Easiest way to prove that the operator is zero
$begingroup$
If an operator $T in L(V)$ satisfies $Tv = 0$ for every $v in V$, then $T = 0$.
By the way, $V$ is a finite dimensional vector space.
Though, I am pretty certain that the statement is true, but I haven't proved it. Is there any way to prove that statement in an easy way?
(What I thought was, if we can represent that T in a basis of V, then if we put vectors $(1,0,...0), (0,1,...0), ... (0,0,...1)$, it is easy to show that all the entries of the matrix is zero which makes the operator $T$ an zero operator. However, I just wonder is there any shorter way to prove it)
linear-algebra matrices
asked Mar 19 '14 at 2:31
JaebumJaebum
1064
$endgroup$
add a comment |
$begingroup$
If an operator $T in L(V)$ satisfies $Tv = 0$ for every $v in V$, then $T = 0$.
By the way, $V$ is a finite dimensional vector space.
Though, I am pretty certain that the statement is true, but I haven't proved it. Is there any way to prove that statement in an easy way?
(What I thought was, if we can represent that T in a basis of V, then if we put vectors $(1,0,...0), (0,1,...0), ... (0,0,...1)$, it is easy to show that all the entries of the matrix is zero which makes the operator $T$ an zero operator. However, I just wonder is there any shorter way to prove it)
linear-algebra matrices
asked Mar 19 '14 at 2:31
JaebumJaebum
1064
$endgroup$
$begingroup$
How about using Contradiction? It works on the same idea but backwards. Assume that the operator is nonzero and design a $v$ such that $Tvneq0$?
$endgroup$
– Nitish
Mar 19 '14 at 2:36
$begingroup$
@Nitish I thought about that but I dont know how to come up a vector that $Tv neq 0$.
$endgroup$
– Jaebum
Mar 19 '14 at 2:39
$begingroup$
Is it a CIRCULAR question ?.
$endgroup$
– Felix Marin
Mar 19 '14 at 2:43
1
$begingroup$
What is the definition of two operators being equal? Check your textbook.
$endgroup$
– Omnomnomnom
Mar 19 '14 at 2:45
add a comment |
$begingroup$
If an operator $T in L(V)$ satisfies $Tv = 0$ for every $v in V$, then $T = 0$.
By the way, $V$ is a finite dimensional vector space.
Though, I am pretty certain that the statement is true, but I haven't proved it. Is there any way to prove that statement in an easy way?
(What I thought was, if we can represent that T in a basis of V, then if we put vectors $(1,0,...0), (0,1,...0), ... (0,0,...1)$, it is easy to show that all the entries of the matrix is zero which makes the operator $T$ an zero operator. However, I just wonder is there any shorter way to prove it)
linear-algebra matrices
asked Mar 19 '14 at 2:31
JaebumJaebum
1064
$endgroup$
If an operator $T in L(V)$ satisfies $Tv = 0$ for every $v in V$, then $T = 0$.
By the way, $V$ is a finite dimensional vector space.
Though, I am pretty certain that the statement is true, but I haven't proved it. Is there any way to prove that statement in an easy way?
(What I thought was, if we can represent that T in a basis of V, then if we put vectors $(1,0,...0), (0,1,...0), ... (0,0,...1)$, it is easy to show that all the entries of the matrix is zero which makes the operator $T$ an zero operator. However, I just wonder is there any shorter way to prove it)
linear-algebra matrices
linear-algebra matrices
asked Mar 19 '14 at 2:31
JaebumJaebum
1064
asked Mar 19 '14 at 2:31
JaebumJaebum
1064
asked Mar 19 '14 at 2:31
JaebumJaebum
1064
asked Mar 19 '14 at 2:31
JaebumJaebum
1064
asked Mar 19 '14 at 2:31
JaebumJaebum
1064
1064
$begingroup$
How about using Contradiction? It works on the same idea but backwards. Assume that the operator is nonzero and design a $v$ such that $Tvneq0$?
$endgroup$
– Nitish
Mar 19 '14 at 2:36
$begingroup$
@Nitish I thought about that but I dont know how to come up a vector that $Tv neq 0$.
$endgroup$
– Jaebum
Mar 19 '14 at 2:39
$begingroup$
Is it a CIRCULAR question ?.
