# Is there an orthogonal matrix that is not unitary?

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?

• 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? – Omnomnomnom Feb 16 '16 at 20:19
• yes, exactly, can u help? – Eduardo Silva Feb 16 '16 at 20:23
• 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 $A^*=A^T$). – Ian Feb 16 '16 at 20:38

## 2 Answers

The matrix $$A = \pmatrix{ \sqrt{2}&i\\ i&-\sqrt{2} }$$ satisfies $$AA^T = I$$ but $$AA^* = \pmatrix{5&-2i\sqrt{2}\\2i\sqrt{2}&5}$$

• 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 $2i\sqrt{2}$ – Danilo Gregorin Nov 20 '18 at 18:46
• @DaniloGregorin well spotted – Omnomnomnom Nov 20 '18 at 19:51

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.