1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.