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There seems consensus that the number of free real parameters of an $n \times n$ Hermitian matrix $M$ is $n^2$; see e.g. this post and also this short note.

On the other hand, such a matrix has a spectral decomposition $M = U D(\lambda) U^*$, where $U$ is an $n\times n$ unitary matrix, $U^*$ its adjoint, and $D(\lambda)$ is the diagonal matrix with entries $\lambda \in \mathbb{R}^n$. There are also posts, such as this one, suggesting that the number of free real parameters of an $n\times n$ unitary matrix is also $n^2$. Yet for "generic" Hermitian matrices $M$, the entries of $\lambda$ are distinct, and the spectral decomposition $M = U D(\lambda) U^*$ will be unique. Thus, specifying $M$ would seem to uniquely specify $n^2 + n$ real parameters, implying that the number of free real parameters of $M$ should be $n^2 + n$.

I suspect I'm missing something basic that would resolve this apparent discrepancy. Any suggestions would be appreciated!

(I'm also interested in the rank $r$ version of this question; specifying $r$ orthonormal vectors (by the same logic as the above post), and specifying a rank $r$ Hermitian matrix both appear to require $2nr - r^2$ real parameters, yet the Hermitian matrix also specifies $r$ eigenvalues.)

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  • $\begingroup$ The decomposition $UDU^\ast$ is not unique. In fact, $UDU^\ast=U\Lambda D\Lambda^\ast U^\ast$ for every diagonal matrix $\Lambda$ whose diagonal entries lie on the unit circle. So, one can always assume that the top row of $U$ is real. This saves up $n$ real parameters for the imaginary parts of the entries on the top row. $\endgroup$
    – user1551
    Jan 19, 2022 at 21:22
  • $\begingroup$ Aha, I had failed to see that eigenvectors are only unique up to a phase. (For others: see this post.) Thank you! $\endgroup$
    – Rafael Frongillo
    Jan 20, 2022 at 3:37

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I think you can refer to the literature about the Lie group and Lie algebra. Particularly, any unitary $U$ can be uniquely represented by its derivative $H$ in the Riemman space in the exponential formula, namely, $U=e^{-itH}$, where $t$ is a fixed constant and $H$ is a Hermitian matrix. In this regard, it is easy to see that a unitary $U$ has the same free real parameters as the Hermitian matrix $H$.

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