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Let $M$ be a square $n\times d$ matrix (i.e., $n\times d$ row and $n\times d$ column). 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 $0\in \mathbb{R}^{d}$ and $\lambda_j\in \mathbb{R}^{d}$. 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 $$\alpha_{i_k,l}>\alpha_0\qquad\text{for}\;k=1,\,.\,.\,.\,,(n-1)\times d\;\text{and}\; l=1,\,.\,.\,.\,n-1.$$

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 $$\alpha_{i_k,l}>\alpha_0\qquad\text{for}\;k=1,\,.\,.\,.\,,(n-1)\times d\;\text{and}\; l=1,\,.\,.\,.\,n-1.$$

Let $M$ be a square $n\times d$ matrix (i.e., $n\times d$ row and $n\times d$ column). 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 $\xi^{i}=(\lambda_1,0,\,.\,.\,.\,,0)+.\,.\,.\,+(0,\,.\,.\,.\,,0,\lambda_n)$, where $0\in \mathbb{R}^{d}$ and $\lambda_j\in \mathbb{R}^{d}$. 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 $$\alpha_{i_k,l}>\alpha_0\qquad\text{for}\;k=1,\,.\,.\,.\,,(n-1)\times d\;\text{and}\; l=1,\,.\,.\,.\,n-1.$$

Let $M$ be a square $n\times d$ matrix (i.e., $n\times d$ row and $n\times d$ column). 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 $0\in \mathbb{R}^{d}$ and $\lambda_j\in \mathbb{R}^{d}$. 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 $$\alpha_{i_k,l}>\alpha_0\qquad\text{for}\;k=1,\,.\,.\,.\,,(n-1)\times d\;\text{and}\; l=1,\,.\,.\,.\,n-1.$$

Let $M$ be a square $n\times d$ matrix (i.e., $n\times d$ row and $n\times d$ column). Let $\xi^{1},\,.\,.\,.\,,\xi^{n\times d}$ be an orthonormal basis of $\mathbb{R}^{n\times d}$ formed of eigenvectors of $M$. We have $\xi^{i}=(\lambda_1,0,\,.\,.\,.\,,0)+.\,.\,.\,+(0,\,.\,.\,.\,,0,\lambda_n)$, where $0\in \mathbb{R}^{d}$ and $\lambda_j\in \mathbb{R}^{d}$. 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 $$\alpha_{i_k,l}>\alpha_0\qquad\text{for}\;k=1,\,.\,.\,.\,,(n-1)\times d\;\text{and}\; l=1,\,.\,.\,.\,n-1.$$

Let $M$ be a square $n\times d$ matrix (i.e., $n\times d$ row and $n\times d$ column). 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 $\xi^{i}=(\lambda_1,0,\,.\,.\,.\,,0)+.\,.\,.\,+(0,\,.\,.\,.\,,0,\lambda_n)$, where $0\in \mathbb{R}^{d}$ and $\lambda_j\in \mathbb{R}^{d}$. 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 $$\alpha_{i_k,l}>\alpha_0\qquad\text{for}\;k=1,\,.\,.\,.\,,(n-1)\times d\;\text{and}\; l=1,\,.\,.\,.\,n-1.$$