What is the n-fold composition of a linear map “T”, if n is zero?
$begingroup$
I am doing an assignment on Groups of Linear Transformations. Here is a defnition with regards to n-fold compositions:
"Let $T in mathcal{L}(V)$ denote the n-fold compositions of $T$ with itself as $T^{n}$. For instance, the two-fold composition, $Tcirc T$ is denoted $T^{2}$. Let $T^{-n}$ denote the n-fold composition of $T^{-1}$."
So, with the above definition in mind, $T^{1}$ is just T, but what would $T^{0}$ yeild?
linear-algebra group-theory linear-transformations function-and-relation-composition
$endgroup$
add a comment |
$begingroup$
I am doing an assignment on Groups of Linear Transformations. Here is a defnition with regards to n-fold compositions:
"Let $T in mathcal{L}(V)$ denote the n-fold compositions of $T$ with itself as $T^{n}$. For instance, the two-fold composition, $Tcirc T$ is denoted $T^{2}$. Let $T^{-n}$ denote the n-fold composition of $T^{-1}$."
So, with the above definition in mind, $T^{1}$ is just T, but what would $T^{0}$ yeild?
linear-algebra group-theory linear-transformations function-and-relation-composition
$endgroup$
1
$begingroup$
Identity linear function.
$endgroup$
– Will M.
Dec 1 '18 at 5:46
$begingroup$
@WillM. Thanks but can you explain how/why? Or is this just the way it is defined?
$endgroup$
– Jaigus
Dec 1 '18 at 5:50
$begingroup$
If $T$ were invertible and you wanted the relation $T^{p+q}=T^p circ T^q$ to hold for all integers $p$ and $q,$ then $T^0$ has to be the identity.
$endgroup$
– Will M.
Dec 1 '18 at 5:54
$begingroup$
@WillM. Thanks. If you write this as an answer I can accept it...
$endgroup$
– Jaigus
Dec 1 '18 at 5:55
$begingroup$
You are welcome. I am OK like this. I'm lazy to write it down, :P
$endgroup$
– Will M.
Dec 1 '18 at 5:59
add a comment |
$begingroup$
I am doing an assignment on Groups of Linear Transformations. Here is a defnition with regards to n-fold compositions:
"Let $T in mathcal{L}(V)$ denote the n-fold compositions of $T$ with itself as $T^{n}$. For instance, the two-fold composition, $Tcirc T$ is denoted $T^{2}$. Let $T^{-n}$ denote the n-fold composition of $T^{-1}$."
So, with the above definition in mind, $T^{1}$ is just T, but what would $T^{0}$ yeild?
linear-algebra group-theory linear-transformations function-and-relation-composition
$endgroup$
I am doing an assignment on Groups of Linear Transformations. Here is a defnition with regards to n-fold compositions:
"Let $T in mathcal{L}(V)$ denote the n-fold compositions of $T$ with itself as $T^{n}$. For instance, the two-fold composition, $Tcirc T$ is denoted $T^{2}$. Let $T^{-n}$ denote the n-fold composition of $T^{-1}$."
So, with the above definition in mind, $T^{1}$ is just T, but what would $T^{0}$ yeild?
linear-algebra group-theory linear-transformations function-and-relation-composition
linear-algebra group-theory linear-transformations function-and-relation-composition
asked Dec 1 '18 at 5:42
JaigusJaigus
2218
2218
1
$begingroup$
Identity linear function.
$endgroup$
– Will M.
Dec 1 '18 at 5:46
$begingroup$
@WillM. Thanks but can you explain how/why? Or is this just the way it is defined?
$endgroup$
– Jaigus
Dec 1 '18 at 5:50
$begingroup$
If $T$ were invertible and you wanted the relation $T^{p+q}=T^p circ T^q$ to hold for all integers $p$ and $q,$ then $T^0$ has to be the identity.
$endgroup$
– Will M.
Dec 1 '18 at 5:54
$begingroup$
@WillM. Thanks. If you write this as an answer I can accept it...
