AAA recovers a scalar transfer-function pole from four samples

The data in order are $(-1\mathrm{/}2,-1,1,1\mathrm{/}2)$, with mean zero. Thus ${\eta }_0=1>\tau =0$, so the initial tolerance test does not return. Their largest absolute residuals are tied at $z=1$ and $z=3$; [F1] selects the earlier sample $z=1$. The first Loewner matrix is the single column $(-1\mathrm{/}2,1,1\mathrm{/}2{)}^T$, since its entries are $\frac{H(0)-H(1)}{0-1}=-\frac{1}{2},\frac{H(3)-H(1)}{3-1}=1,\frac{H(4)-H(1)}{4-1}=\frac{1}{2}\mathrm{.}$ Since its right domain is one-dimensional, the selected unit weight is $1$ under the phase rule. Its Householder reduction has a single diagonal entry; by [F5] there is only a singleton block, so the SVD call converges without using a sweep. The one-support quotient is the constant $-1$. Its remaining residual magnitudes are $1\mathrm{/}2,2,3\mathrm{/}2$ at $z=0,3,4$, respectively, so ${\eta }_1=2>\tau$ and the second support is $z=3$.

The remaining sample points are $0$ and $4$. In that row and support order, the Loewner matrix is $L=\left[\begin{array}{cc}-1\mathrm{/}2& 1\mathrm{/}2\\ 1\mathrm{/}2& -1\mathrm{/}2\end{array}\right]\mathrm{.}$ The rows are negatives of one another and $L(1,1{)}^T=0$, so $L$ has rank one, its smallest singular value is zero, and its right nullspace is spanned by $(1,1)$. The orthogonal projection of $e_1$ onto this line is $(1\mathrm{/}2,1\mathrm{/}2)$; normalizing it and applying the positive-real phase rule [F1] gives $w=(1,1)\mathrm{/}\sqrt{2}$, whose components are both nonzero.

At the second support addition the Loewner matrix in step 1.2 is $L=\left[\begin{array}{cc}-1\mathrm{/}2& 1\mathrm{/}2\\ 1\mathrm{/}2& -1\mathrm{/}2\end{array}\right]\mathrm{.}$ Its first column has norm $1\mathrm{/}\sqrt{2}$ and its second is its negative. By [F4] and [F7], the left Householder sends the first column to $(1\mathrm{/}\sqrt{2},0{)}^T$ and the second to $(-1\mathrm{/}\sqrt{2},0{)}^T$. The one-dimensional right active scalar is $-1\mathrm{/}\sqrt{2}$; its phase is $-1$, its reflector vector is $-1$, and its Householder factor is $[-1]$. Thus the bidiagonal reduction returns $B_0=\left[\begin{array}{cc}1\mathrm{/}\sqrt{2}& 1\mathrm{/}\sqrt{2}\\ 0& 0\end{array}\right]\mathrm{.}$ Then $B_0^{\ast }{B}_0=\left[\begin{array}{cc}1\mathrm{/}2& 1\mathrm{/}2\\ 1\mathrm{/}2& 1\mathrm{/}2\end{array}\right]\mathrm{.}$ Its eigenvalues are $0$ and $1$, each at distance $1\mathrm{/}2$ from the trailing diagonal entry $1\mathrm{/}2$, so [F5] chooses shift $0$. The initial shift vector is $(1\mathrm{/}2,1\mathrm{/}2{)}^T$. The real Givens factor $G=\frac{1}{\sqrt{2}}\left[\begin{array}{cc}1& -1\\ 1& 1\end{array}\right]$ satisfies $G^{\ast }(1\mathrm{/}2,1\mathrm{/}2{)}^T=(1\mathrm{/}\sqrt{2},0{)}^T$, and $B_{0}G=\mathrm{diag}(1,0)$. The left rotation is the identity because the subdiagonal entry is already zero. Thus one block sweep makes the superdiagonal zero; the next zero-tolerance pass deflates it and sees only singleton blocks before checking the exhausted budget, so the SVD call succeeds with $S=1$.

Substitution into the barycentric quotient [F3] gives $r(z)=\frac{-1\mathrm{/}(z-1)+1\mathrm{/}(z-3)}{1\mathrm{/}(z-1)+1\mathrm{/}(z-3)}=\frac{1}{z-2}\mathrm{.}$ The denominator reduces to a nonzero constant times $z-2$, so its sole finite pole is $2$. The values at $z=0,4$ are $-1\mathrm{/}2,1\mathrm{/}2$; the support values at $z=1,3$ are $-1,1$ by [F2]. Thus the quotient matches all four data, and has no pole at either remaining sample point.

All four sample values match, so the second candidate has maximum sampled residual ${\eta }_2=0=\tau$. By [F1], the tolerance branch returns the exact quotient.