T-Annihilators and Minimal polynomial
Let $T$ be a linear operator on a finite-dimensional vector space $V$.
Then there exists a vector $v$ in $V$ such that the $T$-annihilator of
$v$ is the minimal polynomial for $T$.
Definition: $T$-annihilator of a vector $alpha$(denoted as $p_alpha$) is the unique monic polynomial which generates the ideal
such that $g(T)alpha = 0$ for all $g$ in this ideal.
I'm trying to prove the above proposition without invoking Cyclic Decomposition Theorem.
My attempt : Assume that such a $v$ doesn't exist. Then every vector has a $T $ annihilator of degree less than that of minimal polynomial. Then define a monic polynomial $h$ which is the sum of $T$ annihilators of a given basis elements. Then $h(T)v=0 forall vin V.$ But this contradicts the definition of minimal polynomial since degree of $hlt $ degree of minimal polynomial. Hence our assumption is wrong.
Can someone verify my argument?Thanks.
linear-algebra proof-verification linear-transformations
asked Nov 23 '18 at 8:35
Thomas ShelbyThomas Shelby
1,982219
add a comment |
Let $T$ be a linear operator on a finite-dimensional vector space $V$.
Then there exists a vector $v$ in $V$ such that the $T$-annihilator of
$v$ is the minimal polynomial for $T$.
Definition: $T$-annihilator of a vector $alpha$(denoted as $p_alpha$) is the unique monic polynomial which generates the ideal
such that $g(T)alpha = 0$ for all $g$ in this ideal.
I'm trying to prove the above proposition without invoking Cyclic Decomposition Theorem.
My attempt : Assume that such a $v$ doesn't exist. Then every vector has a $T $ annihilator of degree less than that of minimal polynomial. Then define a monic polynomial $h$ which is the sum of $T$ annihilators of a given basis elements. Then $h(T)v=0 forall vin V.$ But this contradicts the definition of minimal polynomial since degree of $hlt $ degree of minimal polynomial. Hence our assumption is wrong.
Can someone verify my argument?Thanks.
linear-algebra proof-verification linear-transformations
asked Nov 23 '18 at 8:35
Thomas ShelbyThomas Shelby
1,982219
What is the definition of ``$T$-annihilator''?
– daw
Nov 23 '18 at 13:04
@daw I've included the definition in the question. Thanks.
– Thomas Shelby
Nov 23 '18 at 13:33
2
No that proof doesn't work. There is no reason that $h(T)v = 0$. If $h$ is the sum $g_1+g_2+cdots+g_n$ then $h(T)v$ equals $g_1(T)v+g_2(T)v+cdots + g_n(T)v$ with possibly only ONE term missing.
– user25959
Nov 23 '18 at 13:51
@user25959 Thank you for pointing out the mistake. I'll try some other way.
– Thomas Shelby
Nov 23 '18 at 14:05
add a comment |
Let $T$ be a linear operator on a finite-dimensional vector space $V$.
Then there exists a vector $v$ in $V$ such that the $T$-annihilator of
$v$ is the minimal polynomial for $T$.
Definition: $T$-annihilator of a vector $alpha$(denoted as $p_alpha$) is the unique monic polynomial which generates the ideal
such that $g(T)alpha = 0$ for all $g$ in this ideal.
I'm trying to prove the above proposition without invoking Cyclic Decomposition Theorem.
My attempt : Assume that such a $v$ doesn't exist. Then every vector has a $T $ annihilator of degree less than that of minimal polynomial. Then define a monic polynomial $h$ which is the sum of $T$ annihilators of a given basis elements. Then $h(T)v=0 forall vin V.$ But this contradicts the definition of minimal polynomial since degree of $hlt $ degree of minimal polynomial. Hence our assumption is wrong.
Can someone verify my argument?Thanks.
linear-algebra proof-verification linear-transformations
asked Nov 23 '18 at 8:35
Thomas ShelbyThomas Shelby
1,982219
Let $T$ be a linear operator on a finite-dimensional vector space $V$.
Then there exists a vector $v$ in $V$ such that the $T$-annihilator of
$v$ is the minimal polynomial for $T$.
Definition: $T$-annihilator of a vector $alpha$(denoted as $p_alpha$) is the unique monic polynomial which generates the ideal
such that $g(T)alpha = 0$ for all $g$ in this ideal.
