Question about Minimal polynomial and diagonalizable matrix












0












$begingroup$


Let $A$ $in$ $M_n$($mathbb{C}$) such that $A^k=I$ for some $k in mathbb{Z}$ . Prove that $A$ is diagonalizable.



Ok, I am trying to prove this way:



If $k=1$, is trivial.



If $k>1$



$A^k = I iff A^k- I = 0 iff A^{k-1}- I= (A-I)(I + A + cdots+ A^{k-1})=0.$



I know that the polynomial $p(x)= x^{k-1}+ cdots + x^2 + 1 = displaystyleprod_{j=1}^{n-1}(x-u_j)$ , $u_j$ is root of unity, $u_j neq 1
hspace{0.8cm} forall j in$
{$1,...,n-1$}.



Let $q(x) = p(x)(x-1)$ we knows that $q(A)=0$ and the minimal polynomial must divide q(x). Since q(x) have only simple roots them the minimal have too so we can conclude that A is diagonalizable?










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$endgroup$








  • 2




    $begingroup$
    Yes. A square matrix over $mathbb{C}$ is diagonalisable if and only if its minimal polynomial is square-free.
    $endgroup$
    – Theo Bendit
    Feb 2 at 0:08










  • $begingroup$
    Furthermore, a polynomial can be confirmed to be squarefree by taking its formal derivative ( actual derivative if the field is the reals or complexes) and confirming $gcd(f,f') = 1,$ which in turn is equivalent to the existence of constants $a,b$ such that $af + b f' = 1$ is the constant polynomial $1.$
    $endgroup$
    – Will Jagy
    Feb 2 at 2:07












  • $begingroup$
    Alternatively, you can just check that a Jordan block (of size $>1$) with eigenvalue one can not satisfy your condition $A^k=1$.
    $endgroup$
    – user138530
    Feb 2 at 2:11


















0












$begingroup$


Let $A$ $in$ $M_n$($mathbb{C}$) such that $A^k=I$ for some $k in mathbb{Z}$ . Prove that $A$ is diagonalizable.



Ok, I am trying to prove this way:



If $k=1$, is trivial.



If $k>1$



$A^k = I iff A^k- I = 0 iff A^{k-1}- I= (A-I)(I + A + cdots+ A^{k-1})=0.$



I know that the polynomial $p(x)= x^{k-1}+ cdots + x^2 + 1 = displaystyleprod_{j=1}^{n-1}(x-u_j)$ , $u_j$ is root of unity, $u_j neq 1
hspace{0.8cm} forall j in$
{$1,...,n-1$}.



Let $q(x) = p(x)(x-1)$ we knows that $q(A)=0$ and the minimal polynomial must divide q(x). Since q(x) have only simple roots them the minimal have too so we can conclude that A is diagonalizable?










share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Yes. A square matrix over $mathbb{C}$ is diagonalisable if and only if its minimal polynomial is square-free.
    $endgroup$
    – Theo Bendit
    Feb 2 at 0:08










  • $begingroup$
    Furthermore, a polynomial can be confirmed to be squarefree by taking its formal derivative ( actual derivative if the field is the reals or complexes) and confirming $gcd(f,f') = 1,$ which in turn is equivalent to the existence of constants $a,b$ such that $af + b f' = 1$ is the constant polynomial $1.$
    $endgroup$
    – Will Jagy
    Feb 2 at 2:07












  • $begingroup$
    Alternatively, you can just check that a Jordan block (of size $>1$) with eigenvalue one can not satisfy your condition $A^k=1$.
    $endgroup$
    – user138530
    Feb 2 at 2:11
















0












0








0





$begingroup$


Let $A$ $in$ $M_n$($mathbb{C}$) such that $A^k=I$ for some $k in mathbb{Z}$ . Prove that $A$ is diagonalizable.



Ok, I am trying to prove this way:



If $k=1$, is trivial.



If $k>1$



$A^k = I iff A^k- I = 0 iff A^{k-1}- I= (A-I)(I + A + cdots+ A^{k-1})=0.$



I know that the polynomial $p(x)= x^{k-1}+ cdots + x^2 + 1 = displaystyleprod_{j=1}^{n-1}(x-u_j)$ , $u_j$ is root of unity, $u_j neq 1
hspace{0.8cm} forall j in$
{$1,...,n-1$}.



