Hi guys I was reading a paper and it was left as easy to show that (for $y_k,s_k \in \mathbb R^n$ and $B_k \in \mathbb R^{n \times n}$)
$$B_{k+1} = B_k + \frac{ y_k (y_k-B_k s_k)^T+ (y_k-B_k s_k) y_k^T }{y_k^Ts_k } -\frac{\langle y_k- B_k s_k, s_k \rangle}{(y_k^Ts_k)^2 } y_k y_k^T $$
implies
$$B_{k+1} = B_k -\frac{B_ks_ks_k^TB_k}{s_k^TB_ks_k}+ \gamma y_k y_k^T$$
For $\gamma = \frac{1}{y_k^T s_k}$. Where this has the assumptions that $s_k = x_{k+1}-x_k$ and $y_k = \nabla f(x_{k+1})- \nabla f(x_{k})$. $B_k$ is symmetric positive definite matrix. Also I think we can assume $B_{k+1}s_k = y_k$
I want to show this is true: My attempt
$$B_{k+1} = B_k + \frac{ y_k y_k^T-y_ks_k^TB_k+ y_k y_k^T-B_k s_k y_k^T }{y_k^Ts_k } -\frac{\langle y_k, s_k \rangle}{(y_k^Ts_k)^2 } y_k y_k^T +\frac{\langle B_k s_k, s_k \rangle}{(y_k^Ts_k)^2 } y_k y_k^T$$
Where I just distributed and used the prop of inner product. Now
$$B_{k+1} = B_k + \frac{ y_k y_k^T-y_ks_k^TB_k+ y_k y_k^T-B_k s_k y_k^T }{y_k^Ts_k } -\frac{ y_k^Ts_k }{(y_k^Ts_k)^2 } y_k y_k^T +\frac{ s_k^T B_k s_k }{(y_k^Ts_k)^2 } y_k y_k^T$$
Using the inner product. $$B_{k+1} = B_k + \frac{ y_k y_k^T-y_ks_k^TB_k+ y_k y_k^T-B_k s_k y_k^T - y_k y_k^T}{y_k^Ts_k } +\frac{ s_k^T B_k s_k }{(y_k^Ts_k)^2 } y_k y_k^T$$
Now I am left with $$B_{k+1} = B_k + \frac{ y_k y_k^T-y_ks_k^TB_k+ -B_k s_k y_k^T }{y_k^Ts_k } +\frac{ s_k^T B_k s_k }{(y_k^Ts_k)^2 } y_k y_k^T$$
Now I am unsure how to cancel the last term because it has a square term I need to get rid off.
The first update formula is not BFGS. It is DFP update. They are different algorithms, it is not possible to rewrite one to another. However, there is a certain symmetry between them: if you look at the DFP update for the inverse Hessian $H_k=B_k^{-1}$ in the link above, you may notice that the BFGS update for $B_k$ looks very similar - you just need to replace all $B$ with $H$, $y$ with $s$ and $s$ with $y$. In this sense, it is easy to get the BFGS update, but from the inverse Hessian DFP update.