$u, v \in \R^n, u \neq 0, v \neq 0 \implies \rank{ u \transpose{ v }} = 1 $.
Given a quasi-Newton methods' approximation of the inverse Hessian $\b{H}_k $. We compute $\b{H}_{k+1} $ by adding a correction to $\b{H}_k $.
We introduce the symmetric correction term $a_k \b{z} ^{(k)} \transpose{ \b{z} ^{(k)}}$, where $a_k \in \R$ and $\b{z} \in \R^n $. Therefore, the update equation is
$$\b{H}_{k+1} = \b{H}_k + a_k \b{z} ^{(k)} \transpose{ \b{z} ^{(k)}}$$
Note that
$$\rank{ \b{z}^{(k)} \transpose{ \b{z}^{(k)}}} = \rank{{\Meq{ z_1^{(k)}}{ \vdots }{ z_n^{(k)}}} {\Mqe{ z_1^{(k)}}{ … }{ z_n^{(n)}}}} = 1 $$
If $\b{H} _k $ is symmetric, $\b{H}_{k+1} $ is also symmetric.
For the rank one algorithm applied to the quadratic with Hessian $Q = \transpose{ Q }$, we have $\b{H}_{k+1} \Delta \b{g}^{(i)} = \Delta \b{x}^{(i)}, 0 \leq i \leq k $.