Leverage scores are basis-independent and sum to the dimension

Let $Q^{\mathrm{\prime }}$ be another orthonormal basis matrix of $U$, and set $W=Q^{\ast }{Q}^{\mathrm{\prime }}$. Since each column of $Q^{\mathrm{\prime }}$ lies in $U=\mathrm{Ran}(Q)$, the orthogonal projection $Q{Q}^{\ast }$ fixes it, so $Q^{\mathrm{\prime }}=Q{Q}^{\ast }{Q}^{\mathrm{\prime }}=QW$. Thus $W^{\ast }W=Q^{\mathrm{\prime }\ast }Q{Q}^{\ast }{Q}^{\mathrm{\prime }}=Q^{\mathrm{\prime }\ast }{Q}^{\mathrm{\prime }}=I_d$. The square matrix $W$ is unitary, and right multiplication by $W$ preserves the Euclidean norm of every row; hence $\mathrm{\parallel }{e}_i^{\ast }{Q}^{\mathrm{\prime }}{\mathrm{\parallel }}_2^2=\mathrm{\parallel }{e}_i^{\ast }Q{\mathrm{\parallel }}_2^2$ for every $i$.

Writing $q_j$ for column $j$ of $Q$, finite-sum reordering gives ${\sum }_{i=0}^{n-1}{\mathrm{\ell }}_i={\sum }_i{\sum }_{j=0}^{d-1}\mathrm{\mid }{Q}_{ij}{\mathrm{\mid }}^2={\sum }_j\mathrm{\parallel }{q}_j{\mathrm{\parallel }}_2^2=d$, because every column is a unit vector.

Put $P=Q{Q}^{\ast }$. Then $P=P^{\ast }$ and $P^2=P$ by [F2], so $e_i=P{e}_i+(I-P)e_i$ is an orthogonal decomposition. Also ${\mathrm{\ell }}_i=\mathrm{\parallel }{Q}^{\ast }{e}_i{\mathrm{\parallel }}_2^2=\mathrm{\parallel }Q{Q}^{\ast }{e}_i{\mathrm{\parallel }}_2^2=\mathrm{\parallel }P{e}_i{\mathrm{\parallel }}_2^2$, using $Q^{\ast }Q=I_d$. Therefore $0\le {\mathrm{\ell }}_i\le \mathrm{\parallel }{e}_i{\mathrm{\parallel }}_2^2=1$, and ${\mathrm{\ell }}_i=0$ exactly when $Q^{\ast }{e}_i=0$, which is exactly when row $i$ of $Q$ is zero.