MATH221 April 2013
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Question 07 (a)
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Let be the linear transformation that reflects points through the line 3x = 4y.
a) Find the eigenvalues and eigenvectors of the standard matrix A of T. (Note: In order to do this, you do not need to evaluate A.)
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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?
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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.
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Hint 1
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Draw a picture which illustrates what the reflection does to a few example vectors.
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Hint 2
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- What does the reflection do to a vector that is on the line ?
- What does the reflection do to a vector that is orthogonal to the line ?
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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.
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Solution
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Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies.
- For any non-trivial reflection, the eigenvalues are always 1 and -1.
- For the eigenvalue 1, we can pick any vector on the line itself. Such a vector will not be moved by the linear transformation. So, for us, we pick as an eigenvector to the eigenvalue 1. Note that (x,y) = (4,3) is a point on the line 3x=4y since 3(4)=4(3).
- The eigenvalue of -1 comes from picking a vector in the line that is perpendicular (i.e. orthogonal) to the line . Any such vector will simply change its sign, thus will correspond to the eigenvalue -1. We choose as an eigenvector to the eigenvalue -1.
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