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I have an endomorphism T(x) = Ax where

$A = \begin{pmatrix} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \end{pmatrix}$ in the canonical basis. I want the transformation to go from $B'$ to $B'$. How do I do that?

This is what I've tried to do so far:

I tell myself that I want the vector columns of A to be expressed in B' = {(1,0,0),(0,1,0),(1,1,1)). Thus, I know that, for example, for the first column of $A$, we have

$\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} * [x_1]_B' = \begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix}$

And now, I tell myself that we need to left multiply both sides by $A^{-1}$, and we get

$[x_1]_B' = A^{-1}\begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix}$.

I do this calculation for the three columns, and I would get a linear transformation taking a vector in basis $B'$ to a vector in basis $B'$. Is this right? Thank you.

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  • $\begingroup$ I don't understand the method you have applied. What is the matrix $A'$ you obtain the basis $B'$? if you show that I can check it. $\endgroup$ – user May 13 '18 at 19:25
  • $\begingroup$ @gimusi This is what I'm trying to find. I want to find the matrix which represents T(x) not from canonical to canonical, but from B' to B'. Am I doing this wrong? Thanks. $\endgroup$ – iaskdumbstuff May 13 '18 at 19:27
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    $\begingroup$ The maethod I've shown works fine, you can try and check the result directly. $\endgroup$ – user May 13 '18 at 19:30
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The matrix

$$M = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}$$

represents the change of basis form the new basis to the canonical then the matrix A' in the new basis is

$$A'=M^{-1}AM$$

Indeed since in the canonical

  • $T(x)=y=Ax$

and

  • $x=Mx'$
  • $y=My'$

we have that

$$y=Ax\implies My'=AMx'\implies y'=M^{-1}AMx'$$

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  • $\begingroup$ I'm having trouble understanding this. $M * [x]_{B'} = [x]$ in the canonical basis, right? $\endgroup$ – iaskdumbstuff May 13 '18 at 19:32
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    $\begingroup$ Note that in the new basis the vector (1,1,1) is represented by (0,0,1) indeed M(0,0,1)=(1,1,1)and so on for the others, thus the matrix M transform vectors in basis B' to the canonical. That's the key point to understand the method. $\endgroup$ – user May 13 '18 at 19:35
  • $\begingroup$ I almost understand, but I have one last question. Why does $y=Ax\implies My'=AMx'$ and not $y=Ax\implies My'=MAx'$? $\endgroup$ – iaskdumbstuff May 13 '18 at 19:39
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    $\begingroup$ @user494405 We simply substitute $y=My'$ and $x=Mx'$ into $y=Ax$ $\endgroup$ – user May 13 '18 at 19:41
  • $\begingroup$ Oh, that makes sense. I have a follow up question, if you don't mind. How would this change if I want it to go from $B'$ to another $B''$ which is represented by the change of basis matrix $C$. Would $A'' = C^{-1}AM$? $\endgroup$ – iaskdumbstuff May 13 '18 at 19:56

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