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The question that I am trying to solve is as follows:

Find the matrix $A$ of the linear transformation $T(M)= \begin{bmatrix} 7 & 3 \\ 0 & 1 \end{bmatrix} M$ from $U^{2×2}$ to $U^{2×2}$ (upper triangular matrices) with respect to the basis $\left\{ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix},\begin{bmatrix}1&1\\ 0&0\end{bmatrix},\begin{bmatrix}0&0\\0&1\end{bmatrix}\right\}$

$A=\begin{bmatrix} \cdot&\cdot&\cdot\\\cdot&\cdot&\cdot\\ \cdot&\cdot&\cdot\end{bmatrix}$

I don't know how to interpret this question. I would have thought that the matrix $A$ is $\begin{bmatrix} 7 & 3 \\ 0 & 1 \end{bmatrix}$ because that is the matrix which defines the linear transformation $T$. I am surprised that the required answer is $A\in M_{3\times 3}$. How can a basis with elements $\in M_{2\times 2}$ form a matrix $\in M_{3\times 3}$?

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  • $\begingroup$ Upper triangular matrices are a three dimensional vector space. But I don't understand your question. Usually $U$ indicates unitary matrices. $\endgroup$ – Emilio Novati Apr 11 '15 at 9:49
  • $\begingroup$ Has three dimensions got anything to do with $A\in M_{3\times 3}$? $\endgroup$ – ahorn Apr 11 '15 at 11:21
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Upper triangular matrices

$ \begin{bmatrix} a&b\\ 0&c \end{bmatrix} $

form a vector space with canonical basis:

$ e_1=\begin{bmatrix} 1&0\\ 0&0 \end{bmatrix} \quad e_2=\begin{bmatrix} 0&1\\ 0&0 \end{bmatrix} \quad e_3=\begin{bmatrix} 0&0\\ 0&1 \end{bmatrix} $

that is isomorphic to $\mathbb{R}^3$ by:

$ e_1\rightarrow\vec e_1=\begin{bmatrix} 1\\0\\0 \end{bmatrix} \quad e_2\rightarrow\vec e_2=\begin{bmatrix} 0\\1\\0 \end{bmatrix} \quad e_3\rightarrow \vec e_3=\begin{bmatrix} 0\\0\\1 \end{bmatrix} $

Your linear transformation $T$ is defined as a matrix multiplication:

$ T(M)=\begin{bmatrix} 7&3\\ 0&1 \end{bmatrix} \begin{bmatrix} a&b\\ 0&c \end{bmatrix} = \begin{bmatrix} 7a&7b+3c\\ 0&c \end{bmatrix} $ so, in this canonical representation, it is given by the matrix $T_e$ such that:

$ T_e\vec M=\begin{bmatrix} 7&0&0\\ 0&7&3\\ 0&0&1 \end{bmatrix} \begin{bmatrix} a\\ b\\ c \end{bmatrix}= \begin{bmatrix} 7a\\ 7b+3c\\ c \end{bmatrix} $

Now you want a representation of the same transformation in a new basis:

$ e'_1= \begin{bmatrix} 1&0\\ 0&0 \end{bmatrix} \rightarrow\vec e'_1=\begin{bmatrix} 1\\0\\0 \end{bmatrix} \quad e'_2=\begin{bmatrix} 1&1\\ 0&0 \end{bmatrix}\rightarrow\vec e'_2=\begin{bmatrix} 1\\1\\0 \end{bmatrix} \quad e'_3=\begin{bmatrix} 0&0\\ 0&1 \end{bmatrix}\rightarrow \vec e'_3=\begin{bmatrix} 0\\0\\1 \end{bmatrix} $

This transformation of basis is represented by the matrices

$ S= \begin{bmatrix} 1&1&0\\ 0&1&0\\ 0&0&1 \end{bmatrix} \qquad S^{-1}= \begin{bmatrix} 1&-1&0\\ 0&1&0\\ 0&0&1 \end{bmatrix} $

So the matrix that represents the transformation $T$ in the new basis is:

$ T_{e'}=S^{-1}T_eS= \begin{bmatrix} 7&0&-3\\ 0&7&3\\ 0&0&1 \end{bmatrix} $

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  • $\begingroup$ When you said "This transformation of basis is represented by the matricies", did you simply put the column vectors $\vec e'_1$, $\vec e'_2$ and $\vec e'_3$ next to each other to get $S$? Or was there a more nuanced way of getting $S$, such as by saying that $\vec e_j= (a_j+b_j+c_j)\vec e'_j$ where $j$ is the number of the column? $\endgroup$ – ahorn Apr 11 '15 at 15:49
  • $\begingroup$ Symply I put the columns ( it's a general rule that you can easely verify by linearity). $\endgroup$ – Emilio Novati Apr 11 '15 at 16:31
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Let $(X,Y,Z)$ be the basis of $U$, compute $T(X)$, $T(Y)$, $T(Z)$ and express them as a linear combination of $X$, $Y$ and $Z$.

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  • $\begingroup$ T(X)=7X; T(Y)=7Y; T(Z)=Z. How does that relate to A? $\endgroup$ – ahorn Apr 11 '15 at 11:20
  • $\begingroup$ A={(7,0,0),(0,7,0),(0,0,1)} take the coefficients as elements of the matrix A. $\endgroup$ – 2ndYearFreshman Apr 11 '15 at 11:39
  • $\begingroup$ I have tried to input all sorts of combinations, such as $\begin{bmatrix} 7 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 1 \end{bmatrix}, \begin{bmatrix} 7 & 0 & 0 \\ 7 & 7 & 0 \\ 0 & 0 & 1 \end{bmatrix}, \text{and} \begin{bmatrix} 7 & 0 & 0 \\ 7 & 3 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ but these are still incorrect (I am answering an online test). Btw, welcome to math.stackexchange! What brought you to the site? $\endgroup$ – ahorn Apr 11 '15 at 14:25
  • $\begingroup$ Try A={(7,0,0),(0,7,0),(-3,3,1)} Thank you,hopefully my answer will help.Your question brought me here. $\endgroup$ – 2ndYearFreshman Apr 11 '15 at 14:34

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