# Inequality involving tensor product of orthonormal unit vectors

Let $$e_1,...,e_r$$ be the first $$r$$ standard basis of $$\mathbb{R}^n, r. Let $$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 $$r=1$$.

• Sure. Let $C = 1$ and $T = \{1, \ldots, n\}.$ Aug 15, 2019 at 12:37
• @VítTuček: they said $r < n$. Aug 15, 2019 at 13:40
• looks to be true for r=1 Aug 15, 2019 at 17:07
• Take $C$ large enough to make $1-C\delta$ negative! Aug 16, 2019 at 4:02
• @MeisamSoleimaniMalekan C should be a constant not depending on $\delta$... I'll make this clearer in the post. Aug 16, 2019 at 12:29

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, $$D$$ 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.