An exact Bfgs secant update on a two-variable quadratic

The gradient is $\mathrm{\nabla }f(x)=Ax$, so $g_0=(2,3{)}^T$ and the direction in [F2] is $p_0=-g_0=(-2,-3{)}^T$. Along this ray, ${\varphi }^{\mathrm{\prime }}(t)=g_0^T{p}_0+t{p}_0^{T}A{p}_0=-13+35t$, so the exact minimizer is ${\alpha }_0=13\mathrm{/}35$.

At this step ${\varphi }^{\mathrm{\prime }}({\alpha }_0)=0$, which satisfies the strong curvature test for $c_2=2\mathrm{/}5$. The quadratic identity $\varphi (\alpha )=\varphi (0)+\alpha {\varphi }^{\mathrm{\prime }}(0)+\frac{1}{2}{\alpha }^2{p}_0^{T}A{p}_0$ gives $\varphi ({\alpha }_0)-\varphi (0)=-169\mathrm{/}70\le -({10}^{-4})(169\mathrm{/}35)={10}^{-4}{\alpha }_0{\varphi }^{\mathrm{\prime }}(0)$, so the strong Wolfe Armijo test also holds.

The step and gradient change are $s_0={\alpha }_0{p}_0$ and $y_0=A{s}_0={\alpha }_0(-4,-9{)}^T$. Hence $s_0^T{y}_0=35{\alpha }_0^2>0$ and $s_0^T{B}_0{s}_0=13{\alpha }_0^2>0$.

Substitution in [F1] gives $B_1=I-\frac{1}{13}\left(\begin{array}{cc}4& 6\\ 6& 9\end{array}\right)+\frac{1}{35}\left(\begin{array}{cc}16& 36\\ 36& 81\end{array}\right)=\frac{1}{455}\left(\begin{array}{cc}523& 258\\ 258& 1193\end{array}\right)\mathrm{.}$ Multiplication yields $B_1{s}_0={\alpha }_0(-4,-9{)}^T=y_0$. The positive curvature from step 1.3 and [F3] also imply $B_1$ is SPD.