Assume a binary matrix $A\in\{0,1\}^{N\times K}$, where $K\leq N$, and there is one and only one non-zero element in each row of $A$, i.e., $A1^{K\times 1}=1^{N\times 1}$. We obtain a matrix $W=AA^T$.
To recover $A$ from $W$, we relax the binary constrain in $A$, and use $H\in [0,1]^{N\times K}$ denote the relaxed matrix, and try to obtain $H^*$ using the following two formulations:
1)\begin{align} H_1^*=\ &\mathop{\mathrm{arg\,min}}_H \|W-HH^T\|_F^2\\ \text{s.t.}\quad&H\in[0,1]^{N\times K}\\ & H1^{K\times1}=1^{N\times 1} \end{align}
2)\begin{align} H_2^*=\ &\mathop{\mathrm{arg\,max}}_H \langle W,HH^T \rangle\\ \text{s.t.}\quad&H\in[0,1]^{N\times K}\\ &H1^{K\times1}=1^{N\times 1} \end{align} where $\langle A, B \rangle=\sum\limits_{i,j} A(i,j)B(i,j)$.
Are the solutions to the two optimization problem, i.e., $H_1^*$ and $H_2^*$, equivalent? If yes, how to prove it?
Hints: Let $B=HH^T$. Use the following identities and the constraints in your optimization problem. Try to prove that every feasible solution in first one is a feasible solution in second one and vice-versa $$||A||_F^2 = \operatorname{trace}(A^TA)$$ $$<A,B> = \sum_{i,j}A(i,j)B(i,j) = \operatorname{trace}(A^TB)$$ $$||W-B||_F^2=\operatorname{trace}((W-B)^T(W-B))=\operatorname{trace}(W^TW)-2<B,W>+\operatorname{trace}(A^TA)$$