Finding a matrix $Q \in \mathbb{R}^{d\times r}$ such that $Q^\top Q=I_r$ and $(QQ^\top)_{ii}=h_{ii}$ Given $\{h_{ii}\}_{i=1}^d$, where $h_{ii}\in[0,1],$ and $\displaystyle\sum_{i=1}^d h_{ii}=r<d,$ does there exist a matrix $Q\in\mathbb{R}^{d\times r},$ s.t. 
$$Q^\top Q=I_r, \qquad (QQ^\top)_{ii}=h_{ii}?$$
 A: (For convenience, I write $h_i$ instead of $h_{ii}$.)
You may start with $Q=\pmatrix{I_r\\ 0}$. The idea is to fix the diagonal entries of $QQ^T$ one by one, by applying Givens rotations to $Q$ recursively. More specifically, suppose at some stage, we have $Q^TQ=I_r$ and
$$
QQ^T=\left[\begin{array}{c|c}P&\ast
\\ \hline\ast&\begin{matrix}q_{k+1}\\ &\ddots\\ &&q_d\end{matrix}
\end{array}\right],\tag{$\dagger$}
$$
where the diagonal entries of $P$ are members of $\{h_1,\ldots,h_d\}$. By relabelling the $h_i$s if necessary, we may assume that it is $(h_1,\ldots,h_k)$. We also suppose that the bottom right subblock of $QQ^T$ in $(1)$ is a diagonal matrix $\operatorname{diag}(q_{k+1},\ldots,q_d)$ such that


*

*$\sum_{i=k+1}^dh_i=\sum_{i=k+1}^dq_i$,

*$q_{k+1}\ge\cdots\ge q_s>h_{k+1}\ge\cdots\ge h_d>q_{s+1}\ge\cdots\ge q_d$ for some $s$.


Now, note that
\begin{align}
&\pmatrix{\cos t&-\sin t\\ \sin t&\cos t}
\pmatrix{q_s\\ &q_{s+1}}
\pmatrix{\cos t&\sin t\\ -\sin t&\cos t}\\
=&\pmatrix{q_s\cos^2 t+q_{s+1}\sin^2 t&\ast\\ \ast&q_s\sin^2 t+q_{s+1}\cos^2 t},
\end{align}
Therefore, by applying an appropriate Givens rotation $R$ to the $s$-th and $(s+1)$-th rows of $Q$, we may turn one of the $s$-th or $(s+1)$-th diagonal entries of $(RQ)(RQ)^T$ into any desired convex combination of $q_s$ and $q_{s+1}$. In particular,


*

*if $q_s-h_{k+1}<h_d-q_{s+1}$, let us make the $s$-th diagonal entry of $(RQ)(RQ)^T$ becomes $h_{k+1}$;

*if $q_s-h_{k+1}\ge h_d-q_{s+1}$ instead, let us make the $(s+1)$-th diagonal entry of $(RQ)(RQ)^T$ equal to $h_d$.


Note that the entries in $P$ are unaffected and we still have $(RQ)^T(RQ)=Q^TQ=I_r$. Perform a further permutation to move the newly set diagonal entry to position $(k+1,k+1)$. If the other diagonal entry involved in the rotational transform also equals to some $h_i$, perform one more permutation to move that diagonal entry to the $(k+2,k+2)$-th position. The resulting matrix is still of the form $(\dagger)$ (but $k$ is now incremented by $1$ or $2$), with the bottom right subblock remains diagonal. More importantly, since the trace is preserved and due to the way we set the new $(k+1)$-th diagonal entry, conditions 1 and 2 in the above are also satisfied in the resulting matrix.
So, we have reduced the dimension of the problem by $1$ or $2$. Proceed recursively, we can construct a matrix $Q$ with orthonormal columns so that the diagonal of $QQ^T$ is a permutation of $(h_1,\ldots,h_d)$. Now, apply a final permutation to the rows of $Q$ so that the diagonal of $QQ^T$ is exactly $(h_1,\ldots,h_d)$.
A: 
EDIT: This answer is wrong; the condition $\sum h_{ii}=r$ is not satisfied.


I've been thinking a lot about this, but there's something. Let $\{f_j\}_{j=1}^d$ be the canonical basis of $\mathbb{R}^d$. Then
$$h_{jj}=f_j^TQQ^Tf_j={\left(Q^Tf_j\right)}^T\left(Q^Tf_j\right)=\lVert Q^Tf_j\rVert^2$$
Hence, $h_{jj}$ is the squared norm of $Q^T$'s $j$-th column, or in other words, $Q$'s $j$-th row. In particular $h_{jj}=0$ if and only if $Q$'s $j$-th row is $0$.
If at least $d-(r-1)$ of the $h_{jj}$'s are zero, the number of nonzero rows in $Q$ will be less than $r$. In this case, $\text{rank}(Q)<r$, which would contradict $Q^TQ=I_r$.$^*$
Therefore, the answer is 'not always'.
$^*$: Recall that $\text{rank}(AB)\leq\min\{\text{rank}(A),\text{rank}(B)\}$, or that $\text{rank}(A)=\text{rank}(A^T)=$ $\text{rank}(AA^T)=\text{rank}(A^TA)$.
