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If $A$ is a skew-symmetrical matrix with it's diagonal elements as $0$, Prove that it's exponent $e^A$ is an orthogonal matrix.

Cyanide2002
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1 Answers1

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Recall that

$$e^A := \sum_{n=0}^\infty \frac{1}{n!} A^n$$

Since $(A^t)^n = (A^n)^t$ we get that $(e^A)^t = e^{(A^t)}$. Moreoever we will need the following property of the exponent that $e^A\cdot e^B = e^{A+B}$ for any commuting matrices $A,B$.

We are given that $A$ is skew symmetric (i.e. $A^t=-A$). Therefore $A$ and $A^t$ commutes and so $$e^A\cdot (e^A)^t = e^{A+A^t} = e^0 = I$$ where $0$ denotes the zero matrix.

Yanko
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  • +1, but your mixed uses of $t$ and $T$ for matrix transpose is confusing. – user1551 Feb 25 '19 at 16:14
  • @user1551 you're absolutely right thanks. – Yanko Feb 25 '19 at 16:15
  • Don't you also need to require that $A$ and $B$ commute to have that property of the exponential? – JonathanZ Feb 25 '19 at 16:17
  • @JonathanZ I'm not sure, but it sounds like something I need to require, yes. Anyway I edited, but $A$ definitely commutes with $A^t = -A$ so it doesn't change the answer. – Yanko Feb 25 '19 at 16:19
  • @Yanko: To be honest it's a point that I'm often unsure about and have check frequently. I just know that I can't use it automatically and didn't want the OP to get the idea that it always holds. And if it's not always true that $A$ and $A^t$ commute (which is another thing I'm not immediately sure of) then maybe you could include your comment about them commuting in this case in your answer? – JonathanZ Feb 25 '19 at 16:29