Science:Math Exam Resources/Courses/MATH152/April 2010/Question A 17/Solution 2

To begin with, since P projects on a line, the vector Pv is on the line for any vector v.

Let (x₀,y₀) be a vector on the line that P projects onto. Projecting a vector that is already on the line onto the line, does not change the vector. In mathematical notation, this means that $P\left[\begin{array}{c}{x}_0\\ y_0\end{array}\right]=\left[\begin{array}{c}{x}_0\\ y_0\end{array}\right]$ and hence (x₀,y₀) is an eigenvector with eigenvalue 1.

To find this eigenvector, find the nullspace of

$P-I=\left[\begin{array}{cc}-1/5& -2/5\\ -2/5& -4/5\end{array}\right]$

Row reducing the matrix above yields

$\left[\begin{array}{cc}1& 2\\ 0& 0\end{array}\right]$

and hence (x₀,y₀) = (-2,1) is a vector on the line that P projects onto. Plugging this into the equation of the line, y = mx yields 1 = m(-2) or $m=-\frac{1}{2}$