Show that the Image of $0$ under a Linear Mapping is also $0$
$begingroup$
In my book of Linear Algebra, I have the following exercises:
Let $T: V to W$ be a linear map from one vector space to another. Show that $T(0) = 0$.
I'm somehow having a block.
For me, it is natural to express it like: If $T(0)$ were not in $Ker(T)$ and thus contradiction, because $Ker(T)$ is a subspace by default (the way it is constructed).
How can we show that with more elementarily? (with less abstraction.)
linear-algebra linear-transformations alternative-proof
edited Jan 15 at 23:22
José Carlos Santos
159k22126231
asked Jan 15 at 23:19
freehumoristfreehumorist
349214
$endgroup$
add a comment |
$begingroup$
In my book of Linear Algebra, I have the following exercises:
Let $T: V to W$ be a linear map from one vector space to another. Show that $T(0) = 0$.
I'm somehow having a block.
For me, it is natural to express it like: If $T(0)$ were not in $Ker(T)$ and thus contradiction, because $Ker(T)$ is a subspace by default (the way it is constructed).
How can we show that with more elementarily? (with less abstraction.)
linear-algebra linear-transformations alternative-proof
edited Jan 15 at 23:22
José Carlos Santos
159k22126231
asked Jan 15 at 23:19
freehumoristfreehumorist
349214
$endgroup$
2
$begingroup$
Hint: Try to deduce $T(0)=T(0)+T(0)$ somehow.
$endgroup$
– BigbearZzz
Jan 15 at 23:24
$begingroup$
@BigbearZzz i will
$endgroup$
– freehumorist
Jan 15 at 23:31
1
$begingroup$
By definition of linear mapping we know that for any vector $ v in V $ $ T(0 v) = 0 t(V) $
$endgroup$
– CrabMan
Jan 27 at 13:45
add a comment |
$begingroup$
In my book of Linear Algebra, I have the following exercises:
Let $T: V to W$ be a linear map from one vector space to another. Show that $T(0) = 0$.
I'm somehow having a block.
For me, it is natural to express it like: If $T(0)$ were not in $Ker(T)$ and thus contradiction, because $Ker(T)$ is a subspace by default (the way it is constructed).
How can we show that with more elementarily? (with less abstraction.)
linear-algebra linear-transformations alternative-proof
edited Jan 15 at 23:22
José Carlos Santos
159k22126231
asked Jan 15 at 23:19
freehumoristfreehumorist
349214
$endgroup$
In my book of Linear Algebra, I have the following exercises:
Let $T: V to W$ be a linear map from one vector space to another. Show that $T(0) = 0$.
I'm somehow having a block.
For me, it is natural to express it like: If $T(0)$ were not in $Ker(T)$ and thus contradiction, because $Ker(T)$ is a subspace by default (the way it is constructed).
How can we show that with more elementarily? (with less abstraction.)
linear-algebra linear-transformations alternative-proof
linear-algebra linear-transformations alternative-proof
edited Jan 15 at 23:22
José Carlos Santos
159k22126231
asked Jan 15 at 23:19
freehumoristfreehumorist
349214
edited Jan 15 at 23:22
José Carlos Santos
159k22126231
asked Jan 15 at 23:19
freehumoristfreehumorist
349214
edited Jan 15 at 23:22
José Carlos Santos
159k22126231
edited Jan 15 at 23:22
José Carlos Santos
159k22126231
edited Jan 15 at 23:22
José Carlos Santos
159k22126231
159k22126231
asked Jan 15 at 23:19
freehumoristfreehumorist
349214
asked Jan 15 at 23:19
freehumoristfreehumorist
349214
asked Jan 15 at 23:19
freehumoristfreehumorist
349214
349214
2
$begingroup$
Hint: Try to deduce $T(0)=T(0)+T(0)$ somehow.
$endgroup$
– BigbearZzz
Jan 15 at 23:24
$begingroup$
@BigbearZzz i will
$endgroup$
– freehumorist
Jan 15 at 23:31
1
$begingroup$
By definition of linear mapping we know that for any vector $ v in V $ $ T(0 v) = 0 t(V) $
$endgroup$
– CrabMan
Jan 27 at 13:45
add a comment |
2
$begingroup$
Hint: Try to deduce $T(0)=T(0)+T(0)$ somehow.
$endgroup$
– BigbearZzz
Jan 15 at 23:24
$begingroup$
@BigbearZzz i will
$endgroup$
– freehumorist
Jan 15 at 23:31
1
$begingroup$
By definition of linear mapping we know that for any vector $ v in V $ $ T(0 v) = 0 t(V) $
$endgroup$
– CrabMan
Jan 27 at 13:45
2
2
$begingroup$
Hint: Try to deduce $T(0)=T(0)+T(0)$ somehow.
$endgroup$
– BigbearZzz
Jan 15 at 23:24
$begingroup$
Hint: Try to deduce $T(0)=T(0)+T(0)$ somehow.
$endgroup$
– BigbearZzz
Jan 15 at 23:24
$begingroup$
@BigbearZzz i will
$endgroup$
– freehumorist
Jan 15 at 23:31
$begingroup$
@BigbearZzz i will
$endgroup$
– freehumorist
Jan 15 at 23:31
1
1
$begingroup$
By definition of linear mapping we know that for any vector $ v in V $ $ T(0 v) = 0 t(V) $
$endgroup$
– CrabMan
Jan 27 at 13:45
$begingroup$
By definition of linear mapping we know that for any vector $ v in V $ $ T(0 v) = 0 t(V) $
$endgroup$
– CrabMan
Jan 27 at 13:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: Begin with $0=0+0implies T(0)=T(0+0)$.
answered Jan 15 at 23:21
José Carlos SantosJosé Carlos Santos
159k22126231
$endgroup$
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%2f3075110%2fshow-that-the-image-of-0-under-a-linear-mapping-is-also-0%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
$begingroup$
Hint: Begin with $0=0+0implies T(0)=T(0+0)$.
answered Jan 15 at 23:21
José Carlos SantosJosé Carlos Santos
159k22126231
$endgroup$
add a comment |
$begingroup$
Hint: Begin with $0=0+0implies T(0)=T(0+0)$.
answered Jan 15 at 23:21
José Carlos SantosJosé Carlos Santos
159k22126231
$endgroup$
add a comment |
$begingroup$
Hint: Begin with $0=0+0implies T(0)=T(0+0)$.
answered Jan 15 at 23:21
José Carlos SantosJosé Carlos Santos
159k22126231
$endgroup$
Hint: Begin with $0=0+0implies T(0)=T(0+0)$.
answered Jan 15 at 23:21
José Carlos SantosJosé Carlos Santos
159k22126231
answered Jan 15 at 23:21
José Carlos SantosJosé Carlos Santos
159k22126231
answered Jan 15 at 23:21
José Carlos SantosJosé Carlos Santos
159k22126231
answered Jan 15 at 23:21
José Carlos SantosJosé Carlos Santos
159k22126231
159k22126231
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%2f3075110%2fshow-that-the-image-of-0-under-a-linear-mapping-is-also-0%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
2
$begingroup$
Hint: Try to deduce $T(0)=T(0)+T(0)$ somehow.
$endgroup$
– BigbearZzz
Jan 15 at 23:24
$begingroup$
@BigbearZzz i will
$endgroup$
– freehumorist
Jan 15 at 23:31
1
$begingroup$
By definition of linear mapping we know that for any vector $ v in V $ $ T(0 v) = 0 t(V) $
$endgroup$
– CrabMan
Jan 27 at 13:45