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Universal Property for Cayley-Hamilton theorem
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Let $V$ be a vector space over a field $k$. Let $T$ be a $k$-linear transformation from $V$ to itself. Without using the notion of characteristic polynomial or Cayley-Hamilton theorem from Linear Algebra, how can I show that there exists a unique monic polynomial $p_T(X) in k[X]$ such that $p_T(T)$ is the zero transformation and that whenever $f(X) in k[X]$ satisfies the property that $f(T)$ is the zero transformation, then $p_T(X)$ divides $f(X)$ in $k[X]$?
abstract-algebra polynomials linear-transformations
edited Jul 26 at 8:03
José Carlos Santos
113k1697176
asked Jul 24 at 20:48
user579315
234
add a comment |Â
up vote
2
down vote
favorite
Let $V$ be a vector space over a field $k$. Let $T$ be a $k$-linear transformation from $V$ to itself. Without using the notion of characteristic polynomial or Cayley-Hamilton theorem from Linear Algebra, how can I show that there exists a unique monic polynomial $p_T(X) in k[X]$ such that $p_T(T)$ is the zero transformation and that whenever $f(X) in k[X]$ satisfies the property that $f(T)$ is the zero transformation, then $p_T(X)$ divides $f(X)$ in $k[X]$?
abstract-algebra polynomials linear-transformations
edited Jul 26 at 8:03
José Carlos Santos
113k1697176
asked Jul 24 at 20:48
user579315
234
1
Nakayama's lemma would seem a good launching point, at least for existence of such a polynomial
– qbert
Jul 24 at 20:56
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $V$ be a vector space over a field $k$. Let $T$ be a $k$-linear transformation from $V$ to itself. Without using the notion of characteristic polynomial or Cayley-Hamilton theorem from Linear Algebra, how can I show that there exists a unique monic polynomial $p_T(X) in k[X]$ such that $p_T(T)$ is the zero transformation and that whenever $f(X) in k[X]$ satisfies the property that $f(T)$ is the zero transformation, then $p_T(X)$ divides $f(X)$ in $k[X]$?
abstract-algebra polynomials linear-transformations
edited Jul 26 at 8:03
José Carlos Santos
113k1697176
asked Jul 24 at 20:48
user579315
234
Let $V$ be a vector space over a field $k$. Let $T$ be a $k$-linear transformation from $V$ to itself. Without using the notion of characteristic polynomial or Cayley-Hamilton theorem from Linear Algebra, how can I show that there exists a unique monic polynomial $p_T(X) in k[X]$ such that $p_T(T)$ is the zero transformation and that whenever $f(X) in k[X]$ satisfies the property that $f(T)$ is the zero transformation, then $p_T(X)$ divides $f(X)$ in $k[X]$?
abstract-algebra polynomials linear-transformations
edited Jul 26 at 8:03
José Carlos Santos
113k1697176
asked Jul 24 at 20:48
user579315
234
edited Jul 26 at 8:03
José Carlos Santos
113k1697176
edited Jul 26 at 8:03
José Carlos Santos
113k1697176
edited Jul 26 at 8:03
José Carlos Santos
113k1697176
113k1697176
asked Jul 24 at 20:48
user579315
234
asked Jul 24 at 20:48
user579315
234
asked Jul 24 at 20:48
user579315
234
234
1
Nakayama's lemma would seem a good launching point, at least for existence of such a polynomial
– qbert
Jul 24 at 20:56
add a comment |Â
1
Nakayama's lemma would seem a good launching point, at least for existence of such a polynomial
– qbert
Jul 24 at 20:56
1
1
Nakayama's lemma would seem a good launching point, at least for existence of such a polynomial
– qbert
Jul 24 at 20:56
Nakayama's lemma would seem a good launching point, at least for existence of such a polynomial
– qbert
Jul 24 at 20:56
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
7
down vote
accepted
Let $n$ be the smallest natural number such that $operatornameId,T,T^2,ldots,T^n$ are linearly dependent. Then $T^n$ is a linear combination of the other ones. In other words, there's a monic polynomal $p_T(X)$ such that $p_T(T)$ is the null transformation.
