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Every matrix corresponds to a linear transformation. Equivalently, for every linear map one can find a corresponding matrix, associated to the function through a specific base.

Let's consider for example the linear transformation $f:\mathbb{R}^3\rightarrow\mathbb{R}^3$ defined by \begin{equation} f(1,0,0)=(0,1,0)\\ f(0,1,1)=(1,1,0)\\ f(0,0,1)=(1,0,0) \end{equation} and the base $B=(0,1,1),(0,1,0),(1,0,0)$. The matrix associated to $f$ through $B$ should be (it comes from a prevoius answer I've asked) \begin{equation} M_{f}^B=\begin{pmatrix} 1&0&0\\1&1&1\\0&0&0 \end{pmatrix} \end{equation} How can one verify if the calculations are correct?

Shouldn't it be by multiplying the obtained matrix $M_{f}^B$ for the coordinates of the vector with respect to the base of the domain and checking if the result is the vector of the coordinates with respect to the base of the codomain? Hence, in this case, considering for example the second vector of $B$, \begin{equation} M_{f}^B\cdot(0,1,1)^T= \begin{pmatrix} 0&0&0\\1&1&1\\0&0&1 \end{pmatrix} \begin{pmatrix} 0\\1\\1 \end{pmatrix}=(0,2,1) \end{equation} Anyway $(0,2,1)\neq (1,1,0)=f(0,1,1)=f(b_{2})$ so the matrix is uncorrect. Or is my reasoning that is wrong?

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In order to discovery what is the matrix that represent the linear transformation we have to check what happen wich the image of the basis. Let's do that for a space wich dimension 2 but the idea is the same for any finite space. Let be $E=\{e_1,e_2\}$ the base of the space and $f:\mathbb{R}^2\rightarrow\mathbb{R}^2$ a linear function. So $$f(x,y)=f(xe_1+ye_2)=xf(e_1)+yf(e_2)=\begin{pmatrix} f(e_1)&f(e_2) \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix}$$ where $\begin{pmatrix} f(e_1)&f(e_2) \end{pmatrix}$ is the matrix that represent the linear transformation at the base $E=\{e_1,e_2\}$. Using that idea we can see that on the specific example we have that represent the transformation at the canonical basis is: $$M_f^C=\begin{pmatrix} 0&0&1\\1&1&0\\0&0&0 \end{pmatrix}$$ and we can also see that the matrix $$(C \rightarrow B)=\begin{pmatrix} 0&0&1\\1&1&0\\1&0&0 \end{pmatrix}$$ is the matrix that change from the canonical base to the base $B=\{(0,1,1),(0,1,0),(1,0,0)\}$ and so $$\begin{equation} M_{f}^B=\begin{pmatrix} 0&0&1\\1&1&0\\1&0&0 \end{pmatrix}\begin{pmatrix} 0&0&1\\1&1&0\\0&0&0 \end{pmatrix}=\begin{pmatrix} 0&0&0\\1&1&1\\0&0&1 \end{pmatrix}\end{equation}$$ The problem with the calculation bellow $$\begin{equation} M_{f}^B\cdot(0,1,1)^T= \begin{pmatrix} 0&0&0\\1&1&1\\0&0&1 \end{pmatrix} \begin{pmatrix} 0\\1\\1 \end{pmatrix}=(0,2,1) \end{equation}$$ is that $M_f^B$ is written at the base B and $(0,1,1)$ is written at the canonical base. You have to use the same base to do the calculation. Or change everybody to the canonical base or to the base B.

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  • $\begingroup$ First of all thank you for your answer @Arnaldo. I understood the last part of your answer. But how did you find the matrix $C\rightarrow B$? Since $ (1,0,0)=0(0,1,1)+0(0,1,0)+1(1,0,0)\\ (0,1,0)=0(0,1,1)+1(0,1,0)+0(1,0,0)\\ (0,0,1)=1(0,1,1)-1(0,1,0)+0(1,0,0)\\$ shouldn't the matrix be $(C\rightarrow B)=\begin{pmatrix} 0&0&1\\0&1&0\\1&-1&0 \end{pmatrix}$? $\endgroup$ – M.Giacchello Dec 1 '16 at 15:47
  • $\begingroup$ If you apply $(C\rightarrow B)$ at the canonical base you will get the base $B$. $\endgroup$ – Arnaldo Dec 1 '16 at 16:11

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