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A complex square matrix U is unitary if its conjugate transpose $U^*$ is also its inverse – that is, if $U^*U=UU^*=I$, where I is the identity matrix.

Question is:$UU^T=I$ or not?

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  • $\begingroup$ Only if the matrix has real entries $\endgroup$
    – Triatticus
    Jun 6, 2016 at 1:48

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It is not true even when $n=1$. Take $U = (i)$ be the $1\times 1$ matrix.

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  • $\begingroup$ If $ {u_1,u_2,...,u_n} $ is orthonormal basis of $R^n$, is it true? $\endgroup$ Jun 6, 2016 at 1:57
  • $\begingroup$ Yes if you mean $U = (u_1\ \ u_2\ \ \cdots u_n)$. @VivianZhang $\endgroup$
    – user99914
    Jun 6, 2016 at 2:06

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