If we consider some unitary operator in matrix form given by $$ U = \left[ {\begin{array}{cc} P_{00} & P_{01} \cdot \cdot & P_{0J} \\ P_{10} & \\ \cdot &\\ \cdot\\ P_{J0} \cdot\cdot\cdot& & P_{JJ} \\ \end{array} } \right]$$ Where each $P_{nm}$ are matrices (sub-blocks of $U$). How can we show that the first $J$ columns of $U$ are orthonormal vectors?

My proposed answer let me know what you think about the validity:

Since $U$ is unitary it follows that each of the sub-blocks on the diagonal of $U^{\dagger}U$ must be the identity. Hence we get the relation $I = \sum_{n}P_{n0}^{\dagger}P_{n0}$. Let $U_i$ be the i-th column of $U$ and consider the elements of $U^{\dagger}U$ then: $$[U^{\dagger}U]_{ij} = \sum_{k}P_{ki}^{\dagger}P_{kj} = U_{i}^{\dagger}U_{j} = \delta_{ij} := \begin{cases} I~~~\text{if }i=j\\ \mathbf{0}~~~\text{if } i \neq j \end{cases}$$ Where $I$ is the identity matrix and $0$ is the zero matrix. This is just the inner-product $\langle U_i, U_j \rangle$ (if we defined the inner product in an analogous way for column vectors as we usually do) and the norm $\langle U_i, U_i \rangle = I := || U_i ||^2$.

Thanks for any assistance in checking.

  • 2
    $\begingroup$ I don't understand your question. Isn't the definition of a unitary matrix a complex square matrix with orthonormal columns? $\endgroup$ – user1551 Jun 21 '17 at 12:54
  • $\begingroup$ Question in edition? $\endgroup$ – Widawensen Jun 21 '17 at 13:23
  • 2
    $\begingroup$ The question is unclear. Please consider editing. :) $\endgroup$ – H. R. Jun 26 '17 at 8:48

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