I'm trying to prove the above proposition without invoking Cyclic Decomposition Theorem.
My attempt : Assume that such a $v$ doesn't exist. Then every vector has a $T $ annihilator of degree less than that of minimal polynomial. Then define a monic polynomial $h$ which is the sum of $T$ annihilators of a given basis elements. Then $h(T)v=0 forall vin V.$ But this contradicts the definition of minimal polynomial since degree of $hlt $ degree of minimal polynomial. Hence our assumption is wrong.
Can someone verify my argument?Thanks.
linear-algebra proof-verification linear-transformations
linear-algebra proof-verification linear-transformations
asked Nov 23 '18 at 8:35
Thomas ShelbyThomas Shelby
1,982219
asked Nov 23 '18 at 8:35
Thomas ShelbyThomas Shelby
1,982219
edited Nov 23 '18 at 13:32
Thomas Shelby
asked Nov 23 '18 at 8:35
Thomas ShelbyThomas Shelby
1,982219
asked Nov 23 '18 at 8:35
Thomas ShelbyThomas Shelby
1,982219
asked Nov 23 '18 at 8:35
Thomas ShelbyThomas Shelby
1,982219
1,982219
What is the definition of ``$T$-annihilator''?
– daw
Nov 23 '18 at 13:04
@daw I've included the definition in the question. Thanks.
– Thomas Shelby
Nov 23 '18 at 13:33
2
No that proof doesn't work. There is no reason that $h(T)v = 0$. If $h$ is the sum $g_1+g_2+cdots+g_n$ then $h(T)v$ equals $g_1(T)v+g_2(T)v+cdots + g_n(T)v$ with possibly only ONE term missing.
– user25959
Nov 23 '18 at 13:51
@user25959 Thank you for pointing out the mistake. I'll try some other way.
– Thomas Shelby
Nov 23 '18 at 14:05
add a comment |
What is the definition of ``$T$-annihilator''?
– daw
Nov 23 '18 at 13:04
@daw I've included the definition in the question. Thanks.
– Thomas Shelby
Nov 23 '18 at 13:33
2
No that proof doesn't work. There is no reason that $h(T)v = 0$. If $h$ is the sum $g_1+g_2+cdots+g_n$ then $h(T)v$ equals $g_1(T)v+g_2(T)v+cdots + g_n(T)v$ with possibly only ONE term missing.
– user25959
Nov 23 '18 at 13:51
@user25959 Thank you for pointing out the mistake. I'll try some other way.
– Thomas Shelby
Nov 23 '18 at 14:05
What is the definition of ``$T$-annihilator''?
– daw
Nov 23 '18 at 13:04
What is the definition of ``$T$-annihilator''?
– daw
Nov 23 '18 at 13:04
@daw I've included the definition in the question. Thanks.
– Thomas Shelby
Nov 23 '18 at 13:33
@daw I've included the definition in the question. Thanks.
– Thomas Shelby
Nov 23 '18 at 13:33
2
2
No that proof doesn't work. There is no reason that $h(T)v = 0$. If $h$ is the sum $g_1+g_2+cdots+g_n$ then $h(T)v$ equals $g_1(T)v+g_2(T)v+cdots + g_n(T)v$ with possibly only ONE term missing.
– user25959
Nov 23 '18 at 13:51
No that proof doesn't work. There is no reason that $h(T)v = 0$. If $h$ is the sum $g_1+g_2+cdots+g_n$ then $h(T)v$ equals $g_1(T)v+g_2(T)v+cdots + g_n(T)v$ with possibly only ONE term missing.
– user25959
Nov 23 '18 at 13:51
@user25959 Thank you for pointing out the mistake. I'll try some other way.
– Thomas Shelby
Nov 23 '18 at 14:05
@user25959 Thank you for pointing out the mistake. I'll try some other way.
– Thomas Shelby
Nov 23 '18 at 14:05
add a comment |
1 Answer
1
active
oldest
votes
I referred some books and this is my second attempt to prove the above proposition. If there is any mistake,please let me know.
