So the question is in the title. It's easy to prove when we know that there are real numbers in it and the dot product is standard. But why this works in the general case - when there are complex numbers inside and the dot product is something else?
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$\begingroup$Let $C_i$ the $i^{\text{th}}$ column of the orthogonal matrix $O$ then we have
$$\langle C_i,C_j\rangle=\delta_{ij}$$ and we have $$O^T=(C_1\;\cdots\; C_n)^T=(C_1^T\;\cdots\; C_n^T)$$ so we get
$$O^TO=(\langle C_i,C_j\rangle)_{1\le i,j\le n}=I_n$$
$\endgroup$ 3 $\begingroup$A is othogonal means A'A = I. That says that A' is the inverse of A!
$\endgroup$ 3 $\begingroup$Represent your orthogonal matrix $O$ as element of the Lie Group of Orthogonal Matrices. You get:
$$O = \exp(\Omega),$$ where $\exp$ means the matrix exponential and $\Omega$ is an element of the corresponding Lie Algebra, which is skew-symmetric, i.e. $\Omega^T = -\Omega$. Now transpose it to get: $$O^T=\exp(\Omega)^T=\exp(\Omega^T)=\exp(-\Omega),$$ which is the inverse of $O$: Since $\Omega$ and $-\Omega$ commute, i.e. $[\Omega,-\Omega]_-=0$ we can write $$O^TO=\exp(-\Omega)\exp(\Omega)=\exp(-\Omega+\Omega)=\exp(0)=1$$
$\endgroup$ 1 $\begingroup$ΩT=−Ω. Now transpose it to get: OT=exp(Ω)T=exp(ΩT)=exp(−Ω), which is the inverse of O: Since Ω and −Ω commute, i.e. [Ω,−Ω]−=0 we can write OTO=exp(−Ω)exp(Ω)=exp(−Ω+Ω)=exp(0)+ 0+1 -1 transpose 1+0 +Y -X +0=1
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