Proving the dimension of the cyclic subspace is an even number if there are no eigenvalues for the operator












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$V$ 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







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    Every odd degree polynomial over $Bbb R$ has a zero over $Bbb R$.
    – Lord Shark the Unknown
    Jan 1 at 12:18
















1














$V$ 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







share|cite|improve this question




















  • 1




    Every odd degree polynomial over $Bbb R$ has a zero over $Bbb R$.
    – Lord Shark the Unknown
    Jan 1 at 12:18














1












1








1







$V$ 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







share|cite|improve this question















$V$ 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






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edited Jan 4 at 17:30









Davide Giraudo

125k16150261




125k16150261










asked Jan 1 at 12:15









TalTal

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133








  • 1




    Every odd degree polynomial over $Bbb R$ has a zero over $Bbb R$.
    – Lord Shark the Unknown
    Jan 1 at 12:18














  • 1




    Every odd degree polynomial over $Bbb R$ has a zero over $Bbb R$.
    – Lord Shark the Unknown
    Jan 1 at 12:18








1




1




Every odd degree polynomial over $Bbb R$ has a zero over $Bbb R$.
– Lord Shark the Unknown
Jan 1 at 12:18




Every odd degree polynomial over $Bbb R$ has a zero over $Bbb R$.
– Lord Shark the Unknown
Jan 1 at 12:18










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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.






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    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.






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      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.






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        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.






        share|cite|improve this answer












        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.







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        answered Jan 1 at 22:14









        José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda

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