Change of Basis Calculation I've just been looking through my Linear Algebra notes recently, and while revising the topic of change of basis matrices I've been trying something:
"Suppose that our coordinates are $x$ in the standard basis and $y$ in a different basis, so that $x = Fy$, where $F$ is our change of basis matrix, then any matrix $A$ acting on the $x$ variables by taking $x$ to $Ax$ is represented in $y$ variables as: $F^{-1}AF$ "
Now, I've attempted to prove the above, is my intuition right?
Proof: We want to write the matrix $A$ in terms of $y$ co-ordinates.
a) $Fy$ turns our y co-ordinates into $x$ co-ordinates.
b) pre multiply by $A$, resulting in $AFy$, which is performing our transformation on $x$ co-ordinates
c) Now, to convert back into $y$ co-ordinates, pre multiply by $F^{-1}$, resulting in $F^{-1}AFy$
d) We see that when we multiply $y$ by $F^{-1}AF$ we perform the equivalent of multiplying $A$ by $x$ to obtain $Ax$, thus proved.
Also, just to check, are the entries in the matrix $F^{-1}AF$ still written in terms of the standard basis?
Thanks.

Without saying much, here is how I usually remember the statement and also the proof in one big picture:
\begin{array}{ccc}
x_{1},\dots,x_{n} & \underrightarrow{\;\;\; A\;\;\;} & Ax_{1},\dots,Ax_{n}\\
\\
\uparrow F &  & \downarrow F^{-1}\\
\\
y_{1},\dots,y_{n} & \underrightarrow{\;\;\; B\;\;\;} & By_{1},\dots,By_{n}
\end{array}
And
$$By=F^{-1}AFy$$
