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Let $T: R^2 \rightarrow R^2$ be given by:

$$T(x_1,x_2) = (4x_1 -2x_2, 2x_1 +x_2)$$

And let $$B = \{(1,1), (-1,0)\}$$ be a basis for $R^2$.

First, I write down the matrix of $$T = \left[\begin{array}{cc}4 & -2 \\2 & 1\end{array}\right]$$.

Then let $$C = \left[\begin{array}{cc}1 & -1 \\1 & 0\end{array}\right]$$.

So $C^{-1}x = [x]_B$.

So why doesn't $T_b([x]_B) = CTC^{-1}x$?

If it did, then shouldn't $CTC^{-1}$ give me the right answer of the $T_B$ (the transformation w.r.t basis $B$). It doesn't; rather it gives $$CTC^{-1} = \left[\begin{array}{cc}3 & -1 \\2 & 2\end{array}\right]$$ and the answer is supposed to be

$$Solution = \left[\begin{array}{cc}3 & -2 \\1 & 2\end{array}\right]$$

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Use $(C^t)^{-1}$ as base change matrix instead. $C^{-1} = (b_1, b_2)$ with $b_1 = (1,1)^t$ and $b_2 = (-1,0)^t$.

Then $C T C^{-1}$ gives the result.

octave> T
T =

   4  -2
   2   1

octave> CI
CI =

   1  -1
   1   0

octave> C = inv(CI)
C =

   0   1
  -1   1

octave> C*T*CI
ans =

   3  -2
   1   2

octave> 
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  • $\begingroup$ (a) why?, (b) I get the wrong answer when I use $C^t$ as change-of-basism matrix. $\endgroup$ – larry Apr 16 '15 at 18:17
  • $\begingroup$ I've updated $C$ in my post to what I originally wanted it. Is that what you are talking about? $\endgroup$ – larry Apr 16 '15 at 18:19
  • $\begingroup$ My real question is why can't I use $CTC^{-1}$. Doesn't doesn't $T_b([x]_B) = CTC^{-1}x$? $\endgroup$ – larry Apr 16 '15 at 18:23
  • $\begingroup$ Then why does $CTC^{-1}$ give the wrong answer? $\endgroup$ – larry Apr 16 '15 at 18:28
  • $\begingroup$ What is $b'$ in this example? $\endgroup$ – larry Apr 16 '15 at 18:34

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