Given complex vector spaces $V$, and antilinear $T:V \rightarrow V$, then if we fix a basis of $V$, we can represent $T$ by the matrix of the linear $T \circ J$, where $J$ is complex conjugation.

I would really like to know if we can always find a basis of $V$ where this matrix is in some nice form like the JNF.

If $T$ has matrix $M$, and $A$ is a change of basis matrix, then with respect to the new basis $T$ has matrix $AM\overline{A^{-1}}$, so an equivalent question would be: Given a complex square matrix $M$, can we always find an invertible $A$ such that $AM\overline{A^{-1}}$ is in a nice form.

Many thanks in advance, please let me know if anything is unclear.

  • $\begingroup$ Right that was me being stupid, I'll edit the question! $\endgroup$ – James Jul 20 at 22:17
  • $\begingroup$ A few further comments - when you say "fix a basis for $V$", do you mean a real basis or a complex one? If you mean a complex basis, then you can't really represent $T$ with respect to such a basis as $T$ is not complex linear -- maybe you can do that but you need to specify what you want. If you mean a real basis, then you can always forget the fact that $T$ is antilinear and use the real jordan canonical form. If $V = \mathbb{C}^n$ then you have a "canonical conjugation" $\sigma$ $\endgroup$ – levap Jul 20 at 22:23
  • $\begingroup$ (don't use $J$ for compex conjugation, $J$ is used for the complex structure -- multiplication by $i$) and you can consider $T \circ \sigma$ which is $\mathbb{C}$-linear and represent $T \circ \sigma$ with respect to a complex basis in Jordan form. $\endgroup$ – levap Jul 20 at 22:23

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