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How do I prove that "if $(e_1, ..., e_k)$ is an orthonormal basis for a subspace $L$, then so is $(Ue_1, ..., Ue_k)$" implies that the matrix of $U$ satisfies $UU^{*}=I=U^{*}U$? The columns have to be linearly independent and the vectors that columns form should be of length one, but that's where I get stuck. Also one direction $UU^{*} = I$ looks to be easier to prove than $U^{*}U=I$.

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Take the standard basis $\{e_1, ..., e_n\}$ for $\mathbb{C}^n$. This is orthonormal. Then the columns of the matrix representing $U$ in the standard basis are given by $$ U = [Ue_1 ... Ue_n]$$ By the definition of matrix multiplication for any matrix $A$ we have $(A^*A)_{ij} = \langle A_i, A_j \rangle$ where $A_i$ is the $i$-th column of $A$. But since these are orthonormal in the case of $U$ by assumption we get $(U^*U)_{ij} = \delta_{ij}$, and thus $U^*U = I$.

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@jackdevilo $UU^* = I$ tells you that $U^* = U^{-1}$ hence it follows immediately that $U^*U=I$. –  user13838 Oct 10 '11 at 14:31
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