AAA weights minimize the Loewner residual and interpolate support data

Let $v_1,\dots ,v_m$ be a full orthonormal right singular-vector basis for $L$, with singular values ${\sigma }_1\ge \cdots \ge {\sigma }_m\ge 0$ and ${\sigma }_m={\sigma }_{\min }(L)$. For any unit vector $v={\sum }_j{c}_j{v}_j$, orthonormality gives ${\sum }_j\mathrm{\mid }{c}_j{\mathrm{\mid }}^2=1$, and the SVD gives $\mathrm{\parallel }Lv{\mathrm{\parallel }}_2^2={\sum }_{j=1}^m{\sigma }_j^2\mathrm{\mid }{c}_j{\mathrm{\mid }}^2\ge {\sigma }_m^2\mathrm{.}$ The selected $w$ lies in the right singular subspace for ${\sigma }_m$, so equality holds. Hence $\mathrm{\parallel }Lw{\mathrm{\parallel }}_2={\sigma }_{\min }(L)$ and no unit vector gives a smaller Loewner residual.

The right singular subspace $E$ is nonzero. If every projection $P_E{e}_j$ of a coordinate basis vector vanished, then every vector of $E$ would be orthogonal to every coordinate basis vector and hence would be zero, a contradiction. Thus the least index with nonzero projection exists; the normalization and phase convention in [F1] choose a unique unit vector in $E$, also when the smallest singular value is repeated or zero.

On the successful-weight branch, every support weight is nonzero. By [F3], the barycentric quotient is defined and its value at each selected support node is the corresponding datum. The candidate is returned only after the algorithm's reduced-denominator pole check succeeds at all remaining sample points, so its finite residual there is defined.

Each iteration removes one previously remaining point and the budget satisfies $B\le \lfloor N\mathrm{/}2\rfloor$. Every branch either returns immediately or proceeds to the next integer $m\le B$, so the algorithm makes at most $B$ iterations and terminates. The minimized quantity in step 1.1 is $\mathrm{\parallel }Lw{\mathrm{\parallel }}_2$; the procedure separately records ${\eta }_m$ and makes no minimax or uniform-error claim.