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Suppose $R=\{A\in GL(n,\mathbb C):A^{-1}=\bar A\}$, where $\bar A$ is the conjugate matrix of $A$. Show that $$b:GL(n,\mathbb C)\to R,\quad A\mapsto A\bar{A}^{-1}$$ is onto.

I don't know where to start. Is it possible to construct directly for any $M\in R$ a matrix $A$ such that $b(A)=M$?

(The problem comes from a description of totally real subspaces of $\mathbb C^n$.)

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  • $\begingroup$ It is notable that for such an $M$, we have $|\det(M)|=1$ and $b(M) = M^2$. $\endgroup$ Jun 16, 2016 at 14:38

2 Answers 2

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cf. https://mathoverflow.net/questions/221340/parameterize-unitary-without-transpose/222121#222121

It is shown that, if $A\in R$, then there is a real matrix $T$ s.t. $A=e^{iT}$. Thus $b(e^{iT/2})=e^{iT/2}e^{iT/2}=A$.

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    $\begingroup$ Where can I find a prove of the fact $U^2+V^2=I$, $UV=VU$ $\implies$ $U=\cos(A)$, $V=\sin(A)$ for some real matrix $A$? $\endgroup$ Jun 17, 2016 at 1:03
  • $\begingroup$ See this file. artofproblemsolving.com/community/q1h352326p1942064 $\endgroup$
    – user91684
    Jun 17, 2016 at 9:41
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Here is an elementary proof. It suffices to show that:

If $C$ is a complex square matrix such that $C\bar{C}=I$, then $C=A\bar{A}^{-1}$ for some matrix $A$.

As all eigenvalues of $C\bar{C}$ are nonnegative, $C$ admits a Youla normal form $C=U^TLU$, where $U$ is unitary and $L$ is triangular (cf. corollary 3 of D.C. Youla, A normal form for a matrix under the unitary congruence group, Canad. J. Math. 13, 1961, pp.694-704).

So, from $C\bar{C}=I$, we get $L\bar{L}=I$. Yet, $L$ is triangular. Hence all its eigenvalues have unit moduli. Therefore, by absorbing an appropriate unit diagonal matrix into $U$, we may further assume that all eigenvalues of $L$ are equal to $1$. It follows that we may construct a matrix square root of $L$ using a real Hermite interpolating polynomial. In other words, we may pick $L^{1/2}=p(L)$ for some polynomial $p$ with real coefficients.

Therefore, $\overline{L^{1/2}} = \overline{p(L)} = p(\bar{L}) = p(L^{-1})$. Since $L^{-1}$ is a polynomial in $L$ and in turn a polynomial in $L^{1/2}$, we see that $\overline{L^{1/2}}$ commutes with $L^{1/2}$. So, from $L\bar{L}=I$, we get $\left(L^{1/2}\overline{L^{1/2}}\right)^2=I$. Yet, $L^{1/2}\overline{L^{1/2}}$ is a triangular matrix with all diagonal entries equal to $1$. So we must have $L^{1/2}\overline{L^{1/2}}=I$ and in turn $\overline{L^{1/2}}^{-1}=L^{1/2}$. Now define $A=U^TL^{1/2}$. Then $$ A\bar{A}^{-1}=U^TL^{1/2}\overline{L^{1/2}}^{-1}\overline{U^T}^{-1}=U^TLU=C. $$

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  • $\begingroup$ What defines elementary? $\endgroup$
    – snulty
    Jun 16, 2016 at 18:16
  • $\begingroup$ @snulty Surely, different people draw the lines differently. I actually haven't any definition of "elementary", but I would definitely consider a proof elementary if all techniques involved are at undergraduate level. And this is the case here. Although Youla form is usually not taught in undergraduate (or even postgraduate) courses, the proof of its existence involves only first-year linear algebra. $\endgroup$
    – user1551
    Jun 16, 2016 at 18:36
  • $\begingroup$ That's fair enough, I was just curious about it. $\endgroup$
    – snulty
    Jun 16, 2016 at 18:42

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