Doubt about the rank-nullity theorem involving a (block) matrix
$begingroup$
Let $Pinmathbb{R}^{ntimes n}$ and $Qinmathbb{R}^{ntimes m}$ and define
$$
S:=[Q, PQ, P^2Q, ldots, P^{n-1}Q].
$$
Then $Sinmathbb{R}^{ntimes(mn)}$. Now, the book says that, if $operatorname{rank}(S)<n$, then there exists a unit vector $xiinmathbb{R}^n$ 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 $xiinmathbb{R}^n$ 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
linear-algebra matrices
asked Dec 9 '18 at 9:06
JejiJeji
33818
$endgroup$
add a comment |
$begingroup$
Let $Pinmathbb{R}^{ntimes n}$ and $Qinmathbb{R}^{ntimes m}$ and define
$$
S:=[Q, PQ, P^2Q, ldots, P^{n-1}Q].
$$
Then $Sinmathbb{R}^{ntimes(mn)}$. Now, the book says that, if $operatorname{rank}(S)<n$, then there exists a unit vector $xiinmathbb{R}^n$ 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 $xiinmathbb{R}^n$ 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
linear-algebra matrices
asked Dec 9 '18 at 9:06
JejiJeji
33818
$endgroup$
1
$begingroup$
Then how about considering $S^{mathrm T}$ instead? If you want the vector to be a unit one, then simply pick $x/vert xvert$.
$endgroup$
– 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 $xinmathbb{R}^n$ 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
add a comment |
$begingroup$
Let $Pinmathbb{R}^{ntimes n}$ and $Qinmathbb{R}^{ntimes m}$ and define
$$
S:=[Q, PQ, P^2Q, ldots, P^{n-1}Q].
$$
Then $Sinmathbb{R}^{ntimes(mn)}$. Now, the book says that, if $operatorname{rank}(S)<n$, then there exists a unit vector $xiinmathbb{R}^n$ 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 $xiinmathbb{R}^n$ 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
linear-algebra matrices
asked Dec 9 '18 at 9:06
JejiJeji
33818
$endgroup$
Let $Pinmathbb{R}^{ntimes n}$ and $Qinmathbb{R}^{ntimes m}$ and define
$$
S:=[Q, PQ, P^2Q, ldots, P^{n-1}Q].
$$
Then $Sinmathbb{R}^{ntimes(mn)}$. Now, the book says that, if $operatorname{rank}(S)<n$, then there exists a unit vector $xiinmathbb{R}^n$ 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 $xiinmathbb{R}^n$ 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
linear-algebra matrices
linear-algebra matrices
asked Dec 9 '18 at 9:06
JejiJeji
33818
asked Dec 9 '18 at 9:06
JejiJeji
33818
edited Dec 9 '18 at 10:22
Jeji
asked Dec 9 '18 at 9:06
JejiJeji
33818
asked Dec 9 '18 at 9:06
JejiJeji
33818
asked Dec 9 '18 at 9:06
JejiJeji
33818
33818
1
$begingroup$
Then how about considering $S^{mathrm T}$ instead? If you want the vector to be a unit one, then simply pick $x/vert xvert$.
$endgroup$
– 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 $xinmathbb{R}^n$ 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
add a comment |
1
$begingroup$
Then how about considering $S^{mathrm T}$ instead? If you want the vector to be a unit one, then simply pick $x/vert xvert$.
$endgroup$
– 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 $xinmathbb{R}^n$ 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
1
1
$begingroup$
Then how about considering $S^{mathrm T}$ instead? If you want the vector to be a unit one, then simply pick $x/vert xvert$.
$endgroup$
– xbh
Dec 9 '18 at 10:26
$begingroup$
Then how about considering $S^{mathrm T}$ instead? If you want the vector to be a unit one, then simply pick $x/vert xvert$.
$endgroup$
– 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 $xinmathbb{R}^n$ 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
$begingroup$
@xbh Ok, you're right. By considering $S^T$, by the rank-nullity theorem we get that there exists at least an $xinmathbb{R}^n$ 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
$begingroup$
@Jean-ClaudeArbaut Yes, now it is clear also with the eigenvector. Thank You too
$endgroup$
– Jeji
Dec 9 '18 at 10:35
add a comment |
1 Answer
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$begingroup$
You have $operatorname{rank}(S^{mathrm T}) = operatorname{rank}(S) < n$, so there is an $xin mathbb{R}^n, xneq 0$ and $xinoperatorname{ker}(S^{mathrm T})$, so $x$ is orthogonal to all columns $sin mathbb{R}^n$ 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.
answered Dec 9 '18 at 10:32
S. M. RochS. M. Roch
674315
$endgroup$
add a comment |
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$begingroup$
You have $operatorname{rank}(S^{mathrm T}) = operatorname{rank}(S) < n$, so there is an $xin mathbb{R}^n, xneq 0$ and $xinoperatorname{ker}(S^{mathrm T})$, so $x$ is orthogonal to all columns $sin mathbb{R}^n$ 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.
answered Dec 9 '18 at 10:32
S. M. RochS. M. Roch
674315
$endgroup$
add a comment |
$begingroup$
You have $operatorname{rank}(S^{mathrm T}) = operatorname{rank}(S) < n$, so there is an $xin mathbb{R}^n, xneq 0$ and $xinoperatorname{ker}(S^{mathrm T})$, so $x$ is orthogonal to all columns $sin mathbb{R}^n$ 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.
answered Dec 9 '18 at 10:32
S. M. RochS. M. Roch
674315
$endgroup$
add a comment |
$begingroup$
You have $operatorname{rank}(S^{mathrm T}) = operatorname{rank}(S) < n$, so there is an $xin mathbb{R}^n, xneq 0$ and $xinoperatorname{ker}(S^{mathrm T})$, so $x$ is orthogonal to all columns $sin mathbb{R}^n$ 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.
answered Dec 9 '18 at 10:32
S. M. RochS. M. Roch
674315
$endgroup$
You have $operatorname{rank}(S^{mathrm T}) = operatorname{rank}(S) < n$, so there is an $xin mathbb{R}^n, xneq 0$ and $xinoperatorname{ker}(S^{mathrm T})$, so $x$ is orthogonal to all columns $sin mathbb{R}^n$ 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.
answered Dec 9 '18 at 10:32
S. M. RochS. M. Roch
674315
answered Dec 9 '18 at 10:32
S. M. RochS. M. Roch
674315
answered Dec 9 '18 at 10:32
S. M. RochS. M. Roch
674315
answered Dec 9 '18 at 10:32
S. M. RochS. M. Roch
674315
674315
add a comment |
add a comment |
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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 $x/vert xvert$.
$endgroup$
– 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 $xinmathbb{R}^n$ 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