For all unitary matrices, i.e. $A \overline{A}^T = I$, there is a skew-Hermitian matrix $X$ so that $A = exp(X)$. So the unitary group has $n^2$ dimensions.

Is there any similar parameterisation of all matrices with $A \overline{A} = I$? What can be said about the number of dimensions? (Btw. is there any name for this type of matrices in literature?)

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    $\begingroup$ Note that If $M$ is a real square matrix of the relevant size, then $A = \exp(iM)$ has the property that $\overline{A} = \exp(-iM)$, so that $A\overline{A} = I$. $\endgroup$ Oct 19, 2015 at 22:56
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    $\begingroup$ @QiaochuYuan: Right, this collection is not closed under multiplication. Does this already exclude any notion of dimensionality and parameterisation? $\endgroup$
    – Jiro
    Oct 20, 2015 at 8:38
  • $\begingroup$ @GeoffRobinson: Right, but are these all matrices with this property? $\endgroup$
    – Jiro
    Oct 20, 2015 at 9:05
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    $\begingroup$ @GeoffRobinson: The answer is yes if we add $A$ invertible, as $\exp: M(N,\mathbb{C}) \rightarrow GL(N,\mathbb{C}) $ is surjective and $\exp(M) \overline{\exp(M)}=I$ follows $M = -\overline{M}$. $\endgroup$
    – Jiro
    Oct 20, 2015 at 9:13
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    $\begingroup$ Well,$A$ certainly needs to be invertible in the question. $\endgroup$ Oct 20, 2015 at 10:00

3 Answers 3


The set $V= \{ A\in M_n(\mathbb{C})\ |\ A\bar A = I\}$ is a smooth submanifold of $M_n(\mathbb{C})$ with real dimension $n^2$.

Proof: Consider the involution $\iota:\mathrm{GL}(n,\mathbb{C})\to \mathrm{GL}(n,\mathbb{C})$ defined by $$ \iota(A) = (\bar A)^{-1}. $$ This is an anti-holomorphic involution of the complex manifold $\mathrm{GL}(n,\mathbb{C})$, and its fixed locus is precisely $V$. Thus, $V$ is a totally real submanifold of $\mathrm{GL}(n,\mathbb{C})$ with (real) dimension $n^2$.

Also: While it's true that the map $f:M_n(\mathbb{R})\to \mathrm{GL}(n,\mathbb{C})$ defined by $$ f(a) = \exp(ia) $$ has its image in $V$, it is not a 'parametrization' everywhere, i.e., $f$ is not a local diffeomorphism everywhere. While this is true on a neighborhood of $0\in M_n(\mathbb{R})$, at other places, the map $f$ definitely is not a local diffeomorphism. For example, if $a$ has $n$ eigenvalues of the form $2k_i\pi$ (where $k_1,\ldots, k_n$ are integers, not all zero), then $f(a) = I$, but the space of such real matrices $a$ has positive dimension for any $n$-tuple $(k_1,\ldots, k_n)$ for which not all of the $k_i$ are equal.

It turns out that $f$ is surjective. The proof uses the fact that, for $A\in V$, we have $A\bar A = \bar A A = I$, so, in particular, $A$ and $\bar A$ commute and hence can be put simultaneously in Jordan normal form by a real conjugation. Then, breaking $\mathbb{C}^n$ into a sum of complexifications of real subspaces according to the eigenvalues of $A$ and using the semi-simple/nilpotent decomposition appropriately, one can reduce to dealing with the upper triangular case, and the result can then be proved using simple facts about power series in commuting nilpotent variables. Details upon request (see below).

Of course, all of this is probably a special case of known facts about affine symmetric spaces. I have now realized that $V$ is simply the Cartan embedding for the affine symmetric space $\mathrm{GL}(n,\mathbb{C})/\mathrm{GL}(n,\mathbb{R})$. This Cartan embedding $$ \sigma:\mathrm{GL}(n,\mathbb{C})/\mathrm{GL}(n,\mathbb{R})\longrightarrow V\subset \mathrm{GL}(n,\mathbb{C}) $$ is given by $\sigma(B\cdot\mathrm{GL}(n,\mathbb{R})) = B\,(\,\overline B\,)^{-1}$. The exponential map we have been discussing is just the geodesic mapping of this affine symmetric space, so, probably all of this follows from general theory.

