Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 21/Solution 2

As in the (third) hint, let $\pi _{1}$ denote the projection onto $L_{1}$ and let $\pi _{2}$ denote the projection onto $L_{2}$. If $x$ is a point on the plane, then (by the geometric interpretation of projection) $\pi _{1}(x)$ is the point in $L_{1}$ such that the line containing $x$ and $\pi _{1}(x)$ is perpendicular to $L_{1}$. It follows that the range of $\pi _{1}$ is $L_{1}$.

Now, let $x$ be a point on the plane like before. To determine $(\pi _{2}\circ \pi _{1})(x)$ (recall that $(\pi _{2}\circ \pi _{1})(x)=\pi _{2}(\pi _{1}(x))$), note the following. $x$ is mapped to $\pi _{1}(x)$, which is in $L_{1}$. Moreover, $\pi _{1}(x)$ is mapped (by the geometric interpretation of projection) to the point $\pi _{2}(\pi _{1}(x))$ in $L_{2}$ such that the line containing $\pi _{1}(x)$ and $\pi _{2}(\pi _{1}(x))$ is perpendicular to $L_{2}$. But $L_{1}$ is perpendicular to $L_{2}$, and $\pi _{1}(x)$ is contained in $L_{1}$, so $\pi _{2}(\pi _{1}(x))$ is contained in $L_{1}$ and in $L_{2}$. As the only point contained in $L_{1}$ and in $L_{2}$ is the origin, the range of $\pi _{2}\circ \pi _{1}$ is the origin.

And the only matrix that maps the entire plane to the origin is the following matrix, which is our answer

$\color {blue}{\begin{pmatrix}0&0\\0&0\end{pmatrix}}$