How to prove that $A^T y=0implies y=0$ if and only if $A$ defines a surjective linear transformation?

1

$begingroup$

I know that in vector spaces, the cokernel of a linear transformation $f$ is isomorphic to the set of all $y$ such that $f^T(y)=0.$ (This is the transpose of $f$.) Then, how can I prove that $text{coker} f={0}$ if and only if $f$ is surjective?

I would apreciate a proof without any heavy machinery (that is, without something that is not taught in a linear algebra course).

Thank you

linear-algebra abstract-algebra category-theory

share|cite|improve this question

asked Jan 10 at 11:33

Gabriel RibeiroGabriel Ribeiro

1,441522

$endgroup$

add a comment | 

1

$begingroup$

I know that in vector spaces, the cokernel of a linear transformation $f$ is isomorphic to the set of all $y$ such that $f^T(y)=0.$ (This is the transpose of $f$.) Then, how can I prove that $text{coker} f={0}$ if and only if $f$ is surjective?

I would apreciate a proof without any heavy machinery (that is, without something that is not taught in a linear algebra course).

Thank you

linear-algebra abstract-algebra category-theory

share|cite|improve this question

asked Jan 10 at 11:33

Gabriel RibeiroGabriel Ribeiro

1,441522

$endgroup$

add a comment | 

1

1

1

$begingroup$

I know that in vector spaces, the cokernel of a linear transformation $f$ is isomorphic to the set of all $y$ such that $f^T(y)=0.$ (This is the transpose of $f$.) Then, how can I prove that $text{coker} f={0}$ if and only if $f$ is surjective?

I would apreciate a proof without any heavy machinery (that is, without something that is not taught in a linear algebra course).

Thank you

linear-algebra abstract-algebra category-theory

share|cite|improve this question

asked Jan 10 at 11:33

Gabriel RibeiroGabriel Ribeiro

1,441522

$endgroup$

I know that in vector spaces, the cokernel of a linear transformation $f$ is isomorphic to the set of all $y$ such that $f^T(y)=0.$ (This is the transpose of $f$.) Then, how can I prove that $text{coker} f={0}$ if and only if $f$ is surjective?

I would apreciate a proof without any heavy machinery (that is, without something that is not taught in a linear algebra course).

Thank you

linear-algebra abstract-algebra category-theory

linear-algebra abstract-algebra category-theory

share|cite|improve this question

asked Jan 10 at 11:33

Gabriel RibeiroGabriel Ribeiro

1,441522

share|cite|improve this question

asked Jan 10 at 11:33

Gabriel RibeiroGabriel Ribeiro

1,441522

share|cite|improve this question

share|cite|improve this question

asked Jan 10 at 11:33

Gabriel RibeiroGabriel Ribeiro

1,441522

asked Jan 10 at 11:33

Gabriel RibeiroGabriel Ribeiro

1,441522

asked Jan 10 at 11:33

Gabriel RibeiroGabriel Ribeiro

1,441522

1,441522

add a comment | 

add a comment | 

3 Answers
3

active

oldest

votes

2

$begingroup$

The implication is equivalent to $A^T$ having a null kernel. This is equivalent to its rank being $m$, where $A$ is $m times n$.

So the implication is equivalent to $rank(A)=m$. Since $im(A)$ is in $K^m$, this is equivalent to $A$ defining a surjective linear transformation.

share|cite|improve this answer

answered Jan 10 at 11:39

MindlackMindlack

3,53717

$endgroup$

add a comment | 

1

$begingroup$

Left multiplication by $A^T$ is injective iff $A^T$ has a pivot in every column iff $A$ has a pivot in every row iff left multiplication by $A$ is surjective.

share|cite|improve this answer

answered Jan 10 at 16:05

André 3000André 3000

12.6k22243

$endgroup$

add a comment | 

0

$begingroup$

So, we have a function $f: Xrightarrow Y$ such that $f(x)=Ax$. Suppose $y$ is in the range of $f$. Then there is some $x$ such that $y=f(x)$, and if $f$ is injective, this $x$ is unique. So given $y$, we can define a function $g$ that returns that $x$. If $By=g(y)$, then $BAx=x$. Thus, $BA$ is the identity (on $X$). So $A^TB^T$ is also the identity (although on $Y$ rather than $X$). Thus, given any $y$, there is some $x$ such that $A^Tx=y$: if we take $x=B^Ty$, then $A^Tx=A^TB^Ty=y$. This show that $A$ being injective implies that $A^T$ is surjective. The other direction follows similarly.

