1. ## Orthogonal Projections

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)

2. Remember that in general, given a linear space $V$ and a subspace $S$ with $
\dim \left( S \right) =n< + \infty
$
the orthogonal projection of a vector $\bold{v}$ over $S$ is given by $
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 $
\left\{ {\bold{w}_1 ,...,\bold{w}_n } \right\}
$
is an orthonormal base of $S$ ( Such a base exists by Gram-Schmidt)

So in our case let $
\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: $
P_L \left( \bold{u} \right) = \left\langle {\bold{u},\bold{w}} \right\rangle \cdot \bold{w}
$
from there: $
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): $
\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)
$