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On many links online [1], [2], [3], etc. they mention the following:

$U$ is a unitary matrix, i.e. $UU^\dagger = U^\dagger U = I$, where $U^\dagger$ is the Hermitian conjugate of $U$.

Then, $$\langle U x, U y \rangle = \langle x, U^\dagger U y \rangle$$

Why is this so?

gust
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    The pure definition of the adjoint of $T$, $T^$, is the operator such that the quality $$\langle v, Tw\rangle = \langle T^v, w \rangle$$ holds for all $v,w$. The swap is only from what the definition of the adjoint is. – Ninad Munshi Jan 16 '21 at 13:01

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Since $\langle u,v\rangle=u^\dagger v$, we generally have $$\langle Tu,\,v\rangle=(Tu)^\dagger v=u^\dagger T^\dagger v=\langle u,\,T^\dagger v\rangle$$ for any vectors $u,v$ and any linear map $T$.

Berci
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  • Why is $\langle u, v \rangle = u^\dagger v$? Shouldn't it only be the transpose, $u^Tv$? – gust Jan 16 '21 at 13:19
  • @gust You need $\langle u,,u\rangle\ge0$ for all $u$. What happens to the two proposed definitions we're comparing if I multiply $u$ by $i$? – J.G. Jan 16 '21 at 13:21
  • @J.G. not completely sure; how does the positive definitivity apply here? – gust Jan 16 '21 at 13:27
  • @gust $(iu)^\dagger=-iu^\dagger$, so $(iu)^\dagger iu=u^\dagger u$, but $(iu)^Tiu=-u^Tu$. – J.G. Jan 16 '21 at 14:05
  • Just verify it using coordinates. What is the definition of the complex inner product you use? – Berci Jan 16 '21 at 14:31