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

MATH152 April 2010

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Question A 17
The matrix $P=\left[{\begin{array}{cc}4/5&-2/5\\-2/5&1/5\end{array}}\right]$ represents projection onto a line in 2D. Write this line in the form y = mx with m determined.

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
Recall that a transformation matrix applies to all vectors including the basis vector.

Hint 2
For an alternative solution: What does the projection do to a vector that is on the line we are looking for? What do we call vectors with this property?

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. All points on the line y=mx are given by [x0,mx0]=x0[1,m]. If we have a general point in space [x,y] then the projection matrix P should map the point [x,y] to x0[1,m]. Since this projection must hold for all points, it certainly must hold for the basis vectors. Therefore we try the vector [1,0]. Since we have, $P={\begin{bmatrix}4/5&-2/5\\-2/5&1/5\end{bmatrix}}$ then $P{\begin{bmatrix}1\\0\end{bmatrix}}={\begin{bmatrix}4/5&-2/5\\-2/5&1/5\end{bmatrix}}{\begin{bmatrix}1\\0\end{bmatrix}}={\begin{bmatrix}4/5\\-2/5\end{bmatrix}}$ which of course is just the first column of the transformation matrix (recall this is exactly how we form transformation matrices). We can write, ${\begin{bmatrix}4/5\\-2/5\end{bmatrix}}=4/5{\begin{bmatrix}1\\-1/2\end{bmatrix}}$ which in the form x0[1,m] tells us that x0=4/5 and m=-1/2. Therefore the slope of the line is -1/2. We can then write that the line is y=-1/2x. To check our answer we could use the second basis vector [0,1] which would produce the second column of the matrix and once again we'd conclude the slope is -1/2.

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. 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 $P{\begin{bmatrix}x_{0}\\y_{0}\end{bmatrix}}={\begin{bmatrix}x_{0}\\y_{0}\end{bmatrix}}$ and hence (x0,y0) is an eigenvector with eigenvalue 1. To find this eigenvector, find the nullspace of $P-I={\begin{bmatrix}-1/5&-2/5\\-2/5&-4/5\end{bmatrix}}$ Row reducing the matrix above yields ${\begin{bmatrix}1&2\\0&0\end{bmatrix}}$ 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 $m=-{\frac {1}{2}}$

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