# Doubt about the rank-nullity theorem involving a (block) matrix

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 $$\operatorname{rank}(S), then there exists a unit vector $$\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 $$\operatorname{rank}(S), then $$\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

• Then how about considering $S^{\mathrm T}$ instead? If you want the vector to be a unit one, then simply pick $x/\vert x\vert$. – xbh Dec 9 '18 at 10:26
• @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 – Jeji Dec 9 '18 at 10:35
• @Jean-ClaudeArbaut Yes, now it is clear also with the eigenvector. Thank You too – Jeji Dec 9 '18 at 10:35

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.