Linear operators $\text{Hi, I am working on a assignment and I came to the solution }\\ \text{but it is not correct according to the book.}\\ \text{Can someone, please, take a look. I would really like to know what am I doing wrong.}\\ \text{Find the matrix of the operator in the canonical basis} \left( i, j, k\right) \text{which maps the vectors } \\a_1=\left(2,3,5 \right)\\a_2=\left( 0,1,2\right)\\ a_3=\left( 1,0,0\right)\\ \text{ in vectors } \\b_1=\left(1,1,1 \right)\\ b_2=\left( 1,1,-1\right)\\ b_3=\left( 2,1,2\right) \text{ respectively. }\\ \\ \text{ I solved it by writing the bases in matrices. } \\ \text{Matrix of base A=} \displaystyle \left( \begin{array}{lll} 2 & 0 & 1 \\ 3 & 1 & 0 \\ 5 & 2 & 0 \\ \end{array}\right)\\ \text{Matrix of base B=} \displaystyle \left( \begin{array}{lll} 1 & 1 & 2 \\ 1 & 1 & 1 \\ 1 & -1 & 2 \\ \end{array}\right)\\ \text{ and by using expression } T= Id_{AB}=Id_{AE}*Id_{EB}=\left( Id_{EA} \right) ^{-1}*Id_{EB}\\ \text{ where E is canonical base. } \\ \text{My solution} \displaystyle \left( \begin{array}{lll} 1 & 3 & 0 \\ -2 & -8 & 1 \\ -1 & -5 & 2 \\ \end{array}\right)\\ \text{Good solution according to the book} \displaystyle \left( \begin{array}{lll} 2 & -11 & 6 \\ 1 & -7 & 4 \\ 2 & -1 & 0 \\ \end{array}\right)\\$
 A: Hint: Let $\mathcal{E}$ be canonical (standard) basis; $\mathcal{A}=\{a_1,a_2.a_3\}$, $\mathcal{B}=\{b_1,b_2,b_3\}$. You found the matrix of the operator in basis $\mathcal{A}$, and this is the matrix $A^{-1}B$. The transformation matrix from basis $\mathcal{E}$ to $\mathcal{A}$ is the matrix $A$. So, to find the matrix of the linear operator in $\mathcal{E}$ we need $A \cdot A^{-1}B \cdot A^{-1} = BA^{-1}$.
A: You computed the product the wrong way around. Your matrix $B$ is the matrix of your operator $T$ for the case where the source vector is in coordinate system $A$ and the result is in canonical coordinates. To find the matrix which goes from canonical coordinates to canonical coordinates, to thus have to first translate a vector from canonical coordinates to coordinate system $A$, and then use matrix $B$ on that. Now, $A$ is the matrix than transform a vector from coordinate system $A$ to the canonical system, so the transform you have to prepend is $A^{-1}$. Therefore, $$
  T = BA^{-1}
$$
Note that this assumes you're working with column vectors.
