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When $A$ is skew-symmetric, then $Q = e^{A t}$ is orthogonal or $Q^T Q = I$.

(For a linear autonomous system $\dot{x} = A x$, the fundamental matrix of the system is $\Phi(t) = e^{A t}$.)

To prove this identity, I make use of the result $$ \left( e^{P} \right)^T = e^{P^T} \tag{1} $$ for any matrix $P$.

Since $e^P$ can be expressed as the uniformly convergent series, $$ e^P = \sum\limits_{k = 0}^\infty \ {P^k \over k!}, $$ it follows that $$ \left( e^P \right)^T = \left( \sum\limits_{k = 0}^\infty \ {P^k \over k!} \right)^T = \sum\limits_{k = 0}^\infty \ {(P^k)^T \over k!} = \sum\limits_{k = 0}^\infty \ {(P^T)^k \over k!} = e^{P^T} $$

Assume that $A$ is skew-symmetric. Then $$ A^T = -A $$

Define $$ Q = e^{A t} $$

Then we have $$ Q^T = e^{( A t)^T} = e^{A^T t} = e^{- A t} = \left( e^{A t} \right)^{-1} = Q^{-1} $$

Thus, it follows that $$ Q^T Q = Q^{-1} Q = I $$ showing that $Q$ is orthogonal.

I hope that this proof is OK!

Dr. Sundar
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2 Answers2

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$e^{A+B} = e^A e^B$ holds when $A$ and $B$ commmute, i.e. $AB = BA$, see for example here.

You correctly derived that $Q^T = e^{-At}$, and it follows that $$ Q^T Q = e^{-At} e^{At} = e^{(-A+A)t} = e^{0} = I $$ since $-A$ and $A$ commute.

Martin R
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A simpler proof.

Define $P(t)=\exp(A^Tt)\exp(At)$. We have that

$$\dot{P}(t)=A^TP(t)+P(t)A=\exp(A^Tt)(A+A^T)\exp(At)$$

where we have used the fact that $A$ (resp. $A^T$) commutes with $\exp(At)$ (resp. $\exp(A^Tt)$).

Since $A$ is skew-symmetric, then $A+A^T=0$ and we get $\dot{P}(t)=0$. Therefore, we have that $P(t)=P(0)=1$.

KBS
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