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The (real) general linear group is defined $GL(n)=\{A \in \mathbb{R}^{n\times n} \mid \operatorname{det}(A) \neq 0\}$. It is a matrix Lie group. Let $J$ be a constant $n$-by-$n$ real matrix. The so-called $J$-orthogonal group is defined $O_J(n)=\{A \in GL(n) \mid A^T J A = J\}$. I'm pretty sure $O_J(n)$ is a matrix Lie group regardless of what $J$ is. The corresponding Lie algebra is $o_J(n) = \{B \in \mathbb{R}^n \mid B^T J + JB = 0\}.$ For $B \in o_J(n)$, the book I am reading defines the Cayley transform of $B$ as

$$A = (I-\alpha B)^{-1}(I+\alpha B). $$

I'm assuming $\alpha$ is some real constant. The book claims that $A$ is $J$-orthogonal, that is, $A^T J A = J$. In the book, it is unclear whether they claim this for $n=4$ and $J = \operatorname{diag}(1, -1, -1, -1)$, or for all real square matrices $J$. I think I have verified that $A$ is $J$-orthogonal if $J$ is symmetric and $J^2 = I$ (which includes $J = \operatorname{diag}(1, -1, -1, -1)$. I guess I have two questions: first, is this correct? And second, are weaker assumptions on $J$ possible?

EDIT: The two factors defining $A$ above commute. If you use this fact and mimic the proof in the answer below, you can show that no assumptions on $J$ are necessary: $J$ can be any $n$-by-$n$ real matrix.

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1 Answer 1

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To get some idea how this works, let us first assume only that $J$ is involutory, i.e., $J^2 = I$. Then $A$ is $J$-orthogonal if and only if $AJA^T = J$. We check that condition:

$$(I - \alpha B)^{-1} (I + \alpha B) J (I + \alpha B)^T (I - \alpha B)^{-T} = J$$

and see that it is equivalent to

$$(I + \alpha B) J (I + \alpha B)^T = (I - \alpha B) J (I - \alpha B)^T.$$

This is equivalent to

$$J + \alpha (BJ + JB^T) + \alpha^2 B J B^T = J - \alpha (BJ + JB^T) + \alpha^2 B J B^T,$$

which is equivalent to

$$\alpha (BJ + JB^T) = 0.$$

Assuming that $\alpha \ne 0$, this is equivalent to

\begin{equation} BJ + JB^T = 0.\tag{*} \end{equation}

If we premultiply and postmultiply $(*)$ with $J$, and keeping in mind that $J^2 = I$, the above is equivalent to

$$0 = J(BJ + JB^T)J = JB + B^TJ,$$

which is equivalent to saying that $B \in o_J(n)$.

Of course, for $\alpha = 0$, any $B$ gives as $J$-orthogonal matrix $A = I$, with no conditions on $J$.

Now, for the general case, let us also lose the involutority of $J$. Then we have to observe the condition $A^T J A = J$. This is equivalent to

$$(I + \alpha B)^T (I - \alpha B)^{-T} J (I - \alpha B)^{-1} (I + \alpha B) = J,$$

which is equivalent to

$$(I - \alpha B)^{-T} J (I - \alpha B)^{-1} = (I + \alpha B)^{-T} J (I + \alpha B)^{-1}.$$

Since $J$ is nonsingular, we see that this is equivalent to

$$\left( (I - \alpha B) J^{-1} (I - \alpha B)^T \right)^{-1} = \left( (I + \alpha B) J^{-1} (I + \alpha B)^T \right)^{-1}.$$

In other words,

$$(I - \alpha B) J^{-1} (I - \alpha B)^T = (I + \alpha B) J^{-1} (I + \alpha B)^T.$$

We have this above, with the only difference that we now have $J^{-1}$, instead of $J$. So, using the same steps, we obtain the following equivalent of $(*)$:

$$BJ^{-1} + J^{-1}B^T = 0.$$

Again, premultiplying and postmultiplying with $J$, we see that this is equivalent to

$$JB + B^T J = 0,$$

which means that the above is equivalent to $B \in o_{J^{\color{red}{-1}}}(n)$.

In other words, the first part of this proof is just an obvious special case ($J = J^{-1}$). This makes sense, since we've used involutority twice, which cancels itself out.

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$J$ defines a nonstandard scalar (inner) product space. This was a subject of most the research I've done in the numerical linear algebra (with some published papers), and I've never seen $J$ being singular. I believe there are few reasons for this lack of interest. To put make the long story short: 1. lack of applications; 2. most such cases, if not all, can be observed as if $J$ was nonsingular, but on some subspace of the "big" space; and 3. loss of definiteness (i.e., you would get a nontrivial subspace orthogonal to the "big" space), which hurts many, if not most proofs. –  Vedran Šego Oct 14 '13 at 1:12
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I've edited my answer, in hopes to clear the confusion. I hope it is more clear now. And yes, by "general case", I mean "$J$ is nonsingular, with no more conditions". My favourite introductory read on the nonstandard scalar product spaces is Gohberg, Lancaster, Rodman, "Indefinite Linear Algebra and Applications", although it covers only indefinite spaces ($J$ is symmetric/Hermitian). For more general spaces, papers by Tisseur, Higham, and few others are quite enlightening. –  Vedran Šego Oct 14 '13 at 1:17
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I've just noticed: the two terms defining $A$ are commutative. Note that if you swap them, your statement will always hold (for any $J$; I don't think you will even need nonsingularity). Just mimic my proof above. –  Vedran Šego Oct 14 '13 at 8:58
    
I don't understand your second-to-last comment, but I used your last comment, swapped the two factors defining $A$, and mimicked your proof, and the conclusion holds for any matrix $J$. –  Stefan Smith Oct 14 '13 at 13:36
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I'm glad we've proved it. :-) –  Vedran Šego Oct 14 '13 at 13:40

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