Let $d = (d\_1,...,d\_k)^t$ with positive entries. Denote $D:=diag(d)$ and let $m > k$. What are sufficient conditions on $d$ and $m$ so that there exists $V \in \mathbb{R}^{m \times k}$ with:


1. $V$ has orthonormal **columns**: $V^tV = I \in \mathbb{R}^{k \times k}$, and
2. $VDV^t \in \mathbb{R}^{m \times m}$ has unit diagonal. Geometrically, The **rows** of $V$ lie on a hyper-ellipse given by $\vec{d}$:
$$
\sum\_{p=1}^k V\_{jp}^2 d\_p = 1, j=1,...,m?
$$


A necessary condition
---------------------


If such $V$ exists, we can extend it to be orthogonal in $\mathbb{R}^{m \times m}$, so denote the extended matrix $\hat{V}$. We can also extend $D$ with zeros, so denote that extended matrix $\hat{D}$. Then $VDV^t = \hat{V}\hat{D}\hat{V}^t$. Thus, it is necessary that $\sum d\_i = m$, by considering traces. Is it also sufficient? This view means that we just need to redistribute the diagonal entries of a matrix using orthogonal similarity, hence the title.


Degrees of freedom
------------------


The matrix $V$ has, a-priori, $mk$ degrees of freedom. The first condition amounts to $k(k+1)/2$ linear constraints. The second condition amounts to $m$ quadratic constraints. So if $mk - m - k(k+1)/2 > 0$ we can at least hope to find a solution. 



Redistribute diagonal entries of a matrix
Yes, this works. Or, to be more honest, I'm fairly confident it does, but I'm only going to give a sketch.


The basic step is: a given symmetric $2\times 2$ matrix $A$ is unitarily equivalent to one with equal diagonal elements. This we can just check by direct computation.
Let's say
$$
A=\begin{pmatrix} a & b \\ b & c \end{pmatrix}, \quad V = \begin{pmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{pmatrix} .
$$
Then setting the diagonal elements of $V^\*AV$ equal to one another gives the equation
$$
(a-c)(\cos^2\alpha-\sin^2\alpha) = 4b\sin\alpha\cos\alpha ,
$$
and, making use of the formulae $\cos^2\alpha-\sin^2\alpha=\cos 2\alpha$, $2\sin\alpha\cos\alpha =\sin 2\alpha$, we see that this always has a solution $\alpha$.


Now in the actual problem, I can take $\widehat{V}$'s that are the identity except for a $2\times 2$ block as above to replace any two diagonal entries by their average. If we repeat this process, we converge to a matrix of the desired form. (The part that's missing is the verification that the product of the $\widehat{V}\_n$ will converge to a single unitary matrix, but I think this will follow from a more careful analysis since diagonal elements that were already close to one another will only require a $\widehat{V}\_n$ that is close to the identity.)


