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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.

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3 Answers 3

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Your approach seems correct.

I don't know if the following helps, but anyway: So you have a vector space $V$ (over say the complex numbers) and you have say two basis $E$ and $D$.

That $F$ is a change of basis matrix means that if as column vector $y = (y_i)$ written with respect to the basis $E$, then you get the coordinates with respect to $D$ by $x = Fy$.

Now you have a linear transformation $T: V \to V$. With respect to each basis, this transformation is given by two matrices, say $A_E$, $A_D$. So if $y = (y_i)$ (wrt. basis $E$) then $Ty = A_Ey$ and the result with be the coordinates in the basis $E$. (And likewise for the basis $D$ using $A_D$).

So given a vector $y = (y_i)$ written in the basis $E$, you could then first transform the coordinates to the basis $D$, then you the matrix $A_D$ and then transform the coordinated back to the basis $E$. So you get $A_E(y) = F^{-1}A_DFy$.

One can actually write out all of this (If you have never done so I recommend that you do it) with coordinates. So you would start with the vector $v$ and write it as a linear combination of the basis $E$: $v = y_1e_1 + \dots y_ne_n$ and continue from there...

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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 $$By=F^{-1}AFy$$

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Your statements a) – d) are correct as stated.

Concerning the sentence "Also, just to check, are the entries in the matrix $F^{-1}AF$ still written in terms of the standard basis?", one can say the following: The entries in this matrix are just numbers. The given matrix $A$ describes a certain linear map acting on points $x=(x_1,\ldots, x_n)$. The matrix $F^{-1}AF$ describes the same linear transformation when the points $x$ get new coordinates $(y_1,\ldots, y_n)$.

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