Was hoping you'd answer this, very nice!

Complete and insightful answer! Thank you a lot!

@A.Γ. It seems that the derivation is applicable for any choice of weight matrix that satisfies the secant equation. What is the role of the average Hessian in the BFGS or DFP update? As per the book by Nocedal, the average Hessian ensures that the solution does not depend on the unit of the problem. But, I do not see where the unit of the problem plays a role in your derivative. Please elaborate on this point.

For at least one symmetric matrix $\hat{B}$ so that $\hat{B} u = u $, I am not sure how one can guarantee the existence of a matrix $u_{\perp}$ with orthonormal columns so that $$u_{\perp}(\hat{B} - B_{k}) u_{\perp} = 0 \quad \text{ and } \quad u_{\perp}^{T} u = 0.$$ You used this assumption to prove that the red term is zero in the solution set.

@R.W.Prado In this optimization, $\hat B_k$ and $u$ are fixed, the basis $u_\bot$ is picked as an orthogonal complement to $u$ and, thus, fixed as well, and $\hat B$ is the optimization variable. The red term being zero is the optimal choice of $\hat B$, not $u_\bot$. It says that $\hat B$ should coincide with $\hat B_k$ on the $u_\bot$-subspace.

You should make a tiny 1×1 example to see the problem. I can't elaborate more than saying “not necessarily the non-constant and positive term of the objective function vanishes in the constrained set, even if the non-constant term vanishes outside the constrained set”. I tried to be very specific here. However, there is no need for it. My first comment was too elaborate because I REALLY was trying to fix your proof.

@R.W.Prado That's a pity that you cannot elaborate more as the situation is pretty simple here: given $u$ one complements it to an orthonormal basis, no doubts that it exists. In the new basis, the optimization looks like $$ \min_{X}\left\|\begin{bmatrix}A & B\\C & D\end{bmatrix}-\begin{bmatrix}0 & 0\\0 & X\end{bmatrix}\right\|_F $$ with the optimal $X$ being $X=D$.

Your problem is not analogous to the original problem.To be analogous,it is necessary to consider the optimization problem: $$\text{min}_{X}\quad \left\| \begin{bmatrix} A & B \\ C & D \end{bmatrix} - \begin{bmatrix} 0 & 0 \\0 & X \end{bmatrix} \right\| \quad \text{s.t} \quad X = X^T \quad \text{and} \quad X \hat{s} = \hat{s},$$ with $\hat{s}, A, B, C, D$ fixed. Compare with your second unnumbered problem. Again, not necessarily the non-constant and positive term of the objective function vanishes in the constrained set, even if the non-constant term vanishes outside the feasible set.

@R.W.Prado No $X\hat s=\hat s$, this constraint is in the $A$-part that corresponds to the $u$-subspace. The $D$-part corresponds to the $u_\bot$-subspace that is totally free. $\|D-X\|_F$ is to be minimized by the totally free (symmetric) $X$.

I doubt that anyone is understanding what you are saying. Change of coordinates in matrix space is a linear transformation which takes a matrix and send to a matrix and this transformation does not preserve symmetry of a matrix.

As change of coordinates does not preserve the symmetry of a matrix, the $X = X^T$ is not a constraint anymore in the new associated problem.

I am afraid that you need to complement your answer and explain how exactly one constraint disappears from the original problem.