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Text of the exercise:

Consider the linear transformation $\varphi:\mathbb{R}^3 \rightarrow \mathbb{R}^3$ defined by:

$\varphi(1,0,0) = (0,1,0)\\ \varphi(0,1,1) = (1,1,0) \\\varphi(0,0,1) =(1,0,0)$

Is $B={(0,1,1),(0,1,0),(1,0,0)}$ a basis of $\mathbb{R}^3$? If the answer is yes, find the matrix $M_{\varphi}^B$ associated to $\varphi$ through $B$.

Reasoning:

$B$ is a basis of $\mathbb{R}^3$ since it is a set of three linearly independent vectors (proof is given by the fact that the matrix constituted by their coordinates is not singular) in a three-dimensional space.

I know from the definition of $\varphi$ the result of the transformation of two of the vectors of $B$; precisely $\varphi(1,0,0)=(0,1,0)$ and $\varphi(0,1,1)=(1,1,0)$. In order to infer the last one, I could use the definind property of a linear mapping, i.e. that a function $f:V\rightarrow W$ is linear if $\forall \textbf{v}_{1},\textbf{v}_{2}\in V$, $\forall\lambda_{1},\lambda_{2}\in \mathbb{K}$ then $f(\lambda_{1}\textbf{v}_{1}+\lambda_{2}\textbf{v}_{2}=\lambda_{1}f(\textbf{v}_{1})+\lambda_{1}f(\textbf{v}_{1})$, so that: \begin{equation} \varphi(0,1,0)=\varphi(0,1,1)-\varphi(0,0,1)=(1,1,0)-(1,0,0)=(0,1,0) \end{equation} The asked matrix is thus: \begin{equation} M_{\varphi}^B=\begin{bmatrix} 1&0&0\\1&1&1\\0&0&0 \end{bmatrix} \end{equation}

I'd greatly appreciate a feedback. Thank you all!

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  • $\begingroup$ Well, you can check yourself through matrix multiplication. For instance, is $(1,0,0)$ mapped onto $(0,1,0)$ through this multiplication? Same for the other two. Notice that your stated matrix is singular! $\endgroup$ – imranfat Nov 23 '16 at 23:36
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Indeed B is a basis for exactly the reason you say.

$e_1 = (0,1,1), e_2 = (0,1,0), e_3 = (1,0,0)$

$\varphi e_1 = \varphi (0,1,1) = (1,1,0) = e_2 + e_3\\ \varphi e_2 = \varphi (0,1,1)- \varphi (0,0,1)= (1,1,0) - (1,0,0) = (0,1,0) = e_2\\ \varphi e_3 = \varphi (1,0,0)= (0,0,1) = e_1 - e_2$

$M = \begin{bmatrix} 0&0&1\\1&1&-1\\1&0&0\end{bmatrix}$

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  • $\begingroup$ I'm not sure about $\varphi e_2$...maybe it is $\varphi e_2=\varphi (0,1,1)-\varphi(0,0,1)$, but check it please. $\endgroup$ – MattG88 Nov 24 '16 at 0:08
  • $\begingroup$ @MattG88 must be going cross eyed. $\endgroup$ – Doug M Nov 24 '16 at 0:46
  • $\begingroup$ ahahahah ok no problem;-) $\endgroup$ – MattG88 Nov 24 '16 at 1:06

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