$endgroup$
– Felix Marin
Mar 19 '14 at 2:43
1
$begingroup$
What is the definition of two operators being equal? Check your textbook.
$endgroup$
– Omnomnomnom
Mar 19 '14 at 2:45
add a comment |
$begingroup$
How about using Contradiction? It works on the same idea but backwards. Assume that the operator is nonzero and design a $v$ such that $Tvneq0$?
$endgroup$
– Nitish
Mar 19 '14 at 2:36
$begingroup$
@Nitish I thought about that but I dont know how to come up a vector that $Tv neq 0$.
$endgroup$
– Jaebum
Mar 19 '14 at 2:39
$begingroup$
Is it a CIRCULAR question ?.
$endgroup$
– Felix Marin
Mar 19 '14 at 2:43
1
$begingroup$
What is the definition of two operators being equal? Check your textbook.
$endgroup$
– Omnomnomnom
Mar 19 '14 at 2:45
$begingroup$
How about using Contradiction? It works on the same idea but backwards. Assume that the operator is nonzero and design a $v$ such that $Tvneq0$?
$endgroup$
– Nitish
Mar 19 '14 at 2:36
$begingroup$
How about using Contradiction? It works on the same idea but backwards. Assume that the operator is nonzero and design a $v$ such that $Tvneq0$?
$endgroup$
– Nitish
Mar 19 '14 at 2:36
$begingroup$
@Nitish I thought about that but I dont know how to come up a vector that $Tv neq 0$.
$endgroup$
– Jaebum
Mar 19 '14 at 2:39
$begingroup$
@Nitish I thought about that but I dont know how to come up a vector that $Tv neq 0$.
$endgroup$
– Jaebum
Mar 19 '14 at 2:39
$begingroup$
Is it a CIRCULAR question ?.
$endgroup$
– Felix Marin
Mar 19 '14 at 2:43
$begingroup$
Is it a CIRCULAR question ?.
$endgroup$
– Felix Marin
Mar 19 '14 at 2:43
1
1
$begingroup$
What is the definition of two operators being equal? Check your textbook.
$endgroup$
– Omnomnomnom
Mar 19 '14 at 2:45
$begingroup$
What is the definition of two operators being equal? Check your textbook.
$endgroup$
– Omnomnomnom
Mar 19 '14 at 2:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
"$T=0$" means "$Tv=0$ for all $vin V$". There is nothing to prove.
answered Mar 20 '14 at 1:05
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
add a comment |
Your Answer
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%2f717686%2feasiest-way-to-prove-that-the-operator-is-zero%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
$begingroup$
"$T=0$" means "$Tv=0$ for all $vin V$". There is nothing to prove.
answered Mar 20 '14 at 1:05
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
add a comment |
$begingroup$
"$T=0$" means "$Tv=0$ for all $vin V$". There is nothing to prove.
answered Mar 20 '14 at 1:05
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
add a comment |
$begingroup$
"$T=0$" means "$Tv=0$ for all $vin V$". There is nothing to prove.
answered Mar 20 '14 at 1:05
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
"$T=0$" means "$Tv=0$ for all $vin V$". There is nothing to prove.
answered Mar 20 '14 at 1:05
Martin ArgeramiMartin Argerami
129k1184185
answered Mar 20 '14 at 1:05
Martin ArgeramiMartin Argerami
129k1184185
answered Mar 20 '14 at 1:05
Martin ArgeramiMartin Argerami
129k1184185
answered Mar 20 '14 at 1:05
Martin ArgeramiMartin Argerami
129k1184185
129k1184185
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.
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%2f717686%2feasiest-way-to-prove-that-the-operator-is-zero%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
$begingroup$
How about using Contradiction? It works on the same idea but backwards. Assume that the operator is nonzero and design a $v$ such that $Tvneq0$?
$endgroup$
– Nitish
Mar 19 '14 at 2:36
$begingroup$
@Nitish I thought about that but I dont know how to come up a vector that $Tv neq 0$.
$endgroup$
– Jaebum
Mar 19 '14 at 2:39
$begingroup$
Is it a CIRCULAR question ?.
$endgroup$
– Felix Marin
Mar 19 '14 at 2:43
1
$begingroup$
What is the definition of two operators being equal? Check your textbook.
$endgroup$
– Omnomnomnom
Mar 19 '14 at 2:45