I could find a example of a unitary matrix such that is not orthogonal, thats simple in $\mathbb{C}$, but for this exercise of a orthogonal that is not unitary i realize that is possible just on $\mathbb{C}$ because all orthogonal matrix on $\mathbb{R}$ is unitary, so anyone have a exemple of this case?
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2$\begingroup$ Just to be clear: you're asking for a matrix $A$ with complex entries for which $AA^T = I$, but $AA^* \neq I$. Is that correct? $\endgroup$– Ben GrossmannFeb 16, 2016 at 20:19
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$\begingroup$ yes, exactly, can u help? $\endgroup$– Eduardo SilvaFeb 16, 2016 at 20:23
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4$\begingroup$ Note that Omnomnomnom's clarification is very important: if a matrix $A$ with real entries satisfies $A A^T=I$ then certainly $A A^*=I$ (since $A^*=A^T$). $\endgroup$– IanFeb 16, 2016 at 20:38
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2 Answers
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The matrix $$ A = \pmatrix{ \sqrt{2}&i\\ i&-\sqrt{2} } $$ satisfies $AA^T = I$ but $$ AA^* = \pmatrix{5&-2i\sqrt{2}\\2i\sqrt{2}&5} $$
answered Feb 16, 2016 at 20:36
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$\begingroup$ Almost irrelevant, but you missed a factor of $2$ in the terms outside the main diagonal, didn't you? My calculations give $-2i\sqrt{2}$ and $2i\sqrt{2}$ $\endgroup$ Nov 20, 2018 at 18:46
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$\begingroup$ @Omnomnomnom, as it is stated in second answer you are not right, because matrix A can not be orthogonal, by definition orthogonal matrix have real entries only. If I misunderstand something, correct me. $\endgroup$ May 10, 2020 at 20:56
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$\begingroup$ @Dan As I verified in my comment on the question, the asker is using the term orthogonal to refer to matrices with complex entries for which $AA^T=I$. As far as I’m concerned, whether or not this is the correct usage of the word “orthogonal” is irrelevant. $\endgroup$ May 10, 2020 at 21:05
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$\begingroup$ @Omnomnomnom so you've just shown matrix satisfying this condition(not orthogonal). Thanks for clarifying $\endgroup$ May 10, 2020 at 21:09
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Late remark. In general, if $K$ is complex skew-symmetric, then $Q=e^{zK}$ is complex orthogonal for every $z\in\mathbb C$. When $K\ne0$, since $e^{zK}$ is holomorphic and its power series expansion has some nonzero high-order terms, $e^{zK}$ is not a constant function. Therefore it cannot be real all the time (otherwise it will fail to satisfy Cauchy-Riemann equations). Pick a $z$ such that $Q=e^{zK}$ is not real. Then $Q$ is not unitary, because all unitary orthogonal matrices are real.
