# Show that mapping {y}mapsto{r}{e}{f}{l}_{{L}}{y} is a linear transformation.

Show that mapping $y↦ref{l}_{L}y$ is a linear transformation.
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Given information:
$u\ne 0\in {R}^{n}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}L=Span\left\{u\right\}.$
For y in
and hath - y.
Calculation:
Consider $T\left(y\right)=ref{l}_{L}y.$
Substitute
$T\left(y\right)=2\cdot pro{j}_{L}y-y\left(1\right)$
Apply Theorem 1 (b) as shown below.
Let u, v, and w be vectors in ${R}^{n}$, and let c be a scalar. Then
$\left(u+v\right)\cdot w=u\cdot w+v\cdot w$
For any vectors y and z in ${R}^{n}$, and any scalars c and d, the properties of inner product as shown below.
Substitute for y in Equation (1).
$T\left(cy+dz\right)=2\cdot pro{j}_{L}\left(cy+dz\right)-\left(cy+dz\right)$
$=2\left(cpro{j}_{L}y+dpro{j}_{L}z\right)-\left(cy+dz\right)$
$=2cpro{j}_{L}y-cy+2dpro{j}_{L}z-dz$
$=c\left(2pro{j}_{L}y-y\right)+d\left(2pro{j}_{L}z-z\right)$

Therefore, the mapping $y↦ref{l}_{L}y$ y is a linear transformation.