Let $q(x) = p(x)(x-1)$ we knows that $q(A)=0$ and the minimal polynomial must divide q(x). Since q(x) have only simple roots them the minimal have too so we can conclude that A is diagonalizable?










share|cite|improve this question









$endgroup$




Let $A$ $in$ $M_n$($mathbb{C}$) such that $A^k=I$ for some $k in mathbb{Z}$ . Prove that $A$ is diagonalizable.



Ok, I am trying to prove this way:



If $k=1$, is trivial.



If $k>1$



$A^k = I iff A^k- I = 0 iff A^{k-1}- I= (A-I)(I + A + cdots+ A^{k-1})=0.$



I know that the polynomial $p(x)= x^{k-1}+ cdots + x^2 + 1 = displaystyleprod_{j=1}^{n-1}(x-u_j)$ , $u_j$ is root of unity, $u_j neq 1
hspace{0.8cm} forall j in$
{$1,...,n-1$}.



Let $q(x) = p(x)(x-1)$ we knows that $q(A)=0$ and the minimal polynomial must divide q(x). Since q(x) have only simple roots them the minimal have too so we can conclude that A is diagonalizable?







linear-algebra






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share|cite|improve this question











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asked Feb 1 at 23:57









user638057user638057

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  • 2




    $begingroup$
    Yes. A square matrix over $mathbb{C}$ is diagonalisable if and only if its minimal polynomial is square-free.
    $endgroup$
    – Theo Bendit
    Feb 2 at 0:08










  • $begingroup$
    Furthermore, a polynomial can be confirmed to be squarefree by taking its formal derivative ( actual derivative if the field is the reals or complexes) and confirming $gcd(f,f') = 1,$ which in turn is equivalent to the existence of constants $a,b$ such that $af + b f' = 1$ is the constant polynomial $1.$
    $endgroup$
    – Will Jagy
    Feb 2 at 2:07












  • $begingroup$
    Alternatively, you can just check that a Jordan block (of size $>1$) with eigenvalue one can not satisfy your condition $A^k=1$.
    $endgroup$
    – user138530
    Feb 2 at 2:11
















  • 2




    $begingroup$
    Yes. A square matrix over $mathbb{C}$ is diagonalisable if and only if its minimal polynomial is square-free.
    $endgroup$
    – Theo Bendit
    Feb 2 at 0:08










  • $begingroup$
    Furthermore, a polynomial can be confirmed to be squarefree by taking its formal derivative ( actual derivative if the field is the reals or complexes) and confirming $gcd(f,f') = 1,$ which in turn is equivalent to the existence of constants $a,b$ such that $af + b f' = 1$ is the constant polynomial $1.$
    $endgroup$
    – Will Jagy
    Feb 2 at 2:07












  • $begingroup$
    Alternatively, you can just check that a Jordan block (of size $>1$) with eigenvalue one can not satisfy your condition $A^k=1$.
    $endgroup$
    – user138530
    Feb 2 at 2:11










2




2




$begingroup$
Yes. A square matrix over $mathbb{C}$ is diagonalisable if and only if its minimal polynomial is square-free.
$endgroup$
– Theo Bendit
Feb 2 at 0:08




$begingroup$
Yes. A square matrix over $mathbb{C}$ is diagonalisable if and only if its minimal polynomial is square-free.
$endgroup$
– Theo Bendit
Feb 2 at 0:08












$begingroup$
Furthermore, a polynomial can be confirmed to be squarefree by taking its formal derivative ( actual derivative if the field is the reals or complexes) and confirming $gcd(f,f') = 1,$ which in turn is equivalent to the existence of constants $a,b$ such that $af + b f' = 1$ is the constant polynomial $1.$
$endgroup$
– Will Jagy
Feb 2 at 2:07






$begingroup$
Furthermore, a polynomial can be confirmed to be squarefree by taking its formal derivative ( actual derivative if the field is the reals or complexes) and confirming $gcd(f,f') = 1,$ which in turn is equivalent to the existence of constants $a,b$ such that $af + b f' = 1$ is the constant polynomial $1.$
$endgroup$
– Will Jagy
Feb 2 at 2:07














$begingroup$
Alternatively, you can just check that a Jordan block (of size $>1$) with eigenvalue one can not satisfy your condition $A^k=1$.
$endgroup$
– user138530
Feb 2 at 2:11






$begingroup$
Alternatively, you can just check that a Jordan block (of size $>1$) with eigenvalue one can not satisfy your condition $A^k=1$.
$endgroup$
– user138530
Feb 2 at 2:11












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