$endgroup$
– Jaigus
Dec 1 '18 at 5:55
$begingroup$
You are welcome. I am OK like this. I'm lazy to write it down, :P
$endgroup$
– Will M.
Dec 1 '18 at 5:59
add a comment |
1
$begingroup$
Identity linear function.
$endgroup$
– Will M.
Dec 1 '18 at 5:46
$begingroup$
@WillM. Thanks but can you explain how/why? Or is this just the way it is defined?
$endgroup$
– Jaigus
Dec 1 '18 at 5:50
$begingroup$
If $T$ were invertible and you wanted the relation $T^{p+q}=T^p circ T^q$ to hold for all integers $p$ and $q,$ then $T^0$ has to be the identity.
$endgroup$
– Will M.
Dec 1 '18 at 5:54
$begingroup$
@WillM. Thanks. If you write this as an answer I can accept it...
$endgroup$
– Jaigus
Dec 1 '18 at 5:55
$begingroup$
You are welcome. I am OK like this. I'm lazy to write it down, :P
$endgroup$
– Will M.
Dec 1 '18 at 5:59
1
1
$begingroup$
Identity linear function.
$endgroup$
– Will M.
Dec 1 '18 at 5:46
$begingroup$
Identity linear function.
$endgroup$
– Will M.
Dec 1 '18 at 5:46
$begingroup$
@WillM. Thanks but can you explain how/why? Or is this just the way it is defined?
$endgroup$
– Jaigus
Dec 1 '18 at 5:50
$begingroup$
@WillM. Thanks but can you explain how/why? Or is this just the way it is defined?
$endgroup$
– Jaigus
Dec 1 '18 at 5:50
$begingroup$
If $T$ were invertible and you wanted the relation $T^{p+q}=T^p circ T^q$ to hold for all integers $p$ and $q,$ then $T^0$ has to be the identity.
$endgroup$
– Will M.
Dec 1 '18 at 5:54
$begingroup$
If $T$ were invertible and you wanted the relation $T^{p+q}=T^p circ T^q$ to hold for all integers $p$ and $q,$ then $T^0$ has to be the identity.
$endgroup$
– Will M.
Dec 1 '18 at 5:54
$begingroup$
@WillM. Thanks. If you write this as an answer I can accept it...
$endgroup$
– Jaigus
Dec 1 '18 at 5:55
$begingroup$
@WillM. Thanks. If you write this as an answer I can accept it...
$endgroup$
– Jaigus
Dec 1 '18 at 5:55
$begingroup$
You are welcome. I am OK like this. I'm lazy to write it down, :P
$endgroup$
– Will M.
Dec 1 '18 at 5:59
$begingroup$
You are welcome. I am OK like this. I'm lazy to write it down, :P
$endgroup$
– Will M.
Dec 1 '18 at 5:59
add a comment |
0
active
oldest
votes
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%2f3021027%2fwhat-is-the-n-fold-composition-of-a-linear-map-t-if-n-is-zero%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3021027%2fwhat-is-the-n-fold-composition-of-a-linear-map-t-if-n-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
1
$begingroup$
Identity linear function.
$endgroup$
– Will M.
Dec 1 '18 at 5:46
$begingroup$
@WillM. Thanks but can you explain how/why? Or is this just the way it is defined?
$endgroup$
– Jaigus
Dec 1 '18 at 5:50
$begingroup$
If $T$ were invertible and you wanted the relation $T^{p+q}=T^p circ T^q$ to hold for all integers $p$ and $q,$ then $T^0$ has to be the identity.
$endgroup$
– Will M.
Dec 1 '18 at 5:54
$begingroup$
@WillM. Thanks. If you write this as an answer I can accept it...
$endgroup$
– Jaigus
Dec 1 '18 at 5:55
$begingroup$
You are welcome. I am OK like this. I'm lazy to write it down, :P
$endgroup$
– Will M.
Dec 1 '18 at 5:59