In other words, being surjective is equivalent to there being a right inverse, and being injective is equivalent to there being a left inverse. So $A$ is injective $rightarrow$ $A$ has a left inverse $rightarrow$ $A^T$ has a right inverse $rightarrow$ $A$ is surjective.

share|cite|improve this answer

answered Jan 10 at 22:23

AcccumulationAcccumulation

6,9042618

$endgroup$

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
});


}
});

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.

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3068526%2fhow-to-prove-that-at-y-0-implies-y-0-if-and-only-if-a-defines-a-surjective%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

2

$begingroup$

The implication is equivalent to $A^T$ having a null kernel. This is equivalent to its rank being $m$, where $A$ is $m times n$.

So the implication is equivalent to $rank(A)=m$. Since $im(A)$ is in $K^m$, this is equivalent to $A$ defining a surjective linear transformation.

share|cite|improve this answer

answered Jan 10 at 11:39

MindlackMindlack

3,53717

$endgroup$

add a comment | 

2

$begingroup$

The implication is equivalent to $A^T$ having a null kernel. This is equivalent to its rank being $m$, where $A$ is $m times n$.

So the implication is equivalent to $rank(A)=m$. Since $im(A)$ is in $K^m$, this is equivalent to $A$ defining a surjective linear transformation.

share|cite|improve this answer

answered Jan 10 at 11:39

MindlackMindlack

3,53717

$endgroup$

add a comment | 

2

2

2

$begingroup$

The implication is equivalent to $A^T$ having a null kernel. This is equivalent to its rank being $m$, where $A$ is $m times n$.

So the implication is equivalent to $rank(A)=m$. Since $im(A)$ is in $K^m$, this is equivalent to $A$ defining a surjective linear transformation.

share|cite|improve this answer

answered Jan 10 at 11:39

MindlackMindlack

3,53717

$endgroup$

The implication is equivalent to $A^T$ having a null kernel. This is equivalent to its rank being $m$, where $A$ is $m times n$.

So the implication is equivalent to $rank(A)=m$. Since $im(A)$ is in $K^m$, this is equivalent to $A$ defining a surjective linear transformation.

share|cite|improve this answer

answered Jan 10 at 11:39

MindlackMindlack

3,53717

share|cite|improve this answer

share|cite|improve this answer

answered Jan 10 at 11:39

MindlackMindlack

3,53717

answered Jan 10 at 11:39

MindlackMindlack

3,53717

answered Jan 10 at 11:39

MindlackMindlack

3,53717

3,53717

add a comment | 

add a comment | 

1

$begingroup$

Left multiplication by $A^T$ is injective iff $A^T$ has a pivot in every column iff $A$ has a pivot in every row iff left multiplication by $A$ is surjective.

share|cite|improve this answer

answered Jan 10 at 16:05

André 3000André 3000

12.6k22243

$endgroup$

add a comment | 

1

$begingroup$

Left multiplication by $A^T$ is injective iff $A^T$ has a pivot in every column iff $A$ has a pivot in every row iff left multiplication by $A$ is surjective.

share|cite|improve this answer

answered Jan 10 at 16:05

André 3000André 3000

12.6k22243

$endgroup$

add a comment | 

1

1

1

$begingroup$

Left multiplication by $A^T$ is injective iff $A^T$ has a pivot in every column iff $A$ has a pivot in every row iff left multiplication by $A$ is surjective.

share|cite|improve this answer

answered Jan 10 at 16:05

André 3000André 3000

12.6k22243

$endgroup$

Left multiplication by $A^T$ is injective iff $A^T$ has a pivot in every column iff $A$ has a pivot in every row iff left multiplication by $A$ is surjective.

share|cite|improve this answer

answered Jan 10 at 16:05

André 3000André 3000

12.6k22243

share|cite|improve this answer

share|cite|improve this answer

answered Jan 10 at 16:05

André 3000André 3000

12.6k22243

answered Jan 10 at 16:05

André 3000André 3000

12.6k22243

answered Jan 10 at 16:05

André 3000André 3000

12.6k22243

12.6k22243

add a comment | 

add a comment | 

0

$begingroup$

So, we have a function $f: Xrightarrow Y$ such that $f(x)=Ax$. Suppose $y$ is in the range of $f$. Then there is some $x$ such that $y=f(x)$, and if $f$ is injective, this $x$ is unique. So given $y$, we can define a function $g$ that returns that $x$. If $By=g(y)$, then $BAx=x$. Thus, $BA$ is the identity (on $X$). So $A^TB^T$ is also the identity (although on $Y$ rather than $X$). Thus, given any $y$, there is some $x$ such that $A^Tx=y$: if we take $x=B^Ty$, then $A^Tx=A^TB^Ty=y$. This show that $A$ being injective implies that $A^T$ is surjective. The other direction follows similarly.

