Finding a Basis for a Reflection in R^2

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To find a basis in R^2 for the reflection transformation T about the line spanned by [1, 2], the matrix must be diagonal with eigenvalues corresponding to the transformation. The correct reflection matrix is identified as [-5/13, 12/13] for the first column and [12/13, 5/13] for the second. The eigenvectors for this reflection are vectors along the line of reflection, which have an eigenvalue of 1, and vectors perpendicular to that line, which have an eigenvalue of -1. Understanding these eigenvalues and their corresponding eigenvectors is crucial for diagonalizing the matrix. Clarifying the distinction between reflection and projection matrices is essential for accurately solving the problem.
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Find a basis Beta in R^2 such that the beta matrix B of the given linear transformation T is diagonal. The Reflection T about the line R^2 spanned by [1 2], [1 2] is suppose to be verticle.



B=S^-1AS

or

B=[[T(v1)]beta [T(v20]beta]



so i found the reflection matrix to be [4/13 6/13] for the first column and [6/13 4/13] for the second. I'm using e1 and e2 for the v1 and v2. Every time I try solving this I keep getting the same matrix and I don't believe that is right. I think there should be negative somewhere because it is a reflection and I can't figure out what I'm dong wrong. Any help would be greatly appreciated.
 
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I mixed up the reflection matrix with the projection matrix so the reflection matrix should be should be [-5/13 12/13] for the first column and [12/13 5/13] for the second one
 
I think the way you are going about this completely wrong. You are not using the fact that this is a reflection. You are asked to find a basis in which the matrix is diagonal- and so has the eigenvalues of the transformation on its diagonal. Which means, in turn, that the basis must be the eigenvectors. And for a reflection, those eigenvalues and eigenvectors are very simple. Any vector lying on the line of reflection is "transformed" to itself- it is an eigenvector with eigenvalue 1. Any vector perpendicular to that line is reflected to its negative- it is an eigenvector with eigenvalue -1.
 
We haven't come across eigenvalues yet, we have only done transformation. So if I wanted to use eigenvalues or vectors to solve this how would I go about doing it?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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