Finite-dimensional trace minimization under orthonormal constraints

If $k=1$, every feasible $Q$ is a unit column $q$, so $\mathrm{tr}(Q^{T}AQ)=q^{T}Aq=R_A(q)$. By [F2], its minimum is the smallest eigenvalue, which is ${\lambda }_1$ in the increasing order of the Statement.

Suppose $k\ge 2$. By [F1], choose an orthonormal eigenbasis $u_1,\dots ,u_n$ with eigenvalues ${\lambda }_1\le \cdots \le {\lambda }_n$, and let $U$ be the matrix with these columns. For any feasible $Q$, put $C=U^{T}Q$; then $C^{T}C=I_k$ and $Q^{T}AQ=C^T\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)C$. Writing $c_{ij}$ for the entries of $C$ and $p_i={\sum }_{j=1}^k{c}_{ij}^2$, direct expansion of the diagonal sum gives $\mathrm{tr}(Q^{T}AQ)={\sum }_{i=1}^n{\lambda }_i{p}_i$.

The matrix $P=C{C}^T$ is symmetric and idempotent because $C^{T}C=I_k$. Hence it is the orthogonal projection onto the column space of $C$, and $p_i=e_i^{T}P{e}_i=\mathrm{\parallel }P{e}_i{\mathrm{\parallel }}_2^2$ lies in $[0,1]$. Also ${\sum }_i{p}_i={\sum }_{i,j}{c}_{ij}^2={\sum }_j(C^{T}C{)}_{jj}=k$.

Using ${\sum }_i{p}_i=k$, subtract the bottom-$k$ eigenvalue sum to obtain ${\sum }_i{\lambda }_i{p}_i-{\sum }_{i=1}^k{\lambda }_i={\sum }_{i=1}^k({\lambda }_k-{\lambda }_i)(1-p_i)+{\sum }_{i=k+1}^n({\lambda }_i-{\lambda }_k)p_i\ge 0.$ Every term on the right is nonnegative. Taking $Q=(u_1,\dots ,u_k)$ gives equality, so the stated minimum is attained. This proves the claim for $k\ge 2$ as well as the $k=1$ case of step 1.1.