Details: Let $A\in\mathrm{GL}(n,\mathbb{C})$ satisfy $A\overline{A} = I$. Let $\lambda\in\mathbb{C}$ be an eigenvalue of $A$ of multiplicity $m\le n$ and let $$ V_\lambda = \{ v\in\mathbb{C}^n \ |\ (A{-}\lambda)^mv = 0\ \}. $$ be the associated generalized eigenspace. Of course, $\mathbb{C}^n$ is the direct sum of these generalized eigenspaces. Since $\overline{A}$ commutes with $A$, it preserves each of these subspaces. Moreover, one clearly has $$ \overline{V_\lambda} = V_{1/\bar\lambda}, $$ so each of the spaces $V_\lambda+V_{1/\bar\lambda}$ is invariant under conjugation and hence is the complexification of a real subspace $W_\lambda\subset \mathbb{R}^n$. It follows that, by conjugating by a real matrix, we can assume that $A$ (and hence $\overline A$) is in block diagonal form, so it suffices to treat the case where either $A$ has a single eigenvalue $\lambda$ satisfying $\lambda\bar\lambda=1$, so that $\mathbb{C}^n=V_\lambda$, or else $A$ has an eigenvalue $\lambda$ satisfying $\lambda\bar\lambda > 1$ and $\mathbb{C}^n=V_\lambda\oplus V_{1/\bar\lambda}$.

In either case, we can assume that $\lambda = r\ge 1$ is real, since, if $\lambda = r e^{i\theta}$, we can replace $A$ by $e^{-i\theta}A$ and show that $e^{-i\theta}A$ is in the image of $f$ (since $I$ commutes with everything and $e^{i\theta} I = \exp(i\theta I)$.)

First, consider the case $\lambda = 1$. Then $A = C + i S$ where $C$ and $S$ are real matrices and $((C-I) + iS)^n = (A-I)^n = 0$. Moreover, $N=C-I$ and $S$ commute since $I = A\bar A = C^2+S^2 +i(SC-CS) = I + 2N + N^2 + S^2$. Note that $N$ and $S$ are nilpotent commuting matrices. Since they satisfy $$ 2N + N^2 = -S^2, $$ it follows that $N = p(-S^2)$ where $p(t) = -\tfrac12 t^2 + \cdots $ is the (unique) power series (with real coefficients) that satisfies $2p(t) + p(t)^2 = -t^2$. Since $S$ is nilpotent, it follows that $N = p(-S^2)$ expresses $N$ canonically as a polynomial in $S$. Now let $q(t)$ be the (unique) power series with real coefficients that satisfies $\sin q(t) = t$. Note that we must have $p(-q(t)^2) = \cos(t)-1$. Now, because $S$ is nilpotent, we have $S = \sin q(S)$, where $q(S)$ is a real polynomial in $S$. Putting all of this together, we have $$ A = I + p(-S^2) + iS = \cos(q(S)) + i\sin(q(S)) = \exp(iq(S)). $$ Finally, before leaving this special case, let us note that, because $A$ satisfies its characteristic polynomial $(A-I)^n = 0$, it follows that $\overline A = A^{-1}$ is expressed as a universal polynomial in $A$ with real coefficients, so $iS = \tfrac12(A-\overline{A})$ is also expressed as a universal polynomial with real coefficients. Since $q(t)$ is an odd power series, it satisfies $iq(it) = f(t)$ where $f$ has real coefficients, so there is actually a formula of the form $A = \exp( i g_n(A))$ for all $A \in \mathrm{GL}(n,\mathbb{C})$ satisfying $A\bar A = I$ and $(A-I)^n=0$ for some universal polynomial $g_n(t)$ with real coefficients that also has the property that $g_n(A)$ is a real $n$-by-$n$ matrix for all such $A$. (This remark will be used below.)