Now, if $q(X)$ is a polynomial such that $q(T)=0$, divide $q(X)$ by $p_T(X)$: there are polynomials $q^star(X),r(X)in K[X]$ such that $q(X)=p_T(X)q^star(X)+r(X)$ and that $deg r(X)<deg p_T(X)$ or $r(X)=0$. But then$$0=q(T)=p_T(T)q^star(T)+r(T)=r(T).$$So, by the choice of $n$, $r(X)=0$. In other words, $p_T(X)mid q(X)$.
answered Jul 24 at 21:08
José Carlos Santos
113k1697176
4
A higher-level way of expressing exactly the same argument: the set of $p in k[X]$ such that $p(T) = 0$ is an ideal of $k[X]$. Since $k[X]$ is a PID that implies that this "annihilator ideal of $T$" is equal to $langle p_T rangle$ for some $p_T in k[X]$. (In case $V$ is finite-dimensional, that implies that the ideal is nonzero so $p_T ne 0$.)
– Daniel Schepler
Jul 24 at 21:21
@DanielSchepler Yes, that's nice.
– José Carlos Santos
Jul 24 at 21:23
add a comment |Â
up vote
2
down vote
The set of $ntimes n$ matrices over $k$ is a so-called algebra over $k$. It is finite dimensional: namely, the dimension is $n^2$.
So every element is algebraic over the field. The direct proof is that if $A$ is an $ntimes n$ matrix, then the matrices $I, A, A^2, ldots, A^n^2$ are linearly dependent (as there are $n^2+1$ of them, which is bigger than the dimension), so some linear combination of these is zero. That produces a good polynomial: a polynomial $p(x)in k[x]$ of degree at most $n^2$ such that $p(A)=0$.
It is well-known that algebraic elements in an algebra (over a field) have a minimal polynomial.
answered Jul 24 at 21:07
A. Pongrácz
1,804116
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
Let $n$ be the smallest natural number such that $operatornameId,T,T^2,ldots,T^n$ are linearly dependent. Then $T^n$ is a linear combination of the other ones. In other words, there's a monic polynomal $p_T(X)$ such that $p_T(T)$ is the null transformation.
Now, if $q(X)$ is a polynomial such that $q(T)=0$, divide $q(X)$ by $p_T(X)$: there are polynomials $q^star(X),r(X)in K[X]$ such that $q(X)=p_T(X)q^star(X)+r(X)$ and that $deg r(X)<deg p_T(X)$ or $r(X)=0$. But then$$0=q(T)=p_T(T)q^star(T)+r(T)=r(T).$$So, by the choice of $n$, $r(X)=0$. In other words, $p_T(X)mid q(X)$.
answered Jul 24 at 21:08
José Carlos Santos
113k1697176
4
A higher-level way of expressing exactly the same argument: the set of $p in k[X]$ such that $p(T) = 0$ is an ideal of $k[X]$. Since $k[X]$ is a PID that implies that this "annihilator ideal of $T$" is equal to $langle p_T rangle$ for some $p_T in k[X]$. (In case $V$ is finite-dimensional, that implies that the ideal is nonzero so $p_T ne 0$.)
– Daniel Schepler
Jul 24 at 21:21
@DanielSchepler Yes, that's nice.
– José Carlos Santos
Jul 24 at 21:23
add a comment |Â
up vote
7
down vote
accepted
Let $n$ be the smallest natural number such that $operatornameId,T,T^2,ldots,T^n$ are linearly dependent. Then $T^n$ is a linear combination of the other ones. In other words, there's a monic polynomal $p_T(X)$ such that $p_T(T)$ is the null transformation.
