Mixing
Prove injectivity is equivalent to null space equals ${0}$
$begingroup$
I think I got the intuition of the reasoning. Because injectivity means one-to-one, if null space is larger than ${0}$, there are more than one way to transform $V$ to ${0}$. So it is not injective.
But I don't understand the proof.
Proof:
First suppose $T$ is injective. We want to prove that null $T = {0}$. We already know that ${0} subset text{null }T$ (I understand). To prove the inclusion in the other direction (what's that mean?), suppose $v in text{null } T$. Then
$$T(v) = 0 = T(0)$$
Because $T$ is injective, the equation above implies that $v=0$. Thus we can conclude that $text{null }T = {0}$, as desired.
To prove the implication in the other direction (what's that mean?), now suppose $text{null }T = {0}$. We want to prove that $T$ is injective. To do this, suppose $u, v in V$ and $Tu = Tv$. Then
$$0 = Tu - Tv = T(u-v)$$
is this step just coming from $Tu = Tv$? $Tu - Tv = 0$
Thus $u-v$ is in null $T$, which equals to ${0}$. Hence $u-v = 0$, which implies that $u=v$. Hence $T$ is injective, as desired
linear-algebra linear-transformations proof-explanation
asked Jan 24 at 22:56
JOHN JOHN
4279
$endgroup$
add a comment |
$begingroup$
I think I got the intuition of the reasoning. Because injectivity means one-to-one, if null space is larger than ${0}$, there are more than one way to transform $V$ to ${0}$. So it is not injective.
But I don't understand the proof.
Proof:
First suppose $T$ is injective. We want to prove that null $T = {0}$. We already know that ${0} subset text{null }T$ (I understand). To prove the inclusion in the other direction (what's that mean?), suppose $v in text{null } T$. Then
$$T(v) = 0 = T(0)$$
Because $T$ is injective, the equation above implies that $v=0$. Thus we can conclude that $text{null }T = {0}$, as desired.
To prove the implication in the other direction (what's that mean?), now suppose $text{null }T = {0}$. We want to prove that $T$ is injective. To do this, suppose $u, v in V$ and $Tu = Tv$. Then
$$0 = Tu - Tv = T(u-v)$$
is this step just coming from $Tu = Tv$? $Tu - Tv = 0$
Thus $u-v$ is in null $T$, which equals to ${0}$. Hence $u-v = 0$, which implies that $u=v$. Hence $T$ is injective, as desired
linear-algebra linear-transformations proof-explanation
asked Jan 24 at 22:56
JOHN JOHN
4279
$endgroup$
add a comment |
$begingroup$
I think I got the intuition of the reasoning. Because injectivity means one-to-one, if null space is larger than ${0}$, there are more than one way to transform $V$ to ${0}$. So it is not injective.
But I don't understand the proof.
Proof:
First suppose $T$ is injective. We want to prove that null $T = {0}$. We already know that ${0} subset text{null }T$ (I understand). To prove the inclusion in the other direction (what's that mean?), suppose $v in text{null } T$. Then
$$T(v) = 0 = T(0)$$
Because $T$ is injective, the equation above implies that $v=0$. Thus we can conclude that $text{null }T = {0}$, as desired.
To prove the implication in the other direction (what's that mean?), now suppose $text{null }T = {0}$. We want to prove that $T$ is injective. To do this, suppose $u, v in V$ and $Tu = Tv$. Then
$$0 = Tu - Tv = T(u-v)$$
is this step just coming from $Tu = Tv$? $Tu - Tv = 0$
Thus $u-v$ is in null $T$, which equals to ${0}$. Hence $u-v = 0$, which implies that $u=v$. Hence $T$ is injective, as desired
linear-algebra linear-transformations proof-explanation
asked Jan 24 at 22:56
JOHN JOHN
4279
$endgroup$
I think I got the intuition of the reasoning. Because injectivity means one-to-one, if null space is larger than ${0}$, there are more than one way to transform $V$ to ${0}$. So it is not injective.
But I don't understand the proof.
Proof:
First suppose $T$ is injective. We want to prove that null $T = {0}$. We already know that ${0} subset text{null }T$ (I understand). To prove the inclusion in the other direction (what's that mean?), suppose $v in text{null } T$. Then
$$T(v) = 0 = T(0)$$
Because $T$ is injective, the equation above implies that $v=0$. Thus we can conclude that $text{null }T = {0}$, as desired.
