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Assuming that A is the matrix of a linear operator F in S find the matrix B of F in R

\begin{bmatrix}1&2&3\\2&1&0\\3&0&1\end{bmatrix}

A is:

S = {(0,0,1), (0,1,0), (1,0,0)}

R = {(1,1,0), (1,1,1), (0,1,1)}


My calculations and result are:

F(0,0,1) = (3,0,1)

F(0,1,0) = (2,1,0)

F(1,0,0) = (1,2,3)


Now let's start with B:

(1,1,0) = (1,0,0) + (0,1,0)

(1,1,1) = (1,0,0) + (0,1,0) + (0,0,1)

(0,1,1) = (0,1,0) + (0,0,1)

So result is

B = \begin{bmatrix}3&6&5\\3&3&1\\3&4&1\end{bmatrix}

Is it correct and if not can you show me my mistakes. Thanks

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closed as unclear what you're asking by zoli, Leucippus, Namaste, kjetil b halvorsen, user223391 Jan 18 '18 at 19:04

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  • $\begingroup$ The beginning has an error: as $(0,0,1)$ is the first vector of $S$, $F(0,0,1)=(1,2,3)$. $\endgroup$ – Bernard Jan 14 '18 at 22:51
  • $\begingroup$ A quick check on such computations is the trace of the matrix. Since the trace of the matrix is the sum of eigenvalues (with multiplicity), it should be the same for $A$ and $B$ (because they should be similar matrices, given by the change of basis for representations of the same transformation $F$). Now $\mathbf{Tr} A = 3$ but $\mathbf{Tr} B = 7$. $\endgroup$ – hardmath Jan 15 '18 at 1:53
  • $\begingroup$ If someone got the answer i would appreciate posting it, coz i got different result now (maybe the correct one now), but i am not able to post it now as i am on the phone $\endgroup$ – James Smith Jan 15 '18 at 10:22

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