# Matrix representation of transformation in ordered bases

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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Does $T_1$ have $a_1$ in all $n$ entries, or all but the last? – copper.hat Oct 11 '12 at 17:05
All of them. Sorry, I'll fix that. – CodyBugstein Oct 11 '12 at 17:07

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

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

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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Following your lead, I test out all the elements of the standard ordered basis for $R^2$, i.e. $T_1(1,0,...,0):=(0,0,...,1)=0\cdot(1,0,...,0)+0\cdot(0,1,0,...,0)+...+1\cdot(0,‌​0,.,,,.1)$ - Am I going the right direction? – CodyBugstein Oct 11 '12 at 18:17
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. – DonAntonio Oct 12 '12 at 3:17