Now, finally, let us assume that $A$ has eigenvalues $\lambda = e^t$ and $1/\lambda = e^{-t}$ for some $t>0$. Then $V_\lambda$ and $\overline{V_\lambda} = V_{1/\lambda}$ are disjoint, complementary complex subspaces of $\mathbb{C}^n$ and so we must have that there exists a matrix $Q\in M_n(\mathbb{R})$ such that $$ V_\lambda = \{ v + iQv \ | \ v\in \mathbb{R}^n \}. $$ Because $V_\lambda$ is closed under multiplication by $i$, it follows that $Q^2 = -I$. Setting $$ S = \cosh t + i\sinh t\,Q = \exp(i t Q), $$ we see that $Sw= e^t w$ for all $w \in V_\lambda$ and $Sw = e^{-t}w$ for all $w \in V_{1/\lambda}$, so $S$ is the semi-simple part of $A$. Hence $S$ can be written as $S = s_\lambda(A)$, where $s_\lambda(t)$ a polynomial in $t$ with real coefficients (that depend on $\lambda$). Thus, $\bar S = s_\lambda(\bar A)$ can be written as a universal polynomial in $A$, implying that $Q$ itself can be written as a universal polynomial in $A$ and hence, in particular, it commutes with $A$ and $\bar A$. Now, writing $$ A = \exp(i t Q) B, $$ we see that $B\bar B = I$ and that $B$ can be written as a polynomial in $A$. Moreover, the eigenvalues of $B$ are now all equal to $1$, so $(B-I)^n=0$, and so $B = \exp(i g_n(B))$ (as per above), where $g_n(B)$ is a polynomial in $A$, which, therefore, commutes with $Q$ (which is a polynomial in $A$). Finally, we have $$ A = \exp(i t Q)\exp(i g_n(B)) = \exp(i (t Q+g_n(B))), $$ as desired.

  • $\begingroup$ Thank you very much for your insightful answer. The only main missing part for me is the surjectivity of $f$. $\endgroup$
    – Jiro
    Oct 20, 2015 at 16:43
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    $\begingroup$ Robert, I learn so much new material from your answers! Just great! $\endgroup$
    – Suvrit
    Oct 21, 2015 at 13:24
  • $\begingroup$ Robert, great answer. So a bit about the details, for simplicity assuming $A$ is diagonalisable. Commutativity gives as $A = U^{-1} \Lambda U$ for real $U$. $A$ is invertible, hence there is $\exp(i L) = \Lambda$. So, $A = \exp( i U^{-1} L U)$. Why is $U^{-1} L U$ real? $\endgroup$
    – Jiro
    Oct 22, 2015 at 20:05
  • $\begingroup$ If $A$ is diagonal, then $A\bar A = I$ implies that all of the eigenvalues of $A$ are complex numbers of unit modulus, so each of them is of the form $e^{ir}$ where $r$ is real. $\endgroup$ Oct 22, 2015 at 20:40
  • $\begingroup$ I'm confused, I only wanted $A$ diagonalisable and not $A$ diagonal. $\endgroup$
    – Jiro
    Oct 22, 2015 at 21:24

Qiaochu Yuan is a little hard about our variety $V$; Kummer showed that an element of $V$ is in the form $U\bar{U}^{-1}$ (a generalization of this result is the Hilbert's theorem 90). Close to the subject, cf. "Ikramov, On the matrix equation $X\bar{X}=A$" http://link.springer.com/article/10.1134%2FS1064562409010153#page-1

Robert wrote a nice answer; we can give more elementary proofs as follows. $\mathbb{C}$ is seen as $\mathbb{R}^2$.

Proposition 1. $dim(V)=n^2$ when $V$ is considered as a real algebraic set.

Proof. According to Kummer, if $R\bar{R}=I$, then there is $T\in GL_n(\mathbb{C})$ s.t. $R=T\overline{T}^{-1}$. Let $f:T\in GL_n\rightarrow T\overline{T}^{-1}$; note that $f$ is a pseudo-parametrization of our set; in fact $f$ is a submersion. It remains to calculate the rank of $Df_T:H\in M_n\rightarrow (H-T\overline{T}^{-1}\overline{H})\overline{T}^{-1}$. Note that $rank(Df_T)=rank(H\rightarrow T^{-1}H-\overline{T}^{-1}\overline{H})$.

