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Consider an antilinear involution, that is an antilinear map on a complex vector space, whose matrix $M$ obeys $MM^*=1$ where the star denotes complex conjugation. Can we find a change of basis whose matrix $\Lambda$ would be such that $\Lambda^* M \Lambda^{-1} = 1$?

By taking the real components of $M$ and $\Lambda$, this can be reduced to a special case of the following problem: given four real commuting square matrices of the same size $A,B,C,D$ such that $AD=BC$, do real vectors $X$ and $Y$ such that $AX=BY$ and $CX=DY$ span the whole space?

The motivation for this question comes from quantum mechanics, where Hermitian conjugation is a antilinear involution on the space of operators. Trivializing this involution means finding a Hermitian basis of operators.

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  • $\begingroup$ You may be interested in Theorem 3.1 from A Canonical Form for Matrices Under Consimilarity by Hong and Horn, which gives canonical matrix representations of antilinear operators on finite dimensional spaces. A corollary, for nonsingular $M$, $MM^*$ and $M^\prime(M^\prime)^*$ are similar iff there exists $\Lambda$ s.t. $M^\prime=\Lambda^*M\Lambda^{-1}$, where $*$ denotes element-wise conjugation. Also Normal Form of Antiunitary Operators by Wigner gives a canonical form that provides a direct answer to your question because antilinear involutions are antiunitary. $\endgroup$
    – David Sykes
    Commented Nov 7, 2019 at 23:59

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If you let $I$ denote multiplication by $\sqrt{-1}$, then the two operators $I$ and $M$ on your vector space (say, $V$) satisfy $$ I^2 = -1,\qquad M^2 = 1,\qquad\text{and}\qquad IM=-MI. $$ (The former since $M$ is an involution; the latter follows since $M$ is anti-linear.) The operators $1,I,M, IM$ span an algebra isomorphic to $M_2(\mathbb{R})$, the ring of $2$-by-$2$ matrices with real entries, and thus, $V$ is a left module over $M_2(\mathbb{R})$.

It is well-known that any finite dimensional left module over $M_2(\mathbb{R})$ is a direct sum of a finite number of copies of the standard $2$-dimensional module $\mathbb{R}^2$ (with its natural action). Thus, your $V$ has a basis that puts it in standard form as a direct sum of copies of $\mathbb{R}^2$.

I believe this basis provides the matrix $\Lambda$ that you seek.

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  • $\begingroup$ One way to see that left modules over $M_2(\mathbb{R})$ are sums of copies of $\mathbb{R}^2$ is that they are sums of copies of $M_2(\mathbb{R})$ itself, and $M_2(\mathbb{R})$ is a sum of two $\mathbb{R}^2$s, right? $\endgroup$
    – Sylvain Ribault
    Commented Oct 8, 2014 at 9:40
  • $\begingroup$ I'm not sure what you mean: $M_2(\mathbb{R})$ has dimension $4>2$, so, no, not every $M_2(\mathbb{R})$ module is a sum of copies of $M_2(\mathbb{R})$. $\endgroup$
    – Robert Bryant
    Commented Oct 8, 2014 at 9:53
  • $\begingroup$ Right, let me correct myself: left modules are sums of cosets of $M_2(\mathbb{R})$, where $M_2(\mathbb{R})$ is considered as a $M_2(\mathbb{R})$ module. The idea is to consider submodules of the type $M_2(\mathbb{R})v$ for $v\in V$. $\endgroup$
    – Sylvain Ribault
    Commented Oct 8, 2014 at 10:17
  • $\begingroup$ There might be an elementary reformulation of the proof: given any basis vector $v$, either $v$ and $Mv$ are proportional in which case we only need to rescale $v$, or they are linearly independent in which case we can replace them with $v+Mv$ and $i(v-Mv)$. $\endgroup$
    – Sylvain Ribault
    Commented Oct 8, 2014 at 15:58
  • $\begingroup$ Yes, the space is the sum of the kernel of $1-M$ and the kernel of $1+M$, and multiplication by $I$ exchanges these two (so they have equal dimensions), so if you choose a basis of the kernel of $1-M$, you can figure everything out from that. I should have pointed that out; you are right that this is a more elementary thing than the argument I was suggesting. (However, the way the classification of modules works in this case, you are essentially reproving the classification in this special case.) $\endgroup$
    – Robert Bryant
    Commented Oct 8, 2014 at 16:04
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An elementary way to solve your problem is to remark that $u : X \mapsto M\overline{X}$, seen as an endomorphism of the real vector space $\mathbb{C}^n$, is involutory and that the automorphism $X \mapsto iX$ swaps its eigenspaces because it skew-commutes with $u$.
From there, any basis of the real vector space $\mathrm{Ker} (u-\text{id})$ is actually a basis of the complex vector space $\mathbb{C}^n$ that suits your needs.

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