Find the 2x2 matrix A of the orthogonal projection onto the line L in R^2 that consists of all scalar multiples of the vector
(5, 1)
Remember that in general, given a linear space $\displaystyle V$ and a subspace $\displaystyle S$ with $\displaystyle
\dim \left( S \right) =n< + \infty
$ the orthogonal projection of a vector $\displaystyle \bold{v}$ over $\displaystyle S$ is given by $\displaystyle
P_S \left( \bold{v} \right) = \sum\limits_{k = 1}^n {\left\langle {\bold{v},\bold{w}_k } \right\rangle \cdot \bold{w}_k }
$ where $\displaystyle
\left\{ {\bold{w}_1 ,...,\bold{w}_n } \right\}
$ is an orthonormal base of $\displaystyle S$ ( Such a base exists by Gram-Schmidt)
So in our case let $\displaystyle
\bold{w} = \tfrac{1}
{{\sqrt {26} }} \cdot \left( {\begin{array}{*{20}c}
5 & 1 \\
\end{array} } \right)
$ this vector belongs to the line L and its norm is 1, thus ( since the dimension of the line is 1) it forms an orthonormal base of L.
Thus: $\displaystyle
P_L \left( \bold{u} \right) = \left\langle {\bold{u},\bold{w}} \right\rangle \cdot \bold{w}
$ from there: $\displaystyle
P_L \left( {\begin{array}{*{20}c}
x & y \\
\end{array} } \right) = \tfrac{1}
{{26}} \cdot \left( {\begin{array}{*{20}c}
{25x + 5y} & {5x + y} \\
\end{array} } \right)
$ and note that this can be re-written as (with column vectors): $\displaystyle
\left( {\begin{array}{*{20}c}
{\tfrac{{25}}
{{26}}} & {\tfrac{5}
{{26}}} \\
{\tfrac{5}
{{26}}} & {\tfrac{1}
{{26}}} \\
\end{array} } \right) \cdot \left( {\begin{array}{*{20}c}
x \\
y \\
\end{array} } \right) = P_L \left( {\begin{array}{*{20}c}
x \\
y \\
\end{array} } \right)
$