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Let $P\in\mathbb{R}^{n\times n}$ and $Q\in\mathbb{R}^{n\times m}$ and define $$ S:=[Q, PQ, P^2Q, \ldots, P^{n-1}Q]. $$ Then $S\in\mathbb{R}^{n\times(mn)}$. Now, the book says that, if $\operatorname{rank}(S)<n$, then there exists a unit vector $\xi\in\mathbb{R}^n$ which is orthogonal to each column of $S$ (otherwise $n$ columns of $S$ would be linearly independent and then there would be a non singular minor of $S$ of order $n$).

I do not understand why there exists a unit vector $\xi\in\mathbb{R}^n$ which is orthogonal to each column of $S$. I think that, since $\operatorname{rank}(S)<n$, then $\operatorname{dim}(\operatorname{Ker}(S))=n-\operatorname{rank}(S)$ and hence there exists at least one vector $x$ such that $Sx=0$. But such $x$ is in $\mathbb{R}^{nm}$ and it is orthogonal to each row of $S$ and it is not a unit vector.

Can someone help me?

Thank You

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    $\begingroup$ Then how about considering $S^{\mathrm T}$ instead? If you want the vector to be a unit one, then simply pick $x/\vert x\vert$. $\endgroup$ – xbh Dec 9 '18 at 10:26
  • $\begingroup$ @xbh Ok, you're right. By considering $S^T$, by the rank-nullity theorem we get that there exists at least an $x\in\mathbb{R}^n$ such that $x$ is orthogonal to each row of $S^T$ (that is to each column of $S$). Thank You $\endgroup$ – Jeji Dec 9 '18 at 10:35
  • $\begingroup$ @Jean-ClaudeArbaut Yes, now it is clear also with the eigenvector. Thank You too $\endgroup$ – Jeji Dec 9 '18 at 10:35
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You have $\operatorname{rank}(S^{\mathrm T}) = \operatorname{rank}(S) < n$, so there is an $x\in \mathbb{R}^n, x\neq 0$ and $x\in\operatorname{ker}(S^{\mathrm T})$, so $x$ is orthogonal to all columns $s\in \mathbb{R}^n$ of $S$.

$x$ might not be a unit vector (vector of length $1$), but then you can simply take $\tilde{x}=\frac{x}{||x||}$, which has the same property.

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