I am trying to understand the proof given for a quadratic programming problem with quadratic constraints:
The optimization problem is:
where $\tilde{\beta}$ is $p \times 1$ and $X$ is $n \times p$. It states that if $\tilde{\beta}^\top \tilde{\beta} \leq c'$, $Tr(MM^\top)$ can reach its unrestricted minimum $0$ at $\hat{M}$ = $\bf0$.
I don't understand how the constraint has become irrelevant if $\tilde{\beta}^\top \tilde{\beta} \leq c'$.
We know that the trace of such a matrix $$ Tr(\mathbf{M} \mathbf{M}^T) = \sum_{r=1}^{p} \mathbf{m}_{r}^T \mathbf{m}_{r} \geq 0 $$ is nonnegative. As inserting $\hat{\mathbf{M}} = \mathbf{0}$ yields the trace $$ Tr(\mathbf{0} \mathbf{0}^T) = \sum_{r=1}^p 0 = 0 $$ Using the first equation we know that there exists no matrix $\bar{M}$ such that $$ Tr(\bar{M}\bar{M}^T) < Tr(\hat{M} \hat{M}^T) = 0 $$ Therefore, if $\hat{M}$ is allowed by the constraint this must be a solution to the minimization problem. By inserting $\hat{\mathbf{M}} = \mathbf{0}$ into the constraint $$ \begin{align*} \tilde{\beta}^T(\hat{M} X - I_p)^T (\hat{M} X - I_p)\tilde{\beta} &\leq c' \\ \tilde{\beta}^T(\mathbf{0} X - I_p)^T(\mathbf{0} X - I_p)\tilde{\beta} &\leq c' \\ (-1)^2 \tilde{\beta}^T(I_p)^T(I_p)\tilde{\beta} &\leq c' \end{align*} $$ we see that $\hat{\mathbf{M}} = \mathbf{0}$ is allowed if $\tilde{\beta}^T\tilde{\beta} \leq c'$.
PS: I apologize for not being consistent on the use of boldface.