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Suppose

  • $n\in\mathbb{N}$, $n\geq 2$
  • $R_i$ is a $1$ x $n$ matrix for $i=1,...,n$
  • Each $R_i$ is a row of the $n$ x $n$ matrix $R$ and the $R_i$ are linearly independent (det$R\neq0$)
  • $C_i$ is an $n$ x $1$ matrix for $i=1,...,n$
  • Each $C_i$ is a column of the $n$ x $n$ matrix $C$ and the $C_i$ are linearly independent (det$C\neq0$)

Then the matrix product $RC$ gives the $n^2$ inner products of each row of $R$ and each column of $C$ considered as vectors.

Suppose $U$ is an $n$ x $n$ unitary matrix.

Then, according to the properties of unitary matrices listed on Wikipedia:

$$(RU)(UC)=RC$$

However,

$$(RU)(UC)=RC$$ $\implies$

$$R(UU)C=RC$$ $\implies$

$$R^{-1}R(U^2)CC^{-1}=R^{-1}RCC^{-1}$$ $\implies$

$$U^2=I$$

This cannot be the case in general, because not every $n$ x $n$ unitary matrix squares to the $n$ x $n$ identity matrix.

What seem(s) to be the mistake(s) I made?

Have I misinterpreted the inner product preserving property of unitary matrices?


Update: I understand now that the inner product preserving property follows naturally from the fact that $U^{T}U=I$ (where we assume $U$ has only real entries and thus $U^†=U^T$), and that my mistake was indeed misinterpreting Wikipedia's statement

However, if anyone considers posting an answer, would you please explain in your answer why the same logic of Qiaochu Yuan's comment extends to the conjugate transpose $U^†$ in the case where $U$ has complex entries with nonzero imaginary part?

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    $\begingroup$ You need $RU^T$ instead. Then you get $U^T U = I$ in the middle. This is clearer if you work with two column vectors $u, v$ instead and consider $\langle Uu, Uv \rangle = (Uv)^T (Uu) = v^T U^T U u = v^T u = \langle u, v \rangle$. $\endgroup$ Commented Sep 4 at 0:31
  • $\begingroup$ Please avoid posting answers in comments. See meta.stackexchange.com/questions/19756/how-do-comments-work $\endgroup$ Commented Sep 4 at 0:33
  • $\begingroup$ If anyone considers posting an answer, would you please explain in your answer why the same logic of Qiaochu Yuan's comment extends to the conjugate transpose in the case where $U$ has complex entries? $\endgroup$
    – Simon M
    Commented Sep 4 at 1:07

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