Lemma:If the $T$ annihilator of $v_1$and $v_2$ are $p_1$and $p_2$ respectively, and if they are prime,then the $T$ annihilator of $v=v_1+v_2$ is $p_1p_2$
Suppose the minimal polynomial of $T$ is
$m(t)=f^{r}(t)$, where $f$ is irreducible. Let ${v_1,v_2,ldots ,v_n}$ be a basis of $V$ and $p_1,p_2,ldots ,p_n$ be the respective $T$ annihilators. Since each $p_i $ divides $m $,it is easy to see that $m=operatorname{lcm}(p_1,p_2,ldots ,p_n)$. From this we can see that $p_i=m=f^{r}$ for some $v_i $.
Now for the general case, where$$m(t)=f_1^{r_1}(t)cdots f_k^{r_k}(t)$$,and $f_i$'s are distinct monic irreducibles. Here we can find $w_i in W_i:=ker(f_i^{r_i}(T))$ such that $p_{w_i}=f_i^{r_i}$ since primary decomposition theorem guarantees minimal polynomial of $T$ restricted to $W_i$ is equal to $f_i^{r_i}$. Since the $f_i$'s are pairwise prime, $T$ annihilator for $w=w_1+w_2+cdots + w_k$, $p_w = p_{w_1}p_{w_2}cdots p_{w_k}=f_1^{r_1}cdots f_k^{r_k}=m $.
QED.
answered Nov 24 '18 at 7:23
Thomas ShelbyThomas Shelby
1,982219
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3010121%2ft-annihilators-and-minimal-polynomial%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
I referred some books and this is my second attempt to prove the above proposition. If there is any mistake,please let me know.
Lemma:If the $T$ annihilator of $v_1$and $v_2$ are $p_1$and $p_2$ respectively, and if they are prime,then the $T$ annihilator of $v=v_1+v_2$ is $p_1p_2$
Suppose the minimal polynomial of $T$ is
$m(t)=f^{r}(t)$, where $f$ is irreducible. Let ${v_1,v_2,ldots ,v_n}$ be a basis of $V$ and $p_1,p_2,ldots ,p_n$ be the respective $T$ annihilators. Since each $p_i $ divides $m $,it is easy to see that $m=operatorname{lcm}(p_1,p_2,ldots ,p_n)$. From this we can see that $p_i=m=f^{r}$ for some $v_i $.
Now for the general case, where$$m(t)=f_1^{r_1}(t)cdots f_k^{r_k}(t)$$,and $f_i$'s are distinct monic irreducibles. Here we can find $w_i in W_i:=ker(f_i^{r_i}(T))$ such that $p_{w_i}=f_i^{r_i}$ since primary decomposition theorem guarantees minimal polynomial of $T$ restricted to $W_i$ is equal to $f_i^{r_i}$. Since the $f_i$'s are pairwise prime, $T$ annihilator for $w=w_1+w_2+cdots + w_k$, $p_w = p_{w_1}p_{w_2}cdots p_{w_k}=f_1^{r_1}cdots f_k^{r_k}=m $.
QED.
answered Nov 24 '18 at 7:23
Thomas ShelbyThomas Shelby
1,982219
add a comment |
I referred some books and this is my second attempt to prove the above proposition. If there is any mistake,please let me know.
Lemma:If the $T$ annihilator of $v_1$and $v_2$ are $p_1$and $p_2$ respectively, and if they are prime,then the $T$ annihilator of $v=v_1+v_2$ is $p_1p_2$
Suppose the minimal polynomial of $T$ is
$m(t)=f^{r}(t)$, where $f$ is irreducible. Let ${v_1,v_2,ldots ,v_n}$ be a basis of $V$ and $p_1,p_2,ldots ,p_n$ be the respective $T$ annihilators. Since each $p_i $ divides $m $,it is easy to see that $m=operatorname{lcm}(p_1,p_2,ldots ,p_n)$. From this we can see that $p_i=m=f^{r}$ for some $v_i $.
Now for the general case, where$$m(t)=f_1^{r_1}(t)cdots f_k^{r_k}(t)$$,and $f_i$'s are distinct monic irreducibles. Here we can find $w_i in W_i:=ker(f_i^{r_i}(T))$ such that $p_{w_i}=f_i^{r_i}$ since primary decomposition theorem guarantees minimal polynomial of $T$ restricted to $W_i$ is equal to $f_i^{r_i}$. Since the $f_i$'s are pairwise prime, $T$ annihilator for $w=w_1+w_2+cdots + w_k$, $p_w = p_{w_1}p_{w_2}cdots p_{w_k}=f_1^{r_1}cdots f_k^{r_k}=m $.
