Find $\sup _\limits{Q \in M_{4\times 2} (\mathbb{R}), Q^{T} Q=I_{2}} \operatorname{tr}\left(Q^{T} A Q\right)$

Question

Let $$ A:=\left[\begin{array}{llll} 3 & 1 & 0 & 0 \\ 1 & 3 & 0 & 0 \\ 0 & 0 & 6 & 2 \\ 0 & 0 & 2 & 6 \end{array}\right] $$ Find $\sup _\limits{Q \in M_{4\times 2} (\mathbb{R}), Q^{T} Q=I_{2}} \operatorname{tr}\left(Q^{T} A Q\right)$, where $M_{4 \times 2}(\mathbb{R})$ represents the set of all matrices of size $4\times 2$.

I know that $\mathrm{tr}A=\sum _i A_{ii}$, but how can we deal with this upper bound? It is obvious that $Q^T AQ$ is a $2\times 2$ matrix, but I don't know how does the condition $Q^TQ=I_2$ help. Also, are there any backgrounds for this problem? I seldom see (linear algebra) problems asking the uppper bound for a trace and I hope I could get further information about these kind of problems (if possible).

Answer 1

$A$ is positive definite and its four eigenvalues are $2,4,4,8$. Von Neumann's trace inequality gives $$ \operatorname{tr}(Q^TAQ)\le\sum_{i=1}^2\sigma_i(Q^T)\sigma_i(AQ)=\sum_{i=1}^2\sigma_i(A)=\sum_{i=1}^2\lambda_i^\downarrow(A)=8+4=12. $$ Alternatively, note that $Q^TAQ$ is a principal submatrix of $U^TAU$ for some orthogonal matrix $U$. By Cauchy's interlacing inequality for bordered submatrices of Hermitian matrices or by Courant-Fischer minimax inequality, we have $\lambda_i^\downarrow(Q^TAQ)\le\lambda_i^\downarrow(U^TAU)=\lambda_i^\downarrow(A)$. Therefore $\operatorname{tr}(Q^TAQ)=\sum_{i=1}^2\lambda_i^\downarrow(Q^TAQ)\le\sum_{i=1}^2\lambda_i^\downarrow(A)=12$.

Obviously, equalities hold in the above when the two columns of $Q$ are two unit eigenvectors corresponding to the eigenvalues $8$ and $4$ respectively.

Answer 2

Here is a more elementary solution.
Let $Q=\begin{bmatrix}Q_1\\Q_2\end{bmatrix}$. Then $$I_{2\times2}=Q^TQ=\begin{bmatrix}Q_1^T Q_2^T\end{bmatrix}\begin{bmatrix}Q_1\\Q_2\end{bmatrix}=Q_1^TQ_1+Q_2^TQ_2$$ $$Q^TAQ=\begin{bmatrix}Q_1^T Q_2^T\end{bmatrix}\begin{bmatrix}B&0\\0&2B\end{bmatrix}\begin{bmatrix}Q_1\\Q_2\end{bmatrix}=Q_1^TBQ_1+2Q_2^TBQ_2$$ Note that $Q_1^TQ_1$ and $Q_2^TQ_2$ are simultaneously diagonalizable, with non-negative eigenvalues that add to one, i.e., $Q_1^TQ_1=PD_1P^T$, $Q_2^TQ_2=P(I-D_1)P^T$ with $P$ orthogonal.
Since the problem is to maximize the trace of the sum, both terms of which are similar to each other, it is optimal to choose $Q_1=0$. Then $Q_2$ is orthogonal and the maximum trace is $2\mathrm{tr}B=2\times6=12$.