Proving the dimension of the cyclic subspace is an even number if there are no eigenvalues for the operator
$begingroup$
$V$ is a finite vector space above the real numbers. $ fin operatorname{End}(v)$ with no eigenvalues. Prove that for all $vin V$ the dimension of cylic subspace $cal Z(v) $ is an even number.
Somehow this makes to me since every two following vector in the ordered base would be non-linearly dependent, but I can't find a formal proof for this.
linear-algebra eigenvalues-eigenvectors
edited Jan 4 at 17:30
Davide Giraudo
128k17154268
asked Jan 1 at 12:15
TalTal
133
$endgroup$
add a comment |
$begingroup$
$V$ is a finite vector space above the real numbers. $ fin operatorname{End}(v)$ with no eigenvalues. Prove that for all $vin V$ the dimension of cylic subspace $cal Z(v) $ is an even number.
Somehow this makes to me since every two following vector in the ordered base would be non-linearly dependent, but I can't find a formal proof for this.
linear-algebra eigenvalues-eigenvectors
edited Jan 4 at 17:30
Davide Giraudo
128k17154268
asked Jan 1 at 12:15
TalTal
133
$endgroup$
1
$begingroup$
Every odd degree polynomial over $Bbb R$ has a zero over $Bbb R$.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 12:18
add a comment |
$begingroup$
$V$ is a finite vector space above the real numbers. $ fin operatorname{End}(v)$ with no eigenvalues. Prove that for all $vin V$ the dimension of cylic subspace $cal Z(v) $ is an even number.
Somehow this makes to me since every two following vector in the ordered base would be non-linearly dependent, but I can't find a formal proof for this.
linear-algebra eigenvalues-eigenvectors
edited Jan 4 at 17:30
Davide Giraudo
128k17154268
asked Jan 1 at 12:15
TalTal
133
$endgroup$
$V$ is a finite vector space above the real numbers. $ fin operatorname{End}(v)$ with no eigenvalues. Prove that for all $vin V$ the dimension of cylic subspace $cal Z(v) $ is an even number.
Somehow this makes to me since every two following vector in the ordered base would be non-linearly dependent, but I can't find a formal proof for this.
linear-algebra eigenvalues-eigenvectors
linear-algebra eigenvalues-eigenvectors
edited Jan 4 at 17:30
Davide Giraudo
128k17154268
asked Jan 1 at 12:15
TalTal
133
edited Jan 4 at 17:30
Davide Giraudo
128k17154268
asked Jan 1 at 12:15
TalTal
133
edited Jan 4 at 17:30
Davide Giraudo
128k17154268
edited Jan 4 at 17:30
Davide Giraudo
128k17154268
edited Jan 4 at 17:30
Davide Giraudo
128k17154268
128k17154268
asked Jan 1 at 12:15
TalTal
133
asked Jan 1 at 12:15
TalTal
133
asked Jan 1 at 12:15
TalTal
133
133
1
$begingroup$
Every odd degree polynomial over $Bbb R$ has a zero over $Bbb R$.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 12:18
add a comment |
1
$begingroup$
Every odd degree polynomial over $Bbb R$ has a zero over $Bbb R$.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 12:18
1
1
$begingroup$
Every odd degree polynomial over $Bbb R$ has a zero over $Bbb R$.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 12:18
$begingroup$
Every odd degree polynomial over $Bbb R$ has a zero over $Bbb R$.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 12:18
add a comment |
1 Answer
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$begingroup$
Let $vin V$ be a vector and $Z$ its associated cyclic space. Suppose that the dimension of $Z$ is odd, so that the set $leftlbrace v, Tv, ldots, T^kvrightrbrace$ is a base for $Z$ for some $k$ even integer (dimension $k+1$), from this we get a polynomial
$T^{k+1}v+alpha_kT^kv+cdots+alpha_1 Tv+alpha_0 v=0$, so that $T$ restricted to the cyclic space has polynomial equal to $p(x)=x^{k+1}+alpha_kx^{k}+cdots+alpha_1 x+alpha_0$ that vanishes in $T$ and which have a real root (odd degree), we have that the minimal polynomial divides $p(x)$ but if the minimal polynomial is not equal to $p(x)$ it has lower degree, but that contradicts the fact that $leftlbrace v, Tv, ldots, T^kvrightrbrace$ is a base for $Z$. Therefore, $T$ have a real eigenvalue.
