Answer to Question #1452 in Linear Algebra for AA
Find an invertible 2x2 matrix Q such that Q^-1AQ is equal to a rotation followed by scaling by a positive real number, and write down both the rotation matrix and the scaling matrix. What is the angle of rotation of the matrix?
cos(theta) -sin(theta) = sx 0
sin(theta) cos(theta) = 0 sy
Q= 1 0
Scalling matrix is the same as well as rotation matrix. As cos(theta) = 1, rotation angle is equal to zero.
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