Bijection from Skew-Symmetric Matrices to Orthogonal Group

Question

I'm trying to prove that the map $f:U \rightarrow V$, where $U$ consists of anti-symmetric matrices $a$ such that $(I_n+a)$ and $(I_n-a)$ are invertible, and V consists of matrices $b \in O(n)$ such that $(I_n + b)$ is invertible, and $f(a) = (I_n + a)(I_n - a)^{-1}$.

To show that the map is surjective, I let $b \in V$ and consider an $a \in U$ such that $a = (b - I_n)(b + I_n)^{-1}$. I can show that $(I+a)$ and $(I-a)$ are invertible, but if I'm not mistaken, I have to show that $a$ is antisymmetric, which I'm having trouble doing.

$U$ and $V$ have the same dimension, but $f$ is not a linear map, so I can't see a way to get out of showing surjectivity directly. Is there a matrix identity that would make it easier to show that $a$ is antisymmetric? Thanks.

score 1 · Accepted Answer · edited Apr 13 '17 at 12:20

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There is the exponential map $$\exp\colon \mathfrak{so}(n)\rightarrow O(n)$$ from the Lie algebra consisting of skew-symmetric matrices to the orthogonal group. It is well-known that $\exp$ for this group is surjective, e.g., see here.

To answer your question, the map to consider is the the Cayley transform: Given any rotation matrix, $R \in SO(n)$, if $R$ does not admit $−1$ as an eigenvalue, then there is a unique skew-symmetric matrix $S$ so that $$ R = (S − I)(S + I)^{−1}. $$ This is a classical result of Cayley (1846) and $R$ is called the Cayley transform of $S$. The inverse is given by $$ S = (I+R)(I-R)^{-1}. $$

edited Apr 13 '17 at 12:20

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answered Sep 21 '16 at 12:23

Dietrich Burde

130,978

Thank you Dietrich. Alternatively, if you're like me and haven't studied Lie Algebras yet, you can just easily show that $a + a^T = 0$ as defined above. – user1447447 Sep 21 '16 at 12:38
You are right, Cayley's result does not need Lie algebras. – Dietrich Burde Sep 21 '16 at 12:41
Does $ R = (S - I)(I + S)^{-1}$ ? – user1447447 Sep 21 '16 at 12:41
Hmm, maybe I'm making an embarrassing mistake: I have $S = (I_n + a)(I_n - a)^{-1}$ so $S(I_n - a) = (I_n + a)$, $S - Sa = I_n + a$, $S - I_n = a + Sa$, $S - I_n = a(I_n + S)$, $a = (S - I_n)(S + I_n)^{-1}$ – user1447447 Sep 21 '16 at 12:47
You are right. I mixed up the two versions with the sign. – Dietrich Burde Sep 21 '16 at 12:52
Thank you very much for helping me solve the problem. – user1447447 Sep 21 '16 at 12:54

Bijection from Skew-Symmetric Matrices to Orthogonal Group

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