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 $\displaystyle V$ 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): $\displaystyle \left( {\begin{array}{*{20}c} {\tfrac{{25}} {{26}}} & {\tfrac{5} {{26}}} \\ {\tfrac{5} {{26}}} & {\tfrac{1} {{26}}} \\ \end{array} } \right) \cdot \left( {\begin{array}{*{20}c} x \\ y \\ \end{array} } \right) = P_L \left( {\begin{array}{*{20}c} x \\ y \\ \end{array} } \right)$