Now, if $q(X)$ is a polynomial such that $q(T)=0$, divide $q(X)$ by $p_T(X)$: there are polynomials $q^star(X),r(X)in K[X]$ such that $q(X)=p_T(X)q^star(X)+r(X)$ and that $deg r(X)<deg p_T(X)$ or $r(X)=0$. But then$$0=q(T)=p_T(T)q^star(T)+r(T)=r(T).$$So, by the choice of $n$, $r(X)=0$. In other words, $p_T(X)mid q(X)$.
answered Jul 24 at 21:08
José Carlos Santos
113k1697176
4
A higher-level way of expressing exactly the same argument: the set of $p in k[X]$ such that $p(T) = 0$ is an ideal of $k[X]$. Since $k[X]$ is a PID that implies that this "annihilator ideal of $T$" is equal to $langle p_T rangle$ for some $p_T in k[X]$. (In case $V$ is finite-dimensional, that implies that the ideal is nonzero so $p_T ne 0$.)
– Daniel Schepler
Jul 24 at 21:21
@DanielSchepler Yes, that's nice.
– José Carlos Santos
Jul 24 at 21:23
add a comment |Â
up vote
7
down vote
accepted
up vote
7
down vote
accepted
Let $n$ be the smallest natural number such that $operatornameId,T,T^2,ldots,T^n$ are linearly dependent. Then $T^n$ is a linear combination of the other ones. In other words, there's a monic polynomal $p_T(X)$ such that $p_T(T)$ is the null transformation.
Now, if $q(X)$ is a polynomial such that $q(T)=0$, divide $q(X)$ by $p_T(X)$: there are polynomials $q^star(X),r(X)in K[X]$ such that $q(X)=p_T(X)q^star(X)+r(X)$ and that $deg r(X)<deg p_T(X)$ or $r(X)=0$. But then$$0=q(T)=p_T(T)q^star(T)+r(T)=r(T).$$So, by the choice of $n$, $r(X)=0$. In other words, $p_T(X)mid q(X)$.
answered Jul 24 at 21:08
José Carlos Santos
113k1697176
Let $n$ be the smallest natural number such that $operatornameId,T,T^2,ldots,T^n$ are linearly dependent. Then $T^n$ is a linear combination of the other ones. In other words, there's a monic polynomal $p_T(X)$ such that $p_T(T)$ is the null transformation.
Now, if $q(X)$ is a polynomial such that $q(T)=0$, divide $q(X)$ by $p_T(X)$: there are polynomials $q^star(X),r(X)in K[X]$ such that $q(X)=p_T(X)q^star(X)+r(X)$ and that $deg r(X)<deg p_T(X)$ or $r(X)=0$. But then$$0=q(T)=p_T(T)q^star(T)+r(T)=r(T).$$So, by the choice of $n$, $r(X)=0$. In other words, $p_T(X)mid q(X)$.
answered Jul 24 at 21:08
José Carlos Santos
113k1697176
answered Jul 24 at 21:08
José Carlos Santos
113k1697176
answered Jul 24 at 21:08
José Carlos Santos
113k1697176
answered Jul 24 at 21:08
José Carlos Santos
113k1697176
113k1697176
4
A higher-level way of expressing exactly the same argument: the set of $p in k[X]$ such that $p(T) = 0$ is an ideal of $k[X]$. Since $k[X]$ is a PID that implies that this "annihilator ideal of $T$" is equal to $langle p_T rangle$ for some $p_T in k[X]$. (In case $V$ is finite-dimensional, that implies that the ideal is nonzero so $p_T ne 0$.)
– Daniel Schepler
Jul 24 at 21:21
@DanielSchepler Yes, that's nice.
– José Carlos Santos
Jul 24 at 21:23
add a comment |Â
4
A higher-level way of expressing exactly the same argument: the set of $p in k[X]$ such that $p(T) = 0$ is an ideal of $k[X]$. Since $k[X]$ is a PID that implies that this "annihilator ideal of $T$" is equal to $langle p_T rangle$ for some $p_T in k[X]$. (In case $V$ is finite-dimensional, that implies that the ideal is nonzero so $p_T ne 0$.)
– Daniel Schepler
Jul 24 at 21:21
@DanielSchepler Yes, that's nice.
– José Carlos Santos
Jul 24 at 21:23
4
4
A higher-level way of expressing exactly the same argument: the set of $p in k[X]$ such that $p(T) = 0$ is an ideal of $k[X]$. Since $k[X]$ is a PID that implies that this "annihilator ideal of $T$" is equal to $langle p_T rangle$ for some $p_T in k[X]$. (In case $V$ is finite-dimensional, that implies that the ideal is nonzero so $p_T ne 0$.)
