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

MATH152 April 2016

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Question A 21
Consider the two perpendicular lines through the origin given below: ${\begin{aligned}L_{1}:&&x+2y=0\\L_{2}:&&x-y/2=0\end{aligned}}$ Find the matrix for the composition of linear transformations: projection onto $L_{1}$ followed by projection onto $L_{2}$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Note that the lines are orthogonal. (graphing will help.)

Hint 2
The matrix of the linear transformation $P_{L}$ that projects $\mathbb {R} ^{2}$ on the straight line $\ell :y=mx$ has the form: ${\begin{pmatrix}{\frac {1}{1+m^{2}}}&{\frac {m}{1+m^{2}}}\\\quad \\{\frac {m}{1+m^{2}}}&{\frac {m^{2}}{1+m^{2}}}\end{pmatrix}}$.

Hint 3
Let $\pi _{1}$ denote the projection onto $L_{1}$ and let $\pi _{2}$ denote the projection onto $L_{2}$. Use the geometric interpretation of projection onto a line to determine the range of $\pi _{1}$. Then, use the same argument, and the fact that $L_{1}$ and $L_{2}$ are perpendicular, to determine the range of $\pi _{2}\circ \pi _{1}$.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Give the formula in the hint, first we write the lines in the standard form $y=mx$. $L_{1}:y=-{\frac {1}{2}}x\ \Longrightarrow m=-{\frac {1}{2}}$, $L_{2}:y=2x\ \Longrightarrow m=2$. The matrix of projection onto $L_{1}$ is $P_{1}={\begin{pmatrix}{\frac {4}{5}}&-{\frac {2}{5}}\\-{\frac {2}{5}}&{\frac {1}{5}}\end{pmatrix}}$, and the matrix of projection onto $L_{2}$ is $P_{2}={\begin{pmatrix}{\frac {1}{5}}&{\frac {2}{5}}\\\ \\{\frac {2}{5}}&{\frac {4}{5}}\end{pmatrix}}$ Thus projecting any point in 2D plane first on $L_{1}$ and then on $L_{2}$ can be represented by $\mathbb {R} ^{2}\ni {\begin{pmatrix}x\\y\end{pmatrix}}\rightarrow P_{2}{\Big (}P_{1}{\begin{pmatrix}x\\y\end{pmatrix}}{\Big )}$ so the transformation matrix is $T=P_{2}\circ P_{1}$ $P_{2}\circ P_{1}={\begin{pmatrix}{\frac {1}{5}}&{\frac {2}{5}}\\\ \\{\frac {2}{5}}&{\frac {4}{5}}\end{pmatrix}}{\begin{pmatrix}{\frac {4}{5}}&-{\frac {2}{5}}\\-{\frac {2}{5}}&{\frac {1}{5}}\end{pmatrix}}=0$ In fact, since these two lines are orthogonal, projecting a vector (point) first on $L_{1}$ and then orthogonally projecting the new point on $L_{2}$ will return us to $0$.

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 $\color {blue}{\begin{pmatrix}0&0\\0&0\end{pmatrix}}$

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