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Given is $f: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}$ with $f:(x,y,z) \rightarrow (x+2y+z, y+z, -x+3y+4z)$. Determine the transformation matrix in terms of the basis $B= \left\{\begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix}, \begin{pmatrix} 0\\ 1\\ 1 \end{pmatrix},\begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}\right\}$

$f(\begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix})= \begin{pmatrix} 1\\ 0\\ -1 \end{pmatrix}= 0 \cdot \begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix} + 0 \cdot \begin{pmatrix} 0\\ 1\\ 1 \end{pmatrix}- 1 \cdot \begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}$

$f(\begin{pmatrix} 0\\ 1\\ 1 \end{pmatrix})= \begin{pmatrix} 3\\ 2\\ 7 \end{pmatrix}= 8 \cdot \begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix} + 2 \cdot \begin{pmatrix} 0\\ 1\\ 1 \end{pmatrix}+5 \cdot \begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}$

$f(\begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix})= \begin{pmatrix} 0\\ 1\\ 5 \end{pmatrix}= 4 \cdot \begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix} + 1 \cdot \begin{pmatrix} 0\\ 1\\ 1 \end{pmatrix}+ 4 \cdot \begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}$

I'm not sure how to read the transformation matrix now. Either I will read it correctly or I will accidentally read its transposition :s

$T=\begin{pmatrix} 0 & 8 & 4\\ 0 & 2 & 1\\ -1 & 5 & 4 \end{pmatrix}$

Did I do it all correctly? If it's alright, there are maybe faster ways doing this?

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  • $\begingroup$ What would you like the representing matrix of your linear transformation to do? I'd say -- agree with $f$ on the basis vectors. Does this help you to choose between $T$ and its transpose? $\endgroup$ – uniquesolution Mar 27 '17 at 10:41
  • $\begingroup$ @uniquesolution I'm not sure if I understood you correctly but it seems like it's fine like that? $\endgroup$ – cnmesr Mar 27 '17 at 10:52
  • $\begingroup$ You don't seem to have decided whether $\mathbb{R}^3$ is the set of row vectors or the set of column vectors. That may generate confusion. $\endgroup$ – ancientmathematician Mar 27 '17 at 11:14
  • $\begingroup$ Looks right, but the notation of the matrix should be $[T]_B$. $\endgroup$ – Itay4 Mar 27 '17 at 11:19
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Perhaps a more straightforward way of computing the matrix is by using the relation \begin{equation} [T]_\mathcal{B} = P^{-1} T P, \end{equation} where $[T]_\mathcal{B}$ is the matrix that you computed, $$P = (b_1\quad b_2\quad b_3) = \begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 1 & 1\end{pmatrix}, $$ and $$T = \begin{pmatrix} f(e_1) & f(e_2) & f(e_3) \end{pmatrix} = \begin{pmatrix}1 & 2 & 1\\ 0 & 1 & 1 \\ -1 & 3 & 4 \end{pmatrix},$$ where $e_1, e_2, e_3$ are the standard basis vectors. With this approach, all you have to do is invert $P$, which is easy, and then multiply the matrices. (check to see that it yields the same result!)

Heres a nice diagram that illustrates why the relation is true, from Linear Algebra Basis Trick.

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