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I would like to know how to find a lot of isomorphism ( or simply morphisms ) of algebras over $ \mathbb{C} $ : $ \varphi : \mathcal{M}_{3} ( \mathbb{C} ) \to \mathcal{M}_3 ( \mathbb{C} ) $ which respect the following transformations : $ \varphi \Big( \begin{pmatrix} 0 & c & 0 \\ 0 & 0 & a \\ b & 0 & 0 \end{pmatrix} \Big) = \begin{pmatrix} 0 & b & 0 \\ 0 & 0 & c \\ a & 0 & 0 \end{pmatrix} $, and , $ \varphi \Big( \begin{pmatrix} 0 & a & 0 \\ 0 & 0 & b \\ c & 0 & 0 \end{pmatrix} \Big) = \begin{pmatrix} 0 & a & 0 \\ 0 & 0 & b \\ c & 0 & 0 \end{pmatrix} $, and, $ \varphi \Big( \begin{pmatrix} 0 & b & 0 \\ 0 & 0 & c \\ a & 0 & 0 \end{pmatrix} \Big) = \begin{pmatrix} 0 & c & 0 \\ 0 & 0 & a \\ b & 0 & 0 \end{pmatrix} $. $ a , b , c \in \mathbb{C} $ are in principle, three fixed scalars. Thanks a lot for your help.

Edit : We can see, in principle, that : $ \begin{pmatrix} 0 & c & 0 \\ 0 & 0 & a \\ b & 0 & 0 \end{pmatrix} $ and, $ \begin{pmatrix} 0 & a & 0 \\ 0 & 0 & b \\ c & 0 & 0 \end{pmatrix} $, and , $ \begin{pmatrix} 0 & b & 0 \\ 0 & 0 & c \\ a & 0 & 0 \end{pmatrix} $ are linearly independant. Thanks a lot. :-)

Bryan261
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    All those three matrices share the same set of eigenvalues. If it is possible to find a linear transformation $P$ such that it maps the eigenspaces of the second matrix to themselves, and interchanges the eigenspaces of the other two, then conjugating by $P$ should do the trick. – Jyrki Lahtonen May 30 '14 at 22:03
  • Yes, thank you, but i don't know how to do it. can you tell me how to do it please ? thanks a lot. :-) – Bryan261 May 30 '14 at 22:23
  • Brian, you can edit your questions if you want. I am telling you this because you asked the same question here http://math.stackexchange.com/questions/815906/isomorphism-between-algebras-over-c . – Daniel Jun 01 '14 at 11:12

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