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Let $M$ be a square matrix of order $n\times d$. Let $\xi_{1},\dots,\xi_{n\times d}$ be an orthonormal basis of $\mathbb{R}^{n\times d}$ formed of eigenvectors of $M$. We have

$$\xi_{i}=(\lambda_1, 0, \dots, 0) + \cdots + (0, \dots,0,\lambda_n)$$

where $0, \lambda_j\in \mathbb{R}^{d}$. We can write $\xi^{i}$ as $$\xi_{i}=\alpha_{i,1}(\tilde{\lambda}_1,0,\,.\,.\,.\,,0)+\alpha_{i,2}(0,\tilde{\lambda}_2,0,\,.\,.\,.\,,0)+.\,.\,.\,+\sqrt{1-\alpha_{i,1}^2-\,.\,.\,.\,-\alpha_{i,n-1}^2}(0,\,.\,.\,.\,,0,\tilde{\lambda}_n).$$ We want to prove that there exist $\alpha_0>0$ small enough and $i_1,\,.\,.\,.\,,i_{(n-1)\times d}$ such that $$\alpha_{i_k,l}>\alpha_0\qquad\text{for}\;k=1,\,.\,.\,.\,,(n-1)\times d\;\text{and}\; l=1,\,.\,.\,.\,n-1.$$

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    $\begingroup$ What is $\alpha_0$? As it’s currently written you can just choose any positive number smaller than the other $\alpha$s, as long as they are nonzero. $\endgroup$ Commented Jul 29, 2020 at 15:24
  • $\begingroup$ What is a square $n \times d$ matrix? This means nothing to me. $\endgroup$ Commented Jul 29, 2020 at 20:54
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    $\begingroup$ @David Handelman - They mean an $nd \times nd$ matrix. They're using $\times$ to mean multiplication, not separation of matrix dimensions. $\endgroup$ Commented Jul 29, 2020 at 23:10
  • $\begingroup$ (To be fair, there was an explanation that the confusing terminology "square $n \times d$ matrix" meant $n \times d$ rows and $n \times d$ columns, but, for some reason, @vidyarthi edited it out.) $\endgroup$
    – LSpice
    Commented Jul 30, 2020 at 0:39
  • $\begingroup$ The $+$ signs mean $\oplus$, right? (Else the lengths of the vectors don't match.) Also, what does "small enough" mean in an existence statement? $\endgroup$ Commented Aug 3, 2020 at 8:28

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