I'm having trouble understanding how to update this formula (it's Broyden's method in multiple dimensions) by solving the following equations:
$$A^{(m)}(x^{(m)}-x^{(m-1)}) = f(x^{(m)})-f(x^{(m-1)})$$ and $$A^{(m)}v=A^{(m-1)}v,$$ where $A^{(m)}$ is a matrix, $x^{(m)}$ is the $m^{th}$ iteration, and for all vectors v orthogonal to $(x^{(m)}-x^{(m-1)})$.
These two equations should lead to $$A^{(m)}=$$ $$A^{(m-1)} + \frac{f(x^{(m)})-f(x^{(m-1)})-A^{(m-1)}(x^{(m)}-x^{(m-1)})}{(x^{(m)}-x^{(m-1)})^T(x^{(m)}-x^{(m-1)})}(x^{(m)}-x^{(m-1)})^T$$.
How does one find the last equation?
You want $A_{+1}s=d$ while $A_{+1}v=Av$ for $v\perp s$. The second condition means that $A_{+1}$ is a rank-one update of $A$, $$ A_{+1}=A+ws^\top. $$ Now insert this into the first condition $$ d=A_{+1}s=As+ws^\top s\implies w = \frac{d-As}{s^\top s}. $$
The orthogonality comes from the minimization in the Frobenius norm, $A_{+1}$ is there the closest matrix to $A$ such that the first condition holds. As Lagrange function this gives (using the notation $A:B={\rm trace}(AB^T)$) \begin{align} L(X,w)&=\frac12\|X-A\|_F^2-w^\top(Xs-d) \\ &=\frac12(X-A):(X-A)-(ws^\top):X+w^\top d \\ &=\frac12\|X-A-ws^\top\|^2-(ws^\top):A-\frac12(w^\top s)_F^2+w^\top d \end{align} with the stationarity condition $$ 0=(X-A)-ws^\top $$ which implies the invariance on the space orthogonal to $s$.