matrix similarity

matrices \(A\) and \(B\) are similar if there exists an invertible \(P\) such that \(A = P^{-1}BP\)

Say that we have some matrix that represents a rotation. If the axis of rotation was aligned with, e.g., the z-axis, then the transformation would be easy to write: \[S = \begin{bmatrix} \cos \theta & -\sin \theta & 0\\ \sin \theta & \cos \theta & 0\\ 0 & 0 & 1\\ \end{bmatrix}\]

If we were in a basis where the axis of rotation was aligned with the z-axis, then we could have: \[ y' = Sx' \] where \(y'\) and \(x'\) are vectors in such a basis.

Let's say we are trying to transform \(x\) into \(y\) with \(T\). So let's use a change of basis matrix \(P\) to move \(x\) into the new basis and then move \(P^{-1}\) to move \(y'\) to our original basis. So we have: \[ y = (P^{-1}SP)x \]

If \(A\) is similar to a diagonal matrix \(D\), then \(A\) is diagonalizable

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