QED.
answered Nov 24 '18 at 7:23
Thomas ShelbyThomas Shelby
1,982219
add a comment |
I referred some books and this is my second attempt to prove the above proposition. If there is any mistake,please let me know.
Lemma:If the $T$ annihilator of $v_1$and $v_2$ are $p_1$and $p_2$ respectively, and if they are prime,then the $T$ annihilator of $v=v_1+v_2$ is $p_1p_2$
Suppose the minimal polynomial of $T$ is
$m(t)=f^{r}(t)$, where $f$ is irreducible. Let ${v_1,v_2,ldots ,v_n}$ be a basis of $V$ and $p_1,p_2,ldots ,p_n$ be the respective $T$ annihilators. Since each $p_i $ divides $m $,it is easy to see that $m=operatorname{lcm}(p_1,p_2,ldots ,p_n)$. From this we can see that $p_i=m=f^{r}$ for some $v_i $.
Now for the general case, where$$m(t)=f_1^{r_1}(t)cdots f_k^{r_k}(t)$$,and $f_i$'s are distinct monic irreducibles. Here we can find $w_i in W_i:=ker(f_i^{r_i}(T))$ such that $p_{w_i}=f_i^{r_i}$ since primary decomposition theorem guarantees minimal polynomial of $T$ restricted to $W_i$ is equal to $f_i^{r_i}$. Since the $f_i$'s are pairwise prime, $T$ annihilator for $w=w_1+w_2+cdots + w_k$, $p_w = p_{w_1}p_{w_2}cdots p_{w_k}=f_1^{r_1}cdots f_k^{r_k}=m $.
QED.
answered Nov 24 '18 at 7:23
Thomas ShelbyThomas Shelby
1,982219
I referred some books and this is my second attempt to prove the above proposition. If there is any mistake,please let me know.
Lemma:If the $T$ annihilator of $v_1$and $v_2$ are $p_1$and $p_2$ respectively, and if they are prime,then the $T$ annihilator of $v=v_1+v_2$ is $p_1p_2$
Suppose the minimal polynomial of $T$ is
$m(t)=f^{r}(t)$, where $f$ is irreducible. Let ${v_1,v_2,ldots ,v_n}$ be a basis of $V$ and $p_1,p_2,ldots ,p_n$ be the respective $T$ annihilators. Since each $p_i $ divides $m $,it is easy to see that $m=operatorname{lcm}(p_1,p_2,ldots ,p_n)$. From this we can see that $p_i=m=f^{r}$ for some $v_i $.
Now for the general case, where$$m(t)=f_1^{r_1}(t)cdots f_k^{r_k}(t)$$,and $f_i$'s are distinct monic irreducibles. Here we can find $w_i in W_i:=ker(f_i^{r_i}(T))$ such that $p_{w_i}=f_i^{r_i}$ since primary decomposition theorem guarantees minimal polynomial of $T$ restricted to $W_i$ is equal to $f_i^{r_i}$. Since the $f_i$'s are pairwise prime, $T$ annihilator for $w=w_1+w_2+cdots + w_k$, $p_w = p_{w_1}p_{w_2}cdots p_{w_k}=f_1^{r_1}cdots f_k^{r_k}=m $.
QED.
answered Nov 24 '18 at 7:23
Thomas ShelbyThomas Shelby
1,982219
answered Nov 24 '18 at 7:23
Thomas ShelbyThomas Shelby
1,982219
answered Nov 24 '18 at 7:23
Thomas ShelbyThomas Shelby
1,982219
answered Nov 24 '18 at 7:23
Thomas ShelbyThomas Shelby
1,982219
1,982219
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3010121%2ft-annihilators-and-minimal-polynomial%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
What is the definition of ``$T$-annihilator''?
– daw
Nov 23 '18 at 13:04
@daw I've included the definition in the question. Thanks.
– Thomas Shelby
Nov 23 '18 at 13:33
2
No that proof doesn't work. There is no reason that $h(T)v = 0$. If $h$ is the sum $g_1+g_2+cdots+g_n$ then $h(T)v$ equals $g_1(T)v+g_2(T)v+cdots + g_n(T)v$ with possibly only ONE term missing.
– user25959
Nov 23 '18 at 13:51
@user25959 Thank you for pointing out the mistake. I'll try some other way.
– Thomas Shelby
Nov 23 '18 at 14:05