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# 6.2: Linear Functions on Hyperplanes

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It is not always so easy to write a linear operator as a matrix. Generally, this will amount to solving a linear systems problem. Examining a linear function whose domain is a hyperplane is instructive.

Example 63

Let $$V=\left\{ c_{1}\begin{pmatrix}1\\1\\0\end{pmatrix} +c_{2}\begin{pmatrix}0\\1\\1\end{pmatrix} \middle| c_{1},c_{2}\in \Re \right\}$$ and consider $$L:V\to \Re^{3}$$ defined by
$$L\begin{pmatrix}1\\1\\0\end{pmatrix} = \begin{pmatrix}0\\1\\0\end{pmatrix},\qquad L\begin{pmatrix}0\\1\\1\end{pmatrix} = \begin{pmatrix}0\\1\\0\end{pmatrix}.$$
By linearity this specifies the action of $$L$$ on any vector from $$V$$ as
$$L\left[ c_{1}\begin{pmatrix}1\\1\\0\end{pmatrix} + c_{2} \begin{pmatrix}0\\1\\1\end{pmatrix} \right]= (c_{1}+c_{2})\begin{pmatrix}0\\1\\0\end{pmatrix}.$$
The domain of $$L$$ is a plane and its range is the line through the origin in the $$x_{2}$$ direction. It is clear how to check that $$L$$ is linear.

It is not clear how to formulate $$L$$ as a matrix;
since
\begin{eqnarray*}
L\begin{pmatrix}c_{1}\\\!\!c_{1}+c_{2}\!\!\\c_{2}\end{pmatrix} =
\begin{pmatrix}
0&0&0\\
1&0&1\\
0&0&0
\end{pmatrix}
\begin{pmatrix}c_{1}\\\!\!c_{1}+c_{2}\!\!\\c_{2}\end{pmatrix} =(c_{1}+c_{2})\begin{pmatrix}0\\1\\0\end{pmatrix}\, ,
\end{eqnarray*}
or since
\begin{eqnarray*}
L\begin{pmatrix}c_{1}\\\!\!c_{1}+c_{2}\!\!\\c_{2}\end{pmatrix} =
\begin{pmatrix}
0&0&0\\
0&1&0\\
0&0&0
\end{pmatrix}
\begin{pmatrix}c_{1}\\\!\!c_{1}+c_{2}\!\!\\c_{2}\end{pmatrix} =(c_{1}+c_{2})\begin{pmatrix}0\\1\\0\end{pmatrix}
\end{eqnarray*}
you might suspect that $$L$$ is equivalent to one of these $$3\times3$$ matrices. It is not. All $$3\times3$$ matrices have $$\Re^{3}$$ as their domain, and the domain of $$L$$ is smaller than that. When we do realize this $$L$$ as a matrix it will be as a $$3\times2$$ matrix. We can tell because the domain of $$L$$ is 2 dimensional and the codomain is $$3$$ dimensional