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An example question asks me to determine $[T]_{\beta}^\gamma$ where $\beta,\ \gamma$ are standard ordered bases of $\mathbb{R}^n$ and $\mathbb{R}^m$ respectively, of $T_1: \mathbb{R}^n \rightarrow \mathbb{R}^n,\ \ T_1(a_1, a_2, ...a_n) = (a_1, a_1, a_1,....., a_1)$ and also of $T_2: \mathbb{R}^n \rightarrow \mathbb{R}^n,\ \ T_2(a_1, a_2, ...a_n) = (a_n, a_{n-1}, a_{n-2},....., a_1)$

I understand that in $T_1$, $T$ needs to be an $n\times n$ matrix consisting of ones in the first column and nowhere else. In the second one, I can deduce that the ones must be on the negative diagonal.

My problem is, I don't know how I can formally present that as a solution to $[T]_{\beta}^\gamma$

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    All of them. Sorry, I'll fix that.2012-10-11

1 Answers 1

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$T_1(1,0,...,0):=(1,1,...,1)=1\cdot(1,0,...,0)+1\cdot(0,1,0,...,0)+...+1\cdot(0,0,.,,,.1)$ $T_1(0,1,...,0):=(0,0,...0)=0\cdot (1,0,...,0)+\cdot(0,1,...,)+...+0\cdot (0,0,...,1)\\.................$

$T_1(0,0,...,1):=(0,0,...,0)=0\cdot (1,0,...,0)+...+0\cdot(0,0,...,1)$

Thus, being $\,e:=\,$ the standard basis of $\,\Bbb R^n\,$

$[T_1]_e^e=[T_1]_e=\begin{pmatrix} 1&0&...&0\\1&0&...&0\\...&...&...&...\\1&0&...&0\end{pmatrix}$

Pay attention to the fact that the matrix above is the transpose of the coefficient matrix in the first part.

Now you try with $\,T_2\,$ following the same model.

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    Yes: you have to apply $\,T\,$ to each and every element of basis $\,\beta\,$ and then you must write the outcome as a linear combination of the basis $\,\gamma\,$ . In this case the situation is pretty comfortable as $\,\beta=\gamma=e=\,$the standard basis.2012-10-12