Let $T^{-1}=U+iV,H=X+iY$, where $U,V,X,Y$ are real matrices. Then $rank(Df_A)=rank(g:(X,Y)\in \mathbb{R}^{2n^2}\rightarrow UY+VX\in \mathbb{R}^{n^2}$; we show that $g$ is onto. Let $A\in M_n$; since $U+iV$ is invertible, there is $\lambda\in \mathbb{R}$ s.t. $U+\lambda V$ is invertible. Finally $g(\lambda(U+\lambda V)^{-1}A,(U+\lambda V)^{-1}A)=A$ and we are done.

Proposition 2. Any $R\in V$ can be written as $exp(iA)$ where $A$ is a real matrix.

Proof. Let $R=U+iV$, where $U,V$ are real. $R\bar{R}=I$ is equivalent to $U^2+V^2=I,VU=UV$; that is equivalent to: there is a real matrix $A$ s.t. $U=\cos(A),V,=\sin(A)$ and $R=\exp(iA)$.

EDIT. (cf. Sebastian's comment). $e^Me^{\bar{M}}=I$ does not imply that $e^{M+\bar{M}}=I$.

Proof. Take $M=\begin{pmatrix}i\pi&1\\0&-i\pi\end{pmatrix}$.

  • $\begingroup$ That is very sleek! Do you need Prop1 for Prop2? Do you get Prop1 also from Prop2? $\endgroup$
    – Jiro
    Oct 29, 2015 at 11:59
  • $\begingroup$ And do you have any suggestion what to name $V$? I understand that there might be no agreed name. Unfortunately, I cannot access right now the Ikramov paper to have a look. $\endgroup$
    – Jiro
    Oct 29, 2015 at 12:01
  • $\begingroup$ I would like to see the details for your Proposition 2. I don't believe that it is as easy as you seem to think. I thought of exactly this line, right up to your 'that is equivalent to there is a real matrix A...". I didn't see how to prove that directly, and I would be curious to see what method you use to prove it. Also, the smoothness of $V$ follows, as I wrote, easily from the fact that it is the fixed point set of an anti-holomorphic involution. I don't see why it needs anything from Kummer, unless you are quoting the surjectivity of the Cartan embedding, a nontrivial fact. $\endgroup$ Oct 29, 2015 at 14:11
  • $\begingroup$ @ Sebastian SchlechtClearly , Prop. 1 and 2 can be independently proved. About a name for $V$, I don't know. In general, the authors consider elements of $V$ and not the whole set. cf also Horn and Johnson; Matrix Analysis, Section 4.6 $\endgroup$
    – loup blanc
    Oct 29, 2015 at 14:59
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    $\begingroup$ @loupblanc: I suspected as much. So it's somewhat misleading to claim that your proof is 'more elementary'; it just leaves out many details and even essential ideas. In fact, while I mention Cartan's results, I don't use them (or any result of Kummer) in the surjectivity argument, which, in fact, uses nothing more than than the semi-simple/nilpotent decomposition (i.e., Jordan normal form), and even the result over $\mathbb{C}$ needs this. $\endgroup$ Oct 29, 2015 at 15:52

I totally lied about this not being a natural thing to ask! As loup blanc alludes to, in fact $n \times n$ matrices such that $M^{-1} = \overline{M}$ can be interpreted as Galois descent data for descending the complex vector space $\mathbb{C}^n$ to a real vector space, or equivalently as a $1$-cocycle in

$$Z^1(B \text{Gal}(\mathbb{C}/\mathbb{R}), GL_n(\mathbb{C})).$$

The statement that any such matrix must have the form $U \overline{U}^{-1}$ says that any such $1$-cocycle is cohomologous to zero, or equivalently that the Galois cohomology set

$$H^1(B \text{Gal}(\mathbb{C}/\mathbb{R}), GL_n(\mathbb{C}))$$

is trivial. This just says that there is only one real form of $\mathbb{C}^n$ up to isomorphism, namely $\mathbb{R}^n$. The space $GL_n(\mathbb{C})/GL_n(\mathbb{R})$ appearing in Robert Bryant's answer can then be interpreted as the space of real forms of $\mathbb{C}^n$.


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