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Consider the Lie subalgebra of $\mathfrak{u}(n)$ given by $L = \{A \in \mathfrak{u}(n): \sum_{j=1}^n A_{ij} = 0 \text{ for all } i \in [n]\}$. What is its dimension? What does the corresponding Lie subgroup of $U(n)$ look like?

Edit: I believe the dimension is $(n-1)^2$, by explicitly determining the number of independent constraints. I highly suspect the subgroup is $U(n-1)$, acting on the subspace $\{z_1 + \ldots + z_n = 0\}$.

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Yup. Your condition is $M \vec v = 0$ for $\vec v = [1 1 1\ldots 1]$. Exponentiating, that's $\exp(M) \vec v = \vec v$.

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