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Finding minimal polynomial with given operator

1

$begingroup$

Given the operator $T:mathbb C_le n[x]→mathbb C_le n[x]$ such that $T(p) = p' + p$ find the minimal polynomial.

What I tried:

I found the representing matrix $$A = beginpmatrix
1 & 1 & 0 & cdots & 0 \
0 & 1 & 2 & cdots & 0 \
vdots & 0 & ddots & ddots & 0\
0 & cdots & 0 & 1 & n\
0 & 0 & 0 & cdots & 1\
endpmatrix$$
and then I found the characteristic polynomial:
$f_T(x) = (x-1)^n+1$.

Now I know that the minimal polynomial is $m_T in (x-1),(x-1)^2,space...space,space(x-1)^n+1$

My guess is that $m_T = (x-1)^n+1$ but I don't know how to find which one it is.

linear-algebra linear-transformations characteristic-functions minimal-polynomials

share|cite|improve this question

asked Mar 30 at 12:43

Guysudai1Guysudai1

18911

$endgroup$

add a comment | 

1

$begingroup$

Given the operator $T:mathbb C_le n[x]→mathbb C_le n[x]$ such that $T(p) = p' + p$ find the minimal polynomial.

What I tried:

I found the representing matrix $$A = beginpmatrix
1 & 1 & 0 & cdots & 0 \
0 & 1 & 2 & cdots & 0 \
vdots & 0 & ddots & ddots & 0\
0 & cdots & 0 & 1 & n\
0 & 0 & 0 & cdots & 1\
endpmatrix$$
and then I found the characteristic polynomial:
$f_T(x) = (x-1)^n+1$.

Now I know that the minimal polynomial is $m_T in (x-1),(x-1)^2,space...space,space(x-1)^n+1$

My guess is that $m_T = (x-1)^n+1$ but I don't know how to find which one it is.

linear-algebra linear-transformations characteristic-functions minimal-polynomials

share|cite|improve this question

asked Mar 30 at 12:43

Guysudai1Guysudai1

18911

$endgroup$

add a comment | 

$begingroup$

Given the operator $T:mathbb C_le n[x]→mathbb C_le n[x]$ such that $T(p) = p' + p$ find the minimal polynomial.

What I tried:

I found the representing matrix $$A = beginpmatrix
1 & 1 & 0 & cdots & 0 \
0 & 1 & 2 & cdots & 0 \
vdots & 0 & ddots & ddots & 0\
0 & cdots & 0 & 1 & n\
0 & 0 & 0 & cdots & 1\
endpmatrix$$
and then I found the characteristic polynomial:
$f_T(x) = (x-1)^n+1$.

Now I know that the minimal polynomial is $m_T in (x-1),(x-1)^2,space...space,space(x-1)^n+1$

My guess is that $m_T = (x-1)^n+1$ but I don't know how to find which one it is.

linear-algebra linear-transformations characteristic-functions minimal-polynomials

share|cite|improve this question

asked Mar 30 at 12:43

Guysudai1Guysudai1

18911

$endgroup$

share|cite|improve this question

asked Mar 30 at 12:43

Guysudai1Guysudai1

18911

share|cite|improve this question

asked Mar 30 at 12:43

Guysudai1Guysudai1

18911

asked Mar 30 at 12:43

Guysudai1Guysudai1

18911

asked Mar 30 at 12:43

Guysudai1Guysudai1

18911

1 Answer
1

active

oldest

votes

1

$begingroup$

Let's denote by $S$ the derivative operator $S(p) = p'$. You noted that the minimal polynomial of $T$ has the form $(x-1)^k$ for $1 leq k leq n + 1$ so let $m(x) = (x-1)^k$. For $m$ to be the minimal polynomial of $T$, we must have

$$ m(T) = (T - I)^k = S^k = 0. $$

However, if $k leq n$ then
$$ S^k(x^k) = k! neq 0$$

so we must have $k = n + 1$. And indeed, $S^n+1$ acts on polynomials by taking the derivative $n + 1$ times and since all the polynomials $S$ acts on are of degree $ leq n$ we get $S^n+1 = 0$.

share|cite|improve this answer

answered Mar 30 at 14:03

levaplevap

48k33274

$endgroup$

add a comment | 

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1

$begingroup$

Let's denote by $S$ the derivative operator $S(p) = p'$. You noted that the minimal polynomial of $T$ has the form $(x-1)^k$ for $1 leq k leq n + 1$ so let $m(x) = (x-1)^k$. For $m$ to be the minimal polynomial of $T$, we must have

$$ m(T) = (T - I)^k = S^k = 0. $$

However, if $k leq n$ then
$$ S^k(x^k) = k! neq 0$$

so we must have $k = n + 1$. And indeed, $S^n+1$ acts on polynomials by taking the derivative $n + 1$ times and since all the polynomials $S$ acts on are of degree $ leq n$ we get $S^n+1 = 0$.

share|cite|improve this answer

answered Mar 30 at 14:03

levaplevap

48k33274

$endgroup$

add a comment | 

1

$begingroup$

Let's denote by $S$ the derivative operator $S(p) = p'$. You noted that the minimal polynomial of $T$ has the form $(x-1)^k$ for $1 leq k leq n + 1$ so let $m(x) = (x-1)^k$. For $m$ to be the minimal polynomial of $T$, we must have

$$ m(T) = (T - I)^k = S^k = 0. $$

However, if $k leq n$ then
$$ S^k(x^k) = k! neq 0$$

so we must have $k = n + 1$. And indeed, $S^n+1$ acts on polynomials by taking the derivative $n + 1$ times and since all the polynomials $S$ acts on are of degree $ leq n$ we get $S^n+1 = 0$.

share|cite|improve this answer

answered Mar 30 at 14:03

levaplevap

48k33274

$endgroup$

add a comment | 

$begingroup$

Let's denote by $S$ the derivative operator $S(p) = p'$. You noted that the minimal polynomial of $T$ has the form $(x-1)^k$ for $1 leq k leq n + 1$ so let $m(x) = (x-1)^k$. For $m$ to be the minimal polynomial of $T$, we must have

$$ m(T) = (T - I)^k = S^k = 0. $$

However, if $k leq n$ then
$$ S^k(x^k) = k! neq 0$$

so we must have $k = n + 1$. And indeed, $S^n+1$ acts on polynomials by taking the derivative $n + 1$ times and since all the polynomials $S$ acts on are of degree $ leq n$ we get $S^n+1 = 0$.

share|cite|improve this answer

answered Mar 30 at 14:03

levaplevap

48k33274

$endgroup$

share|cite|improve this answer

answered Mar 30 at 14:03

levaplevap

48k33274

answered Mar 30 at 14:03

levaplevap

48k33274

answered Mar 30 at 14:03

levaplevap

48k33274