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T(x,y)=(2x-y,3x-y)

with respect to basis (1,0), and (0,1) Now M(T)=
$$ \left[ \begin{array}{cc|c} 2&3&x\\ -1&-1&y \end{array} \right] $$

gives \begin{pmatrix} 2x+3y \\ -x-y \\ \end{pmatrix}

so this cannot be the transformation matrix

The correct one is

$$ \left[ \begin{array}{cc|c} 2&-1&x\\ 3&-1&y \end{array} \right] $$

Also, how do I find the matrix transformation of (x,y,z)=(x+y,y+z) for basis (1,0) and (0,1). So we are going from 3 dimensions to 2. So my matrix should have 2 rows and 2 columns right?

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1 Answer 1

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Consider linear map ${\sf T}:{\sf R}^3\rightarrow {\sf R}^2$ defined by ${\sf T}(x,y,z)=(x+y,y+z)$, we always use the two bases for the domain and the co-domain, to obtain a matrix representation. For example, use the standard ordered bases for ${\sf R}^3$ and ${\sf R}^2$, respectively. That is, $$\beta=\{e_1,e_2,e_3\}=\left\{\begin{pmatrix}1\\0\\0\end{pmatrix},\begin{pmatrix}0\\1\\0\end{pmatrix},\begin{pmatrix}0\\0\\1\end{pmatrix}\right\}\quad\mbox{and}\quad \gamma=\{\bar{e}_1,\bar{e}_2\}=\left\{\begin{pmatrix}1\\0\end{pmatrix},\begin{pmatrix}0\\1\end{pmatrix}\right\},$$ and do the following works: \begin{align*} {\sf T}(e_1) &=\begin{pmatrix}1\\0\end{pmatrix}=1\cdot\bar{e}_1+0\cdot\bar{e}_2,\\ {\sf T}(e_2) &=\begin{pmatrix}1\\1\end{pmatrix}=1\cdot\bar{e}_1+1\cdot\bar{e}_2,\\ {\sf T}(e_3) &=\begin{pmatrix}0\\1\end{pmatrix}=0\cdot\bar{e}_1+1\cdot\bar{e}_2. \end{align*} Hence we get the matrix representation of ${\sf T}$ with respect to $\beta$ and $\gamma$ as below: $$[{\sf T}]_\beta^\gamma= \begin{pmatrix}1&1&0\\0&1&1\end{pmatrix}.$$

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  • $\begingroup$ And what about the first one? Did I get that right? $\endgroup$ Commented Apr 12, 2016 at 9:57
  • $\begingroup$ @anna_xox Yes, but I emphasize again that the matrix representation of the given transformation is determined by the two bases. $\endgroup$
    – Solumilkyu
    Commented Apr 12, 2016 at 11:00

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