answered Jan 1 at 22:14
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
$endgroup$
add a comment |
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$begingroup$
Let $vin V$ be a vector and $Z$ its associated cyclic space. Suppose that the dimension of $Z$ is odd, so that the set $leftlbrace v, Tv, ldots, T^kvrightrbrace$ is a base for $Z$ for some $k$ even integer (dimension $k+1$), from this we get a polynomial
$T^{k+1}v+alpha_kT^kv+cdots+alpha_1 Tv+alpha_0 v=0$, so that $T$ restricted to the cyclic space has polynomial equal to $p(x)=x^{k+1}+alpha_kx^{k}+cdots+alpha_1 x+alpha_0$ that vanishes in $T$ and which have a real root (odd degree), we have that the minimal polynomial divides $p(x)$ but if the minimal polynomial is not equal to $p(x)$ it has lower degree, but that contradicts the fact that $leftlbrace v, Tv, ldots, T^kvrightrbrace$ is a base for $Z$. Therefore, $T$ have a real eigenvalue.
answered Jan 1 at 22:14
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
$endgroup$
add a comment |
$begingroup$
Let $vin V$ be a vector and $Z$ its associated cyclic space. Suppose that the dimension of $Z$ is odd, so that the set $leftlbrace v, Tv, ldots, T^kvrightrbrace$ is a base for $Z$ for some $k$ even integer (dimension $k+1$), from this we get a polynomial
$T^{k+1}v+alpha_kT^kv+cdots+alpha_1 Tv+alpha_0 v=0$, so that $T$ restricted to the cyclic space has polynomial equal to $p(x)=x^{k+1}+alpha_kx^{k}+cdots+alpha_1 x+alpha_0$ that vanishes in $T$ and which have a real root (odd degree), we have that the minimal polynomial divides $p(x)$ but if the minimal polynomial is not equal to $p(x)$ it has lower degree, but that contradicts the fact that $leftlbrace v, Tv, ldots, T^kvrightrbrace$ is a base for $Z$. Therefore, $T$ have a real eigenvalue.
answered Jan 1 at 22:14
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
$endgroup$
add a comment |
$begingroup$
Let $vin V$ be a vector and $Z$ its associated cyclic space. Suppose that the dimension of $Z$ is odd, so that the set $leftlbrace v, Tv, ldots, T^kvrightrbrace$ is a base for $Z$ for some $k$ even integer (dimension $k+1$), from this we get a polynomial
$T^{k+1}v+alpha_kT^kv+cdots+alpha_1 Tv+alpha_0 v=0$, so that $T$ restricted to the cyclic space has polynomial equal to $p(x)=x^{k+1}+alpha_kx^{k}+cdots+alpha_1 x+alpha_0$ that vanishes in $T$ and which have a real root (odd degree), we have that the minimal polynomial divides $p(x)$ but if the minimal polynomial is not equal to $p(x)$ it has lower degree, but that contradicts the fact that $leftlbrace v, Tv, ldots, T^kvrightrbrace$ is a base for $Z$. Therefore, $T$ have a real eigenvalue.
answered Jan 1 at 22:14
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
$endgroup$
Let $vin V$ be a vector and $Z$ its associated cyclic space. Suppose that the dimension of $Z$ is odd, so that the set $leftlbrace v, Tv, ldots, T^kvrightrbrace$ is a base for $Z$ for some $k$ even integer (dimension $k+1$), from this we get a polynomial
$T^{k+1}v+alpha_kT^kv+cdots+alpha_1 Tv+alpha_0 v=0$, so that $T$ restricted to the cyclic space has polynomial equal to $p(x)=x^{k+1}+alpha_kx^{k}+cdots+alpha_1 x+alpha_0$ that vanishes in $T$ and which have a real root (odd degree), we have that the minimal polynomial divides $p(x)$ but if the minimal polynomial is not equal to $p(x)$ it has lower degree, but that contradicts the fact that $leftlbrace v, Tv, ldots, T^kvrightrbrace$ is a base for $Z$. Therefore, $T$ have a real eigenvalue.
answered Jan 1 at 22:14
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
answered Jan 1 at 22:14
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
answered Jan 1 at 22:14
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
answered Jan 1 at 22:14
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
802110
add a comment |
add a comment |
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$begingroup$
Every odd degree polynomial over $Bbb R$ has a zero over $Bbb R$.
$endgroup$
– Lord Shark the Unknown
Jan 1 at 12:18