Let $e\_1,...,e\_r$ be the first $r$ standard basis of $\mathbb{R}^n, r<n$. Let $u\_1,...,u\_n$ be another orthonormal basis of $\mathbb{R}^n$. Let $\otimes$ be the tensor product on $\mathbb{R}^n$ and define a subspace of $\mathbb{R}^{n\times n}$ as
$$S = span\{e\_1\otimes e\_1,...,e\_r\otimes e\_r\}$$
$\mathbb{R}^{n\times n}$, as a Hilbert space, is also equipped with the usual matrix inner product $\langle,\rangle$ and the induced norm $\|\|$ (which is the Frobenius norm). 


Take any $\Lambda\in S$ with $\|\Lambda\|=1$. Is the following conjecture true?


There exists a constant $C\geq1$, if $u\_1,...,u\_n$ are chosen such that 
$$1 -\delta\leq\sum\_{i=1}^n\left|\langle\Lambda,u\_i\otimes u\_i\rangle\right|^2$$ for some very small $\delta>0$ (for example $0<\delta<\frac{1}{rn}$), then there exists a set $T\subset\{1,2,...,n\}$ with $|T| = r$ satisfying 
$$1-C\delta\leq\sum\_{i\in T}\left|\langle\Lambda,u\_i\otimes u\_i\rangle\right|^2$$


The conjecture can be verified when $r=1$. Note that $C$ cannot depend on $\delta$, which means the conjecture states that there is a $C>0$ to make the assertion true for all $\delta$ small enough. For example, $C$ can be taken as 2 when $r=1$.



Inequality involving tensor product of orthonormal unit vectors
Without loss of generality, $u\_1,\dots,u\_n$ can be taken to be the standard basis of ${\bf R}^n$ (so $e\_1,\dots,e\_r$ is just some arbitrary orthonormal system). We can view $\Lambda$ as a real symmetric $n \times n$ matrix of Frobenius norm $1$ and rank at most $r$. (Conversely, every such matrix has a representation of the desired form for some $e\_1,\dots,e\_r$ by the spectral theorem, so this reformulation has not lost any information.) If $D = \sum\_{i=1}^n \langle \Lambda, u\_i \otimes u\_i \rangle u\_i \otimes u\_i$ is the diagonal component of $\Lambda$, the hypothesis is then $\|D\|^2 \geq 1-\delta$.


Let $\lambda = \mathrm{diag}(\lambda\_1,\dots,\lambda\_r,0,\dots,0)$ be the diagonalisation of $\Lambda$ (the ordering of the eigenvalues is irrelevant), thus $\lambda$ is a unit vector supported on a set of $r$ indices in $\{1,\dots,n\}$. By the [Schur-Horn theorem](https://en.wikipedia.org/wiki/Schur%E2%80%93Horn_theorem), $D$ is a convex combination of the permutations $\sigma(\lambda)$ of $\lambda$, $\sigma \in S\_n$, where the symmetric group $S\_n$ acts on diagonal matrices in the obvious manner, thus
$$ D = \sum\_{\sigma \in S\_n} c\_\sigma \sigma(\lambda)$$
for some non-negative coefficients $c\_\sigma$ summing to $1$. (Again, the Schur-Horn theorem is an if-and-only-if statement, so we have still not lost any information so far.) 


Now we exploit the uniform convexity of the Frobenius norm. We take the norm square
$$ 1-\delta \leq \|D\|^2 = \sum\_{\sigma,\sigma' \in S\_n} c\_\sigma c\_{\sigma'} \langle \sigma(\lambda), \sigma'(\lambda) \rangle$$
and then apply the cosine rule to conclude
$$ \sum\_{\sigma,\sigma' \in S\_n} c\_\sigma c\_{\sigma'} \| \sigma(\lambda) - \sigma'(\lambda)\|^2 \leq 2\delta$$
hence by pigeonholing there exists $\sigma\_0 \in S\_n$ such that
$$ \sum\_{\sigma \in S\_n} c\_\sigma \| \sigma(\lambda) - \sigma\_0(\lambda)\|^2 \leq 2\delta$$
hence by Cauchy-Schwarz
$$ \sum\_{\sigma \in S\_n} c\_\sigma \| \sigma(\lambda) - \sigma\_0(\lambda)\| \leq \sqrt{2\delta}$$
hence by the triangle inequality
$$ \| \sum\_{\sigma \in S\_n} c\_\sigma \sigma(\lambda) - \sigma\_0(\lambda)\| \leq \sqrt{2\delta}$$
thus
$$ \| D - \sigma\_0(\lambda) \| \leq \sqrt{2\delta}.$$
The diagonal matrix $\sigma\_0(\lambda)$ is supported on a set $T \subset \{1,\dots,n\}$ of cardinality $r$, and the above inequality implies that the Frobenius norm of $D$ outside of $T$ is at most $\sqrt{2\delta}$. Thus
$$ \sum\_{i \not \in T} |\langle \Lambda, u\_i \otimes u\_i \rangle|^2 \leq 2\delta$$
so that
$$ \sum\_{i \in T} |\langle \Lambda, u\_i \otimes u\_i \rangle|^2 \geq 1-3\delta$$
as required.