– Daniel Schepler
Jul 24 at 21:21
A higher-level way of expressing exactly the same argument: the set of $p in k[X]$ such that $p(T) = 0$ is an ideal of $k[X]$. Since $k[X]$ is a PID that implies that this "annihilator ideal of $T$" is equal to $langle p_T rangle$ for some $p_T in k[X]$. (In case $V$ is finite-dimensional, that implies that the ideal is nonzero so $p_T ne 0$.)
– Daniel Schepler
Jul 24 at 21:21
@DanielSchepler Yes, that's nice.
– José Carlos Santos
Jul 24 at 21:23
@DanielSchepler Yes, that's nice.
– José Carlos Santos
Jul 24 at 21:23
add a comment |Â
up vote
2
down vote
The set of $ntimes n$ matrices over $k$ is a so-called algebra over $k$. It is finite dimensional: namely, the dimension is $n^2$.
So every element is algebraic over the field. The direct proof is that if $A$ is an $ntimes n$ matrix, then the matrices $I, A, A^2, ldots, A^n^2$ are linearly dependent (as there are $n^2+1$ of them, which is bigger than the dimension), so some linear combination of these is zero. That produces a good polynomial: a polynomial $p(x)in k[x]$ of degree at most $n^2$ such that $p(A)=0$.
It is well-known that algebraic elements in an algebra (over a field) have a minimal polynomial.
answered Jul 24 at 21:07
A. Pongrácz
1,804116
add a comment |Â
up vote
2
down vote
The set of $ntimes n$ matrices over $k$ is a so-called algebra over $k$. It is finite dimensional: namely, the dimension is $n^2$.
So every element is algebraic over the field. The direct proof is that if $A$ is an $ntimes n$ matrix, then the matrices $I, A, A^2, ldots, A^n^2$ are linearly dependent (as there are $n^2+1$ of them, which is bigger than the dimension), so some linear combination of these is zero. That produces a good polynomial: a polynomial $p(x)in k[x]$ of degree at most $n^2$ such that $p(A)=0$.
It is well-known that algebraic elements in an algebra (over a field) have a minimal polynomial.
answered Jul 24 at 21:07
A. Pongrácz
1,804116
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The set of $ntimes n$ matrices over $k$ is a so-called algebra over $k$. It is finite dimensional: namely, the dimension is $n^2$.
So every element is algebraic over the field. The direct proof is that if $A$ is an $ntimes n$ matrix, then the matrices $I, A, A^2, ldots, A^n^2$ are linearly dependent (as there are $n^2+1$ of them, which is bigger than the dimension), so some linear combination of these is zero. That produces a good polynomial: a polynomial $p(x)in k[x]$ of degree at most $n^2$ such that $p(A)=0$.
It is well-known that algebraic elements in an algebra (over a field) have a minimal polynomial.
answered Jul 24 at 21:07
A. Pongrácz
1,804116
The set of $ntimes n$ matrices over $k$ is a so-called algebra over $k$. It is finite dimensional: namely, the dimension is $n^2$.
So every element is algebraic over the field. The direct proof is that if $A$ is an $ntimes n$ matrix, then the matrices $I, A, A^2, ldots, A^n^2$ are linearly dependent (as there are $n^2+1$ of them, which is bigger than the dimension), so some linear combination of these is zero. That produces a good polynomial: a polynomial $p(x)in k[x]$ of degree at most $n^2$ such that $p(A)=0$.
It is well-known that algebraic elements in an algebra (over a field) have a minimal polynomial.
answered Jul 24 at 21:07
A. Pongrácz
1,804116
edited Jul 24 at 21:29
answered Jul 24 at 21:07
A. Pongrácz
1,804116
answered Jul 24 at 21:07
A. Pongrácz
1,804116
answered Jul 24 at 21:07
A. Pongrácz
1,804116
1,804116
add a comment |Â
add a comment |Â
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1
Nakayama's lemma would seem a good launching point, at least for existence of such a polynomial
– qbert
Jul 24 at 20:56