How can I solve this maximization problem with inequality constraint on the Frobenius norm?

Question

$$ \max_{C\in \mathbb{S}^n} x^TCx\\ s.t. \|C\|_{tr}^2\leq b $$ where $\|C\|_{tr}^2=tr(C^TC)=tr(CC)$, $C$ is a symmetric matrix.

Answer 1

Through some calculations, I have an idea as follows: $$ L=x^TCx-\lambda(\|C\|_{tr}^2-b) $$ Calculating $\frac{\partial L}{\partial C}$ and $\frac{\partial L}{\partial \lambda}$, we have $$ \frac{\partial L}{\partial C}=xx^T=\lambda C\\ \frac{\partial L}{\partial \lambda}=tr(CC^T)=b $$ that is $$ C=\frac{xx^T}{\lambda} $$ plugging the term into the second equation, we have $$ tr(xx^T/\lambda)=\sqrt{b}\\ tr(xx^T/\lambda)=1=\sum{x_i^2}/\lambda\\ \lambda = \|x\|^2=\sum{x_i^2} $$ Hence, the solution is $$ C^*=\sqrt{b}*\frac{xx^T}{\|x\|^2} $$

Answer 2

Rephrasing slightly, given vector ${\rm a} \in \Bbb R^n \setminus \{ \Bbb 0_n \}$ and scalar $b >0$,

$$\begin{array}{ll} \text{maximize} & {\rm a}^\top {\rm X} \, {\rm a}\\ \text{subject to} & \| {\rm X} \|_{\text{F}} \leq b\end{array}$$

Using the cyclic property of the trace, the Cauchy-Schwarz inequality and $\| {\rm a} {\rm a}^\top \|_{\text{F}} = \| {\rm a} \|_2^2$,

$${\rm a}^\top {\rm X} \, {\rm a} = \mbox{tr} \left( {\rm a}^\top {\rm X} \, {\rm a} \right) = \mbox{tr} \left( {\rm a} {\rm a}^\top {\rm X} \right) = \langle {\rm a} {\rm a}^\top, {\rm X} \rangle \leq \| {\rm a} {\rm a}^\top \|_{\text{F}} \underbrace{\| {\rm X} \|_{\text{F}}}_{\leq b} \leq b \| {\rm a} \|_2^2$$

where the maximum is attained at

$$\bar{{\rm X}} := b \underbrace{\frac{{\rm a} {\rm a}^\top}{\,\,\,\| {\rm a} {\rm a}^\top \|_{\text{F}}}}_{=: {\rm P}_{\rm a}} = b \left( \frac{{\rm a}}{\,\,\,\| {\rm a} \|_2} \right) \left( \frac{{\rm a}}{\,\,\,\| {\rm a} \|_2} \right)^\top = b \, {\rm P}_{\rm a}$$

where ${\rm P}_{\rm a}$ is a rank-$1$ projection matrix.