Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 0 down vote accepted

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...

share|cite|improve this answer

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 $(y_1,\ldots, y_n)$.

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.