An equality in matrix theory (also interesting in analysis)

Question

Let $M_{n\times n}$ be a real, symmetric matrix and $$ S=\left\{ \left(u_{1},\ldots,u_{k}\right)\Bigl|0\leq k\leq n,\left\{ u_{1},\ldots,u_{k}\right\} \textrm{ is an orthonormal subset in } \mathbb{R}^{n}\right\}. $$ Prove that $$ \sum_{\lambda\textrm{ are positive eigenvalues of }M}\lambda=\max_{S}\left\{ u_{1}Mu_{1}^{T}+\ldots+u_{k}Mu_{k}^{T}\right\} . $$

A very special case is also interesting: Let $\lambda_{1}$ and $\lambda_{2}$ are positive, prove that $$ \lambda_{1}\left(a_{11}^{2}+a_{21}^{2}\right)+\lambda_{2}\left(a_{12}^{2}+a_{22}^{2}\right)\leq\lambda_{1}+\lambda_{2}, $$ where $a_{11}a_{21}+a_{12}a_{22}=0$ and $a_{11}^{2}+a_{12}^{2}=a_{21}^{2}+a_{22}^{2}=1.$

Answer 1

The question in matrix format is to prove that $$ \sum \lambda_+(M)=\max_{U^TU=I}\operatorname{tr}U^TMU. $$ The unitary diagonalization $M=Q\Lambda Q^T$ reduces the problem to proving that $$ \sum \lambda_+(M)=\max_{U^TU=I}\operatorname{tr}U^T\Lambda U=\max_{U^TU=I}\operatorname{tr}\Lambda UU^T.\tag{*} $$ As @Tom Chen has answered, building the matrix $U$ of those columns of the identity matrix that correspond to positive $\lambda$'s we can prove that $$ \sum \lambda_+(M)\le\max_{U^TU=I}\operatorname{tr}U^T\Lambda U. $$ To prove the opposite we introduce the set of $n\times n$ matrices $$ {\cal W}=\{W\colon W=W^T,\, 0\le W\le I\} $$ where the inequalities are understood in the semidefinite sense, i.e. $W$ and $I-W$ are positively semidefinite matrices. Such matrices, in particular, have the diagonal elements being between $0$ and $1$ (as a simple conclusion by definition of positively semidefiniteness). We have $$ \operatorname{tr}\Lambda W=\sum_{i=1}^n\lambda_i W_{ii}. $$ The maximization over ${\cal W}$ is trivial - the best choice is to take $W_{ii}=1$ for a positive $\lambda_i$ and $0$ otherwise for all other $W_{ij}$. Thus, $$ \max_{W\in{\cal W}}\operatorname{tr}\Lambda W=\sum\lambda_+(M). $$ Finally, we rewrite the RHS in $(*)$ as $$ \max_{U^TU=I}\operatorname{tr}\Lambda UU^T $$ and notice that $W=UU^T\in{\cal W}$ if $U^TU=I$. Indeed, $W=W^T$ and $W\ge 0$ is clear. Furthermore, $U^TU=I$ $\Rightarrow$ $\|U\|=1$ $\Rightarrow$ $\|U^T\|=1$ $\Rightarrow$ $UU^T\le I$ $\Rightarrow$ $I-UU^T\ge 0$. Thus, $$ \sum\lambda_+(M)=\max_{W\in{\cal W}}\operatorname{tr}\Lambda W\ge \max_{U^TU=I}\operatorname{tr}\Lambda UU^T $$ Since the maximum over a subset is smaller. It gives the opposite inequality that proves $(*)$.

Answer 2

By properties of eigensystems, we have $Mu_i = \lambda_i u_i$ for orthonormal eigenvectors $\{u_1, \cdots, u_n\}$, hence $u_i^\intercal M u_i = \lambda_i$. For any $U \in S_M$, where $S_M \subseteq S$ is the restriction of $S$ to eigenvectors of $M$, we have $UMU^\intercal$ corresponds to some sum of a subset of eigenvalues of $M$. Clearly the maximum sum is the sum of the positive eigenvalues of $M$, and the conclusion follows.