To prove the implication in the other direction (what's that mean?), now suppose $text{null }T = {0}$. We want to prove that $T$ is injective. To do this, suppose $u, v in V$ and $Tu = Tv$. Then
$$0 = Tu - Tv = T(u-v)$$
is this step just coming from $Tu = Tv$? $Tu - Tv = 0$
Thus $u-v$ is in null $T$, which equals to ${0}$. Hence $u-v = 0$, which implies that $u=v$. Hence $T$ is injective, as desired
linear-algebra linear-transformations proof-explanation
linear-algebra linear-transformations proof-explanation
asked Jan 24 at 22:56
JOHN JOHN
4279
asked Jan 24 at 22:56
JOHN JOHN
4279
asked Jan 24 at 22:56
JOHN JOHN
4279
asked Jan 24 at 22:56
JOHN JOHN
4279
asked Jan 24 at 22:56
JOHN JOHN
4279
4279
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The statement is$$Ttext{ is injective}iffoperatorname{null}T={0}.$$So, there are actually two things to prove:
$Ttext{ is injective}impliesoperatorname{null}T={0}$;- $operatorname{null}T={0}implies Ttext{ is injective}.$
Now, in order to prove the first statement, we have to prove that if $T$ is injective, then the sets ${0}$ and $operatorname{null}T$ are equal. But, since it is trivial that ${0}subsetoperatorname{null}T$, what remains to be proved is the opposite inclusion, which is $operatorname{null}Tsubset{0}$.
The second statement is the implication in the other direction. Finally, the equality $T(u)-T(v)=T(u-v)$ comes from the fact that the map $T$ is linear.
answered Jan 24 at 23:07
José Carlos SantosJosé Carlos Santos
168k23132236
$endgroup$
add a comment |
$begingroup$
Proving the inclusion in the other direction means proving the reverse inclusion: $ker Tsubset{0}$.
Provin the implication in ther direction means this:
The author has just proved that injectivity implies $ker T={0}$. ($Rightarrow$). Now he/she going to prove the reverse implication:
$$text{injectivity}Leftarrow ker T={0}.$$
Fianlly the step you mention comes from the hypothesis $Tu=Tv;$ AND $;$linearity of $T$: $; Tu-Tv=T(u-v)$.
answered Jan 24 at 23:08
BernardBernard
123k741116
$endgroup$
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%2f3086474%2fprove-injectivity-is-equivalent-to-null-space-equals-0%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The statement is$$Ttext{ is injective}iffoperatorname{null}T={0}.$$So, there are actually two things to prove:
$Ttext{ is injective}impliesoperatorname{null}T={0}$;- $operatorname{null}T={0}implies Ttext{ is injective}.$
Now, in order to prove the first statement, we have to prove that if $T$ is injective, then the sets ${0}$ and $operatorname{null}T$ are equal. But, since it is trivial that ${0}subsetoperatorname{null}T$, what remains to be proved is the opposite inclusion, which is $operatorname{null}Tsubset{0}$.
The second statement is the implication in the other direction. Finally, the equality $T(u)-T(v)=T(u-v)$ comes from the fact that the map $T$ is linear.
answered Jan 24 at 23:07
José Carlos SantosJosé Carlos Santos
168k23132236
$endgroup$
add a comment |
$begingroup$
The statement is$$Ttext{ is injective}iffoperatorname{null}T={0}.$$So, there are actually two things to prove:
$Ttext{ is injective}impliesoperatorname{null}T={0}$;- $operatorname{null}T={0}implies Ttext{ is injective}.$
Now, in order to prove the first statement, we have to prove that if $T$ is injective, then the sets ${0}$ and $operatorname{null}T$ are equal. But, since it is trivial that ${0}subsetoperatorname{null}T$, what remains to be proved is the opposite inclusion, which is $operatorname{null}Tsubset{0}$.
The second statement is the implication in the other direction. Finally, the equality $T(u)-T(v)=T(u-v)$ comes from the fact that the map $T$ is linear.
answered Jan 24 at 23:07
José Carlos SantosJosé Carlos Santos
168k23132236
$endgroup$
add a comment |
$begingroup$
The statement is$$Ttext{ is injective}iffoperatorname{null}T={0}.$$So, there are actually two things to prove:
$Ttext{ is injective}impliesoperatorname{null}T={0}$;- $operatorname{null}T={0}implies Ttext{ is injective}.$
Now, in order to prove the first statement, we have to prove that if $T$ is injective, then the sets ${0}$ and $operatorname{null}T$ are equal. But, since it is trivial that ${0}subsetoperatorname{null}T$, what remains to be proved is the opposite inclusion, which is $operatorname{null}Tsubset{0}$.
