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I have two matrices, $A$ and $B$. I was (perhaps naively) expecting them to be more-or-less similar ("more-or-less" because this is in a numerical setting), but instead of having exactly the same eigenvalues, two of the eigenvalues are the negation of the corresponding ones in the other matrix.

$\operatorname{eig}(A) = \{ \cos(\theta) + i*\sin(\theta), \cos(\theta) - i*\sin(\theta), 1, 1\} $

$\operatorname{eig}(B) = \{ -\cos(\theta) - i*\sin(\theta), -cos(\theta) + i*\sin(\theta), 1, 1\} $

What does this imply, besides that they are not similar? Can they be made similar, by negating the two "problematic" eigenvalues and the corresponding eigenvectors, or something?

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  • $\begingroup$ OK, that sounds reasonable. I'll give it a shot, thanks! $\endgroup$ – anjruu Oct 16 '14 at 14:14
  • $\begingroup$ I've put an answer instead of the comment, maybe I can find something when A is not diagonalizable $\endgroup$ – mvggz Oct 16 '14 at 14:19
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If the matrices $A,B\in M_n$ are nxn, then yes.

Since those eigenvectors have the same complex modulus, you can rotate matrix B to be in the same plane as A. That rotation matrix is nonsingular, therefore you can combine it with the matrix that diagonalizes $A$ to get a nonsigular matrix that makes it similar to $B$.

For instance, for $A,B$

$$\operatorname{eig}(A) = \{ \cos(\theta) + i*\sin(\theta), \cos(\theta) - i*\sin(\theta), 1, 1\}$$ and $$ \operatorname{eig}(B) = \{ -\cos(\theta) - i*\sin(\theta), -cos(\theta) + i*\sin(\theta), 1, 1\}$$

Both of the matrices have n distinct eigenvalues, so find nonsingular matrices that diagonalize them.

$$ S^{-1}AS = diag(-\cos(\theta) - i*\sin(\theta), -cos(\theta) + i*\sin(\theta), 1, 1)$$

Then just multiply this matrix by $$ I = \begin{vmatrix} -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{vmatrix} $$

Since similarity is transitive, you have that $A$ is similar to $B$.

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if A is diagonalizable, it's possible I think.

You define the opposite of the projection on the subspace formed by your two eigenvalues with the cos and sin, and you change the base to be in the diagonalisation base:

$E = P*D*P^{−1} , A = P*D1∗P^{−1} , A*E = P*D3∗P^{−1}$ , with the opposite eigenvalues ( the eigenvectors have been changed into their opposite thanks to the opposite of the projection)

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