In other words, being surjective is equivalent to there being a right inverse, and being injective is equivalent to there being a left inverse. So $A$ is injective $rightarrow$ $A$ has a left inverse $rightarrow$ $A^T$ has a right inverse $rightarrow$ $A$ is surjective.

share|cite|improve this answer

answered Jan 10 at 22:23

AcccumulationAcccumulation

6,9042618

$endgroup$

add a comment | 

0

$begingroup$

So, we have a function $f: Xrightarrow Y$ such that $f(x)=Ax$. Suppose $y$ is in the range of $f$. Then there is some $x$ such that $y=f(x)$, and if $f$ is injective, this $x$ is unique. So given $y$, we can define a function $g$ that returns that $x$. If $By=g(y)$, then $BAx=x$. Thus, $BA$ is the identity (on $X$). So $A^TB^T$ is also the identity (although on $Y$ rather than $X$). Thus, given any $y$, there is some $x$ such that $A^Tx=y$: if we take $x=B^Ty$, then $A^Tx=A^TB^Ty=y$. This show that $A$ being injective implies that $A^T$ is surjective. The other direction follows similarly.

In other words, being surjective is equivalent to there being a right inverse, and being injective is equivalent to there being a left inverse. So $A$ is injective $rightarrow$ $A$ has a left inverse $rightarrow$ $A^T$ has a right inverse $rightarrow$ $A$ is surjective.

share|cite|improve this answer

answered Jan 10 at 22:23

AcccumulationAcccumulation

6,9042618

$endgroup$

add a comment | 

0

0

0

$begingroup$

So, we have a function $f: Xrightarrow Y$ such that $f(x)=Ax$. Suppose $y$ is in the range of $f$. Then there is some $x$ such that $y=f(x)$, and if $f$ is injective, this $x$ is unique. So given $y$, we can define a function $g$ that returns that $x$. If $By=g(y)$, then $BAx=x$. Thus, $BA$ is the identity (on $X$). So $A^TB^T$ is also the identity (although on $Y$ rather than $X$). Thus, given any $y$, there is some $x$ such that $A^Tx=y$: if we take $x=B^Ty$, then $A^Tx=A^TB^Ty=y$. This show that $A$ being injective implies that $A^T$ is surjective. The other direction follows similarly.

In other words, being surjective is equivalent to there being a right inverse, and being injective is equivalent to there being a left inverse. So $A$ is injective $rightarrow$ $A$ has a left inverse $rightarrow$ $A^T$ has a right inverse $rightarrow$ $A$ is surjective.

share|cite|improve this answer

answered Jan 10 at 22:23

AcccumulationAcccumulation

6,9042618

$endgroup$

So, we have a function $f: Xrightarrow Y$ such that $f(x)=Ax$. Suppose $y$ is in the range of $f$. Then there is some $x$ such that $y=f(x)$, and if $f$ is injective, this $x$ is unique. So given $y$, we can define a function $g$ that returns that $x$. If $By=g(y)$, then $BAx=x$. Thus, $BA$ is the identity (on $X$). So $A^TB^T$ is also the identity (although on $Y$ rather than $X$). Thus, given any $y$, there is some $x$ such that $A^Tx=y$: if we take $x=B^Ty$, then $A^Tx=A^TB^Ty=y$. This show that $A$ being injective implies that $A^T$ is surjective. The other direction follows similarly.

In other words, being surjective is equivalent to there being a right inverse, and being injective is equivalent to there being a left inverse. So $A$ is injective $rightarrow$ $A$ has a left inverse $rightarrow$ $A^T$ has a right inverse $rightarrow$ $A$ is surjective.

share|cite|improve this answer

answered Jan 10 at 22:23

AcccumulationAcccumulation

6,9042618

share|cite|improve this answer

share|cite|improve this answer

answered Jan 10 at 22:23

AcccumulationAcccumulation

6,9042618

answered Jan 10 at 22:23

AcccumulationAcccumulation

6,9042618

answered Jan 10 at 22:23

AcccumulationAcccumulation

6,9042618

6,9042618

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.

draft saved

draft discarded

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.

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3068526%2fhow-to-prove-that-at-y-0-implies-y-0-if-and-only-if-a-defines-a-surjective%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

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

Name

Email

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

Name

Email

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

Name

Email

Required, but never shown

Name

Name

Email

Required, but never shown

Email

Required, but never shown

Email

Required, but never shown

Email

Required, but never shown

Name

Name

Email

Required, but never shown

Email

Required, but never shown

Email

Required, but never shown

Email

Required, but never shown

This page is only for reference, If you need detailed information, please check here