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:

$V$ has orthonormal

**columns**: $V^tV = I \in \mathbb{R}^{k \times k}$, and$VDV^t \in \mathbb{R}^{m \times m}$ 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.