let $Q$ be the change of basis matrix from $\gamma$ to $\gamma'$. We can write the linear map $T$ as $Id_W \circ T \circ Id_V$. We have further that $P = [Id_V]_{\beta}^{\beta'}$, and $Q = [Id_W]_{\gamma}^{\gamma'}$. The relationship we search is \begin{align*} B = Q A P^{-1} \end{align*} I want to prove this now in the same manner as above, but I'm struggling here. Can someone help? I think I have to start off from $Id_W \circ T \circ Id_V$, but not sure what to do next.

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  • $\begingroup$ What you call the transition matrix is not clear. What are its columns made of? $\endgroup$ – Bernard Mar 30 '15 at 22:32
  • $\begingroup$ Oh, my bad, thought that was clear. I just meant the 'change of basis matrix'. I edited it. $\endgroup$ – Kamil Mar 30 '15 at 22:35
  • $\begingroup$ You misunderstood me. The question is: according to you, what are the columns of this change of basis matrix made up of? I ask the question, because, for me, you formula is false: it should be $T_\gamma=P^{-1}T_\beta P$. $\endgroup$ – Bernard Mar 30 '15 at 22:40
  • $\begingroup$ It depends on wether how you choose the change of basis matrix. If I chose $P$ as the change of basis from $\beta'$ to $\beta$, and $Q$ from $\gamma'$ to $\gamma$, then the equation would be $B = Q^{-1} A P$. The $j$th column of $B$ is the coordinatevector of $x$ with respect to the basis $\gamma'$. $\endgroup$ – Kamil Mar 30 '15 at 23:03
  • $\begingroup$ What is normally called the change of basis matrix from β to β' is the matrix with columns equal to the coordinates of the vectors that belong to β', in the basis β. $\endgroup$ – Bernard Mar 30 '15 at 23:08
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To solve $B = QAP^{-1}$, it suffices to show that $BP = QA$ since $A$ is invertible. So what's wrong with an approach similar to what you did above?

$$BP = [T]^{\gamma'}_{\beta'}[Id_V]_\beta^{\beta'} = [T\circ Id_V]_\beta^{\gamma'} = [Id_W\circ T]_\beta^{\gamma'} = [Id_W]_\gamma^{\gamma'}[T]_\beta^\gamma = QA$$

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