How does $D=C^{-1}AC$ matrix of a linear map with respect to alternate basis. I was watching a Khan video on the topic and wondered how you can use the graph he draws to derive $D=C^{-1}AC$ I see how it follows without using the graph but the graph confuses me.
Looking at this shouldn't it be $CAC^{-1}$ because we start at the bottom left and want to work around clockwise to bottom right?
Here is the graph to see.
http://imgur.com/PaS9Xmb
Many thanks.
 A: No, it should be as described. Maybe this complemented drawing will convince you:

A: Yes the thing is quite tricky. I hope this will clear it out even if at first glance could seem confusing. I used the symbol $[.]_B$ to indicate a representation of the application respect to a certain base. 
Let's suppose we have a linear transformation $$T:\mathbf{X}\rightarrow\mathbf{X}$$ 
This is a linear transformation and so it can be represented as a matrix. More clearly if we have a base $\mathscr{E}=\left\{ \mathbf{e}_{i}\right\} _{1\leq i\leq n}$ of the vectorial space $\mathbf{X}$, the endomorphism $T$ can be represented in the base $\mathscr{E}$ with $A\in\mathfrak{M}_{n}^{n}(\mathbb{K})$ formed by the image by $T$ of the elements of the base. i.e.:
\begin{equation}
_{\mathscr{E}}[T]_{\mathscr{E}}=A=\left(\left[T(\mathbf{e}_{1})\right]_{\mathscr{E}},\left[T(\mathbf{e}_{2})\right]_{\mathscr{E}},..,\left[T(\mathbf{e}_{n})\right]_{\mathscr{E}}\right)\in\mathfrak{M}_{n}^{n}(\mathbb{K})
\end{equation}
Where$\left[T(\mathbf{e}_{1})\right]_{\mathscr{E}}$ is representing the vector $T(\mathbf{e}_{1})$ in the coordinate expressed in the base $\mathscr{E}$. So for a vector $\mathbf{x}\in\mathbf{X}$ where $\mathbf{x}=\underset{i=1..n}{\sum}\xi^{i}\mathbf{e}_{i}$ we can write:
\begin{equation}
T\mathbf{x}={}_{\mathscr{E}}[T]_{\mathscr{E}}[\mathbf{x}]_{\mathscr{E}}=\left(\begin{array}{cccc}
a_{1}^{1} & a_{2}^{1} & \ldots & a_{n}^{1}\\
a_{1}^{2} & a_{2}^{2} & \ldots & a_{n}^{2}\\
\vdots & \vdots &  & \vdots\\
a_{1}^{n} & a_{2}^{n} & \ldots & a_{n}^{n}
\end{array}\right)\left(\begin{array}{c}
\xi^{1}\\
\xi^{2}\\
\vdots\\
\xi^{n}
\end{array}\right)
\end{equation}
Now if we have $\mathscr{F}=\left\{ \mathbf{f}_{i}\right\} _{1\leq i\leq n}$
a new base of the space $\mathbf{X}$, and $C_{\mathscr{EF}}=(c_{j}^{i})$
the matrix with the coordinates of the base  $\mathscr{E}$ in respect of $\mathscr{F}$ : 
$$
\mathbf{e}_{j}={\sum}c_{j}^{i}\mathbf{f}_{i}
$$
And $C_{\mathscr{FE}}=C_{\mathscr{EF}}^{-1}=(d_{j}^{i})$
then we are interested in writing the endomorphism $T$ in the base
$\mathscr{F}$ which means $_{\mathscr{F}}[T]_{\mathscr{F}}$.
So we have
\begin{equation}
_{\mathscr{F}}[T]_{\mathscr{F}}=D=\left(C_{\mathscr{FE}}\left[T(C_{\mathscr{EF}}\mathbf{e}_{1})\right]_{\mathscr{E}},..,C_{\mathscr{FE}}\left[T(C_{\mathscr{EF}}\mathbf{e}_{n})\right]_{\mathscr{E}}\right)=C_{\mathscr{FE}}^{-1}AC_{\mathscr{EF}}
\end{equation}
Finally to be clear we have:
\begin{equation}
T\mathbf{x}={}_{\mathscr{E}}[T]_{\mathscr{E}}[\mathbf{x}]_{\mathscr{E}}={}_{\mathscr{F}}[T]_{\mathscr{F}}[\mathbf{x}]_{\mathscr{F}}
\end{equation}
