We have an SPD problem defined as: $$\begin{aligned} v^*=\max{} & \operatorname{tr}(V X)\\ \text{s.t. } & \|V\|_2\le1 \\ &V,X\in\mathcal{S}^n \end{aligned}$$ where $\mathcal{S}^n$ is a space containing all $n\times n$ symmetry matrices. $X$ is given. $\|V\|_2$ is the spectral norm (http://mathworld.wolfram.com/SpectralNorm.html)
How to prove that the maximum $v^*=\sum_{i=1}^n |\lambda_i|$, while $\lambda_i$ is the eigenvalues of $X$.
My Trying:
As hinted by @Brian Borchers, $X$ can be decomposed as $X=PDP^T$ in which $P$ is an orthogonal matrix.
$$tr(VX)=tr(VPDP^T)=tr(P^TVPD)$$
I guess if we can prove $tr(AD)=\sum_{i=1}^n |\lambda_i|$, the problem will be solved. But how to prove this? and how to connect this with $tr(P^TVPD)$?
Here is a rather tedious answer:
It is clear that we can take $X$ to be diagonal with entries $\lambda_k$. By choosing $V$ to be diagonal with entries $\operatorname{sgn} \lambda_k$ we get $v^* \ge \sum_k |\lambda_k|$.
To establish an upper bound, note that we can take $V$ to have the form $U\Sigma U^T$ where $U$ is orthogonal and $\Sigma$ is diagonal with diagonal entries $|\sigma_k| \le 1$. We can write $V= \sum_i \sigma_i u_i u_i^T$ to get $\operatorname{tr} (XV) = \sum_i \sigma_i \operatorname{tr} (Xu_i u_i^T) = \sum_i \sigma_i u_i^T X u_i = \sum_i \sigma_i \sum_k \lambda_k [u_i]_k^2 = \sum_k \lambda_k \sum_i \sigma_i [u_i]_k^2 $, and so $\operatorname{tr} (XV) \le \sum_k |\lambda_k| \sum_i |\sigma_i| [u_i]_k^2 \le \sum_k |\lambda_k|$.
Hence we have the desired result.