Let $\{e_1,\dots,e_n\}$ be an orthogonal basis of $\mathbb{R}^n$. Let $d_i=\|e_i\|$ and $m\in[\![1,n]\!]$. Then, the following are equivalent Let $\|\cdot\|$ the euclidean norm of $\mathbb{R}^n$ and $\{e_1,\ldots,e_n\}$ be an orthogonal basis of $\mathbb{R}^n$. Let $d_i=\|e_i\|$ and $m$ be an integer between $1$ and $n$. How can I prove that the two following propositions are equivalent?
$(\textrm i)$ There exists a linear subspace $\textsf W$ of $\mathbb{R}^n$ of dimension $m$ such that the orthogonal projections of $e_1,\ldots,e_n$ on $\textsf W$ have the same norm.
$(\textrm{ii})$ $\forall i\in[\![1,n]\!],\,{d_i}^2\left(\displaystyle \sum_{j=1}^n{{d_j}^{-2}}\right)\geq m$.
Thank you for your help.
 A: This might be a start to a full proof.
Proof that (ii) implies (i) for $m=n-1$
Let $u_i$ be an orthonormal basis for $\mathbb{R}^n$ with $e_i=d_iu_i$. Let $W$ be an $m-$dimensional subspace perpendicular to the unit vector $p=\sum_{i=1}^{n}{a_iu_i}$, where $\sum_{i=1}^{n}{a_i^2}=1.$
The projection of $e_i$ onto $W$ has length $$d_i^2-(p.e_i)^2=d_i^2-a_i^2d_i^2.$$
We therefore require there to be a constant $c$ such that, for all $i$,
$$1-a_i^2=\frac {c}{d_i^2}.$$
Summing over all $i$ gives $$c\sum_{i=1}^{n}\frac {1}{d_i^2}=n-1.$$ and we must therefore  solve, for all $i$, $$d_i^2(1-a_i^2)\sum_{i=1}^{n}\frac {1}{d_i^2}=n-1 .$$
For there to be a solution simply requires $d_i^2\sum_{i=1}^{n}\frac {1}{d_i^2}\ge n-1 $ which is precisely condition (ii).
A: After a bit of work, the question essentially boils down to: which vectors can appear as the diagonal of the matrix of a projection operator?
Instead of thinking about the linear subspace $W$, think about the orthogonal projection operator $P_W$. Also, instead of thinking about the orthogonal basis $\{e_1, \dots, e_n\}$, consider the corresponding orthonormal basis $\{e_1', \dots, e_n'\}$ (where $e_i' = e_i/d_i$) and the numbers $d_1, \dots, d_n > 0$ independently. Then we can reformulate (i) as the condition:

(i') There exists a projection operator $P$ of rank $m$ and a constant $r > 0$ such that $\|Pe_i'\| = r/d_i$ for $i = 1, \dots, n$. 

Now, note that since a projection $P$ is self-adjoint and idempotent, the $i$-th diagonal element of $P$ (under the basis $e_1', \dots, e_n'$) is $\langle e_i', P e_i' \rangle = \langle e_i', P^2 e_i' \rangle = \langle Pe_i', Pe_i' \rangle = \|P e_i'\|^2$, hence $\mathrm{tr}(P) = \sum_{i=1}^m \|Pe_i'\|^2$. But for a projection we have $\mathrm{tr}(P) = \mathrm{rank}(P)$, hence any projection $P$ of rank $m$ satisfying condition (i') must have $m = \sum_{i=1}^n \|Pe_i'\|^2 = r^2 \sum_{i=1}^n d_i^{-2}$, meaning $r^2 = m(\sum_{i=1}^n d_i^{-2})^{-1}$. This lets us again reformulate (i') as the condition:

(i'') There exists a projection operator $P$ of rank $m$ such that $\|P e_i'\|^2 = m d_i^{-2} (\sum_{j=1}^n d_j^{-2})^{-1}$ for $i = 1, \dots, n$.

This condition clearly implies (ii), since each $\|Pe_i'\|^2 \leq 1$, so we only need to prove (ii) $\Rightarrow$ (i'').
For convenience, we can set $a_i = m d_i^{-2} (\sum_{j=1}^n d_j^{-2})^{-1}$, so each $a_i > 0$ and $\sum_{i=1}^n a_i = m$, and (ii) simply becomes the condition that each $a_i \leq 1$. Then to show (ii) $\Rightarrow$ (i'') it suffices to show that

If $a_1, \dots, a_n \in [0, 1]$ with $\sum_{i=1}^n a_i = m$, then there is a projection operator of rank $m$ with $\|Pe_i'\|^2 = a_i$ for $i = 1, \dots, n$.

or phrased differently (since the converse is true),

A vector $(a_1, \dots, a_n)$ appears as the diagonal of an idempotent self-adjoint matrix of rank $m$ if and only if each $a_i \in [0, 1]$ and $\sum_{i=1}^n a_i = m$.

This was proved, for example, by Kadison here (see theorems 6 or 7) -- he refers to it as "the Carpenter's theorem." It's also a special case of the Schur-Horn theorem.
A: Partial proof:
Set $c_i=d_i^{-1}$, $u_i=c_ie_i$, $i=1,\ldots,n$. Then $\{u_1,\ldots,u_n\}$ is an orthonormal basis of $\mathbb R^n$.
The statement: 
There exists a linear subspace $W$ of $\mathbb R^n$ of dimension $$, such that the orthogonal projections of $e_1,\ldots,e_n$ on $W$ have the same norm is now equivalent to the statement: 
(i') There exists a linear subspace $W$ of $\mathbb R^n$ of dimension $$, such that the orthogonal projections of $u_1,\ldots,u_n$ on $W$ are proportional to $c_1,\ldots,c_n$.
Whereas the second statement becomes equivalent to:
(ii') For all $i=1,\ldots,m$,
$$
\sum_{j=1}^n c_j^2\ge mc_i^2.  \tag{1}
$$
We have:
$(i')\to(ii'):$ If $w_1,\ldots,w_m$ is an orthonormal basis of $W$, then there exists a $k>0$, such that
$$
1=\|u_i\|^2\ge \sum_{j=1}^m(u_i,w_j)^2=k^2c_i^2 \quad\Longrightarrow\quad k^2\le\frac{1}{c_i^2}, \quad i=1,\ldots,n
$$
and hence for $i=1,\ldots,m$,
$$
\frac{1}{c_i^2}\sum_{i=1}^nc_i^2\ge k^2\sum_{i=1}^nc_i^2=\sum_{i=1}^n\sum_{j=1}^m(u_i,w_j)^2
=\sum_{j=1}^m\big(\sum_{i=1}^n(u_i,w_j)^2\big)=m
$$
which implies $(1)$.
