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.
Yes, as in the comment by @Joppy, consider the matrix $$\begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix} .$$ Its characteristic polynomial is $x^2 + 1$ which has distinct roots $\pm i$ so is not diagonalizable over $\mathbb{R}$ but is diagonalizable over $\mathbb{C}$.
The issue, as you notice, is with non-real eigenvalues; if all the eigenvalues are real then the Jordan form over $\mathbb{C}$ is defined over $\mathbb{R}$ (and is conjugate over $\mathbb{R}$ to the original matrix. This is because the dimension of $E_\lambda = \ker(T - \lambda)$ does not depend on the field as long as $T - \lambda$ is defined over that field.