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Finding a Basis for a Reflection in R^2

  • Thread starter Punkyc7
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  • #1
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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.
 

Answers and Replies

  • #2
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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
 
  • #3
HallsofIvy
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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.
 
  • #4
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We havent 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?
 

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