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I'm supposed to solve this task using the theory behind change of basis. So, I'm given a linear operator $\varphi :{ R }^{ 3 }\rightarrow { R }^{ 3 }$ that maps

$\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}\mapsto \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix},\quad \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix}\mapsto \begin{pmatrix} -1 \\ 0 \\ 3 \end{pmatrix},\quad \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\mapsto \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$.

I'm supposed to find the matrix of this map in the standard basis $f$ that is the set of columns of the identity matrix of $dim=3$

Let $e$ and $e'$ be:

$e=\begin{pmatrix} 1 & -1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 1 \end{pmatrix},\quad e'=\begin{pmatrix} 2 & -1 & 1 \\ 1 & 0 & 1 \\ -1 & 3 & 2 \end{pmatrix}$

Then we can find the transition matrix from $e$ to $e'$:

$e'=eS\quad \Rightarrow \quad S={ e }^{ -1 }e'$

Then we can use elementary operation with rows of the augmented matrix to calculate ${ e }^{ -1 }e'$

$\left( { e }|e' \right) \quad \rightarrow \quad \left( { E }|{ e }^{ -1 }e' \right) \quad \rightarrow \quad S$

Well, I don't know what to do next. I was told to recall that the $j$-th column of the matrix $A$ of the operator $\varphi$ in the basis $f$ is the coordinate column of $\varphi({f}_{j})$ in the basis $f$. Here $f$ denotes the standard basis, i.e. ${f}_{1}=(1,0,0)$, ${f}_{2}=(0,1,0)$ and ${f}_{3}=(0,0,1)$, but I don't see how this is supposed to help me.

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Let $A$ be the map from the standard basis to the standard basis. Let $e,e'$ be the matrices given in the question. Then we have $e'=A e$, so you want to compute $A=e' e^{-1}$.

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  • $\begingroup$ I understand why $A$ is the matrix of $\varphi$, but I'm not sure I really get why $A$ (given by the equation above) is the matrix of $\varphi$ in the standard basis. I would be really glad if you participate on this a little bit. Seems like there is something important about images of linear functions that I completely or partially missed. $\endgroup$ – Eli Korvigo Jun 4 '15 at 18:15
  • $\begingroup$ The matrix $e'$ gives $\phi$ in terms of the domain basis $e$. You want the representation of $\phi$ using the standard basis for the domain. So you need a map that takes a point in the standard basis and gives the representation in the $e$ basis.The matrix $e$ maps from the $e$ basis to the standard basis, so the other way around is given by $e^{-1}$. Hence the map is $e' e^{-1}$ (that is, apply $e^{-1}$ to get the representation in the $e$ basis and then use the $e'$ matrix to compute the value of $\phi$). (It is a little confusing.) $\endgroup$ – copper.hat Jun 4 '15 at 18:22

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