Interactive figure (draft for §1.1). The matrix \(A_\varepsilon=[\,\mathbf a\ \ \ \mathbf a+\varepsilon\mathbf b\,]\) with \(\mathbf a=(1.1,\,0.9,\,2.1,\,1.9)\) and \(\mathbf b=(-1,\,1,\,-1,\,1)\); for \(\varepsilon=0.2\) it is the matrix of Example 11.1. The outcomes are \(\mathbf y=A_\varepsilon\begin{bmatrix}1\\1\end{bmatrix}+\mathbf e\) with the errors \(\mathbf e=(0.1,\,-0.1,\,0.1,\,-0.1)\) of Example 11.1, so the rule \((1,1)\) and the errors stay the same and only the second column changes. Left: the level curves \(\|A_\varepsilon\boldsymbol\beta-\mathbf y\|=0.1,\ 0.2,\ 0.4,\ 0.8,\ 1.6\) (the first three are those of Figure 11.1). Right: the graph of \(\|A_\varepsilon\boldsymbol\beta-\mathbf y\|^2\) above the same curves, cut off at the height \(1.6^2\) and at the distance \(1.4\) from the rule along the floor (red); drag the picture to turn it. The box gives the condition number \(\kappa(A_\varepsilon)=\sigma_1/\sigma_2\): every level curve is an ellipse \(\kappa(A_\varepsilon)\) times as long as it is wide, and the bowl is \(\kappa(A_\varepsilon^{\mathsf T}\!A_\varepsilon)=\kappa(A_\varepsilon)^2\) times as curved across its floor as along it. Since \(\mathbf e=-0.1\,\mathbf b\) and \(A_\varepsilon\begin{bmatrix}1\\-1\end{bmatrix}=-\varepsilon\,\mathbf b\), the lowest point is \(\hat{\boldsymbol\beta}=(1,1)+\dfrac{0.1}{\varepsilon}\,(1,-1)\): the error \(0.1\) is divided by \(\varepsilon\). The box also gives its norm, \(\|\hat{\boldsymbol\beta}\|=\sqrt{2+0.02/\varepsilon^{2}}\); the rule has norm \(\sqrt2\approx1.41\). As \(\varepsilon\) decreases the bowl stretches into a canyon, its floor flattens, and the lowest point moves away along the floor. For \(\varepsilon=0\) the two columns are equal and the bowl has become a trough like the one in Figure 9.10: its floor is flat, and every point of the floor is a lowest point; the shortest of them has norm \(1.42\).