The second statement is the implication in the other direction. Finally, the equality $T(u)-T(v)=T(u-v)$ comes from the fact that the map $T$ is linear.
answered Jan 24 at 23:07
José Carlos SantosJosé Carlos Santos
168k23132236
$endgroup$
The statement is$$Ttext{ is injective}iffoperatorname{null}T={0}.$$So, there are actually two things to prove:
$Ttext{ is injective}impliesoperatorname{null}T={0}$;- $operatorname{null}T={0}implies Ttext{ is injective}.$
Now, in order to prove the first statement, we have to prove that if $T$ is injective, then the sets ${0}$ and $operatorname{null}T$ are equal. But, since it is trivial that ${0}subsetoperatorname{null}T$, what remains to be proved is the opposite inclusion, which is $operatorname{null}Tsubset{0}$.
The second statement is the implication in the other direction. Finally, the equality $T(u)-T(v)=T(u-v)$ comes from the fact that the map $T$ is linear.
answered Jan 24 at 23:07
José Carlos SantosJosé Carlos Santos
168k23132236
edited Jan 25 at 16:02
answered Jan 24 at 23:07
José Carlos SantosJosé Carlos Santos
168k23132236
answered Jan 24 at 23:07
José Carlos SantosJosé Carlos Santos
168k23132236
answered Jan 24 at 23:07
José Carlos SantosJosé Carlos Santos
168k23132236
168k23132236
add a comment |
add a comment |
$begingroup$
Proving the inclusion in the other direction means proving the reverse inclusion: $ker Tsubset{0}$.
Provin the implication in ther direction means this:
The author has just proved that injectivity implies $ker T={0}$. ($Rightarrow$). Now he/she going to prove the reverse implication:
$$text{injectivity}Leftarrow ker T={0}.$$
Fianlly the step you mention comes from the hypothesis $Tu=Tv;$ AND $;$linearity of $T$: $; Tu-Tv=T(u-v)$.
answered Jan 24 at 23:08
BernardBernard
123k741116
$endgroup$
add a comment |
$begingroup$
Proving the inclusion in the other direction means proving the reverse inclusion: $ker Tsubset{0}$.
Provin the implication in ther direction means this:
The author has just proved that injectivity implies $ker T={0}$. ($Rightarrow$). Now he/she going to prove the reverse implication:
$$text{injectivity}Leftarrow ker T={0}.$$
Fianlly the step you mention comes from the hypothesis $Tu=Tv;$ AND $;$linearity of $T$: $; Tu-Tv=T(u-v)$.
answered Jan 24 at 23:08
BernardBernard
123k741116
$endgroup$
add a comment |
$begingroup$
Proving the inclusion in the other direction means proving the reverse inclusion: $ker Tsubset{0}$.
Provin the implication in ther direction means this:
The author has just proved that injectivity implies $ker T={0}$. ($Rightarrow$). Now he/she going to prove the reverse implication:
$$text{injectivity}Leftarrow ker T={0}.$$
Fianlly the step you mention comes from the hypothesis $Tu=Tv;$ AND $;$linearity of $T$: $; Tu-Tv=T(u-v)$.
answered Jan 24 at 23:08
BernardBernard
123k741116
$endgroup$
Proving the inclusion in the other direction means proving the reverse inclusion: $ker Tsubset{0}$.
Provin the implication in ther direction means this:
The author has just proved that injectivity implies $ker T={0}$. ($Rightarrow$). Now he/she going to prove the reverse implication:
$$text{injectivity}Leftarrow ker T={0}.$$
Fianlly the step you mention comes from the hypothesis $Tu=Tv;$ AND $;$linearity of $T$: $; Tu-Tv=T(u-v)$.
answered Jan 24 at 23:08
BernardBernard
123k741116
answered Jan 24 at 23:08
BernardBernard
123k741116
answered Jan 24 at 23:08
BernardBernard
123k741116
answered Jan 24 at 23:08
BernardBernard
123k741116
123k741116
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%2f3086474%2fprove-injectivity-is-equivalent-to-null-space-equals-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
