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

$\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$