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

Contributor

David Cherney, Tom Denton, and Andrew Waldron (UC Davis)