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Show that eigen values of a matrix remain invariant under rotation.

HINT:Consider $QAQ^T$ where $Q^TQ=I$

I am unable to answer this question.I don't know what is meant by rotation of a matrix.

On using the hint I have shown that if $\lambda $ is an eigen value of $A$ corresponding to $v$ then $\lambda $ is an eigen vector of $QAQ^T$ corresponding to $Qv$.

But I don't understand how that solves the problem of rotation of matrices or what is meant by that. Neither do I understand why should I find eigen values of $Q^TAQ$.

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It is not the rotation of a matrix. It is the rotations of the 3-dimensional space; and hence the corresponding changes in co-ordinates are given by a matrix $Q$ satisfying $Q^TQ=I$. So the original matrix $A$ gets transformed to $QAQ^{-1}$. This matrix has the same eigenvalues as $A$ (in fact they have the same characteristic polynomials). However eigenvectors would be most of the times different.

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  • $\begingroup$ Why does it get transferred to $QAQ^T$ ?Can you please explain $\endgroup$ – Learnmore Sep 5 '17 at 10:23
  • $\begingroup$ Matrices corresponding to rotations are orthogonal, that means their transposes are their inverses. IN 2-dimensional case the rotation of the axes by angle $\theta$ means $(1,0)$ goes to $(\cos\theta, \sin\theta)$ and $(0,1)$ goes to $(-\sin\theta, \cos\theta)$. A $2\times2$ matrix with $(\cos\theta, \sin\theta)$ and $(-\sin\theta, \cos\theta)$ as its rows (or columns) can be readily checked to be orthogonal. $\endgroup$ – P Vanchinathan Sep 5 '17 at 10:50
  • $\begingroup$ No my problem is not that ;I am asking how to show that the resultant matrix will be $QAQ^T$ $\endgroup$ – Learnmore Sep 5 '17 at 10:54
  • $\begingroup$ If $(v_1,v_2,\ldots,,v_n)$ is a new basis (as opposed to the standard basis consisting of columns of identity matrix) and if $x_1,x_2,\ldots, x_n$ are the co-ordinates of a vector in the new basis then its co-ordinates with respect to the standard basis is the result of multiplication of the $n\times n$ matrix consisting of new basis vectors (as columns) with the column vector $(x_1,x_2,\ldots,x_n)$. From this one can derive the matrix of a linear transformation after a change of basis as $QAQ^{-1}$ which becomes $QAQ^T$ in the orthogonal case. $\endgroup$ – P Vanchinathan Sep 5 '17 at 11:03

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