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Suppose we have such a linear transformation where $ n \in \mathbb{N} $ and $ \mathcal{V}(F) $ is a vector space over a field $ F $. $$ T : \mathcal{V}(\mathbb{R}) \to \mathcal{V}(\mathbb{R}) $$

Now assume that $ T $ is not diagonalizable. Can there be a case where the complexification of $ T $ given by $ T_{\mathbb{C}} $ defined as follows is diagonalizable? $$ T_{\mathbb{C}} : \mathcal{V}(\mathbb{C}) \to \mathcal{V}(\mathbb{C}) $$

Furthermore, does the choice of field affect the diagonalizability at all given our assumption carries over to this point?

I thought of this for some time intuitively, and for me, it does not make a lot of sense that the field would determine the diagonalizability. My thought process was that if $ T $ were not diagonalizable, then there would exist at least one eigenvalue $ \lambda $ with multiplicity $ m $ such that the dimension of its eigenspace $ \dim E_\lambda < m $.

At this point, I believe that if the eigenvalues were all real, to begin with, the choice of the field would not matter as it would not affect the eigenspace at all and thus its dimension would remain the same, preserving its non-diagonalizability. However, if the eigenvalues were not real would this change? I do not know where to proceed from here.

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    $\begingroup$ A rotation of the plane by 90 degrees is not diagonalisable, but becomes diagonalisable over the complex numbers, with eigenvalues $\pm i$. $\endgroup$ – Joppy Apr 8 '18 at 21:31
  • $\begingroup$ @Joppy, but is i an eigenvalue of the Complexification of rotation by 90 degrees? $\endgroup$ – gary Jun 23 '18 at 23:24

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