The two minimization problems below are equivalent:
$\min\{\mathrm{trace}(AX^TBX): XX^T=I_n\}=\min\{\mathrm{trace}(AQ^T\tilde{B}Q): QQ^T=I_m\}$, where $A,\tilde{B}$ and $Q$ are square matrices of the same size, $A,B$ are also p.s.d and $X$ is a $m \times n, (m >n) $ rectangular matrix.
The solution to this latter minimization problem is well known: the extrema of the trace are the extrema of $\{\sum_{i=1}^m\lambda_i(A)\lambda_{\sigma(i)}(\tilde{B}): \sigma\in S_m\}$, where $\lambda_i(M)$ denotes the $i$-th eigenvalue of a real symmetric matrix $M$ and $S_m$ is the symmetric group of order $m$.
Suppose $A$ and $\tilde{B}$ are orthogonally diagonalized as $A=U\Lambda U^T$ and $\tilde{B}=V\Sigma V^T$, where $\Lambda = \mathrm{diag}(\lambda_1(A), \lambda_2(A), \ldots, \lambda_m(A))$ contains the eigenvalues of $A$ arranged in ascending order and $\Sigma$ is analogously defined, but the eigenvalues are arranged in descending order. That is, if $\lambda_1(B),\ldots,\lambda_n(B)$ are arranged in ascending order, and for $\Sigma=\mathrm{diag}\left(\lambda_n(B),\ldots,\lambda_1(B),0,\ldots,0\right)$
Then am looking for a 'detailed proof' that the minima is reached at an $X^*$ given by:
\begin{align} X^* &= VU^T \begin{bmatrix}I_n\\ 0_{(m-n)\times n}\end{bmatrix}. \end{align}
for my better understanding.