Is this linear operator surjective? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Is there an accepted term for “locally” nilpotent linear operators?Eigenvalues of a linear operator over a K-vector spaceFinding trace and determinant of linear operatorExistence of a surjective compact linear operator on an infinite dimensional Banach space$V$ be a vector space ; $f,g in mathcal L (V)$ ; $fcirc g-g circ f=I$ ; then is the set $g^n: nge 0$ linearly independent?Eigenvalues of an infinite dimensional linear operatorOperator norm of matricesif one linear operator of a set of linear operators sends a vector to zero then the product of linear operators must send the vector to zero?Some insight regarding a difficult problem on Linear Operators.$T$-annihilators divide minimal polynomial of a linear operator $T$
String `!23` is replaced with `docker` in command line
Dating a Former Employee
Identifying polygons that intersect with another layer using QGIS?
What's the meaning of 間時肆拾貳 at a car parking sign
Can I cast Passwall to drop an enemy into a 20-foot pit?
Withdrew £2800, but only £2000 shows as withdrawn on online banking; what are my obligations?
Apollo command module space walk?
prime numbers and expressing non-prime numbers
What to do with chalk when deepwater soloing?
Denied boarding although I have proper visa and documentation. To whom should I make a complaint?
How to find all the available tools in mac terminal?
Do I really need recursive chmod to restrict access to a folder?
How to call a function with default parameter through a pointer to function that is the return of another function?
How discoverable are IPv6 addresses and AAAA names by potential attackers?
Why do people hide their license plates in the EU?
How widely used is the term Treppenwitz? Is it something that most Germans know?
2001: A Space Odyssey's use of the song "Daisy Bell" (Bicycle Built for Two); life imitates art or vice-versa?
How does the particle を relate to the verb 行く in the structure「A を + B に行く」?
Should I discuss the type of campaign with my players?
Book where humans were engineered with genes from animal species to survive hostile planets
Using audio cues to encourage good posture
Extract all GPU name, model and GPU ram
Single word antonym of "flightless"
Is there a (better) way to access $wpdb results?
Is this linear operator surjective?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Is there an accepted term for “locally” nilpotent linear operators?Eigenvalues of a linear operator over a K-vector spaceFinding trace and determinant of linear operatorExistence of a surjective compact linear operator on an infinite dimensional Banach space$V$ be a vector space ; $f,g in mathcal L (V)$ ; $fcirc g-g circ f=I$ ; then is the set $g^n: nge 0$ linearly independent?Eigenvalues of an infinite dimensional linear operatorOperator norm of matricesif one linear operator of a set of linear operators sends a vector to zero then the product of linear operators must send the vector to zero?Some insight regarding a difficult problem on Linear Operators.$T$-annihilators divide minimal polynomial of a linear operator $T$
$begingroup$
Consider the linear operator $f_A$ on $K^n×n$ defined by $f_A(X)=AX-XA$ where $K$ is field. Is $f_A$ is surjective? If no, then under what conditions $f_A$ is surjective?
My attempt: I can see easily that $f_A$ is not surjective if I take $K=mathbbR$. Hence in general $f_A$ is not surjective.
But I am not able to determine the condition under which $f_A$ is surjective.
It is given in a hint that, if $char(K)$ does not divide $n,$ then $f_A$ is not surjective. So does that mean, if I consider $K$ to be a finite field having characteristic which divides $n,$ then $f_A$ is surjective? How? Please help!
linear-algebra linear-transformations
edited Mar 26 at 16:56
J. W. Tanner
4,9371420
asked Mar 26 at 16:36
Akash PatalwanshiAkash Patalwanshi
1,0171820
$endgroup$
add a comment |
$begingroup$
Consider the linear operator $f_A$ on $K^n×n$ defined by $f_A(X)=AX-XA$ where $K$ is field. Is $f_A$ is surjective? If no, then under what conditions $f_A$ is surjective?
My attempt: I can see easily that $f_A$ is not surjective if I take $K=mathbbR$. Hence in general $f_A$ is not surjective.
But I am not able to determine the condition under which $f_A$ is surjective.
It is given in a hint that, if $char(K)$ does not divide $n,$ then $f_A$ is not surjective. So does that mean, if I consider $K$ to be a finite field having characteristic which divides $n,$ then $f_A$ is surjective? How? Please help!
linear-algebra linear-transformations
edited Mar 26 at 16:56
J. W. Tanner
4,9371420
asked Mar 26 at 16:36
Akash PatalwanshiAkash Patalwanshi
1,0171820
$endgroup$
4
$begingroup$
It's surjective iff its injective, but $I$ is always in the kernel.
$endgroup$
– Lord Shark the Unknown
Mar 26 at 18:04
$begingroup$
Thanks to all of you.
$endgroup$
– Akash Patalwanshi
Mar 27 at 2:38
add a comment |
$begingroup$
Consider the linear operator $f_A$ on $K^n×n$ defined by $f_A(X)=AX-XA$ where $K$ is field. Is $f_A$ is surjective? If no, then under what conditions $f_A$ is surjective?
My attempt: I can see easily that $f_A$ is not surjective if I take $K=mathbbR$. Hence in general $f_A$ is not surjective.
But I am not able to determine the condition under which $f_A$ is surjective.
It is given in a hint that, if $char(K)$ does not divide $n,$ then $f_A$ is not surjective. So does that mean, if I consider $K$ to be a finite field having characteristic which divides $n,$ then $f_A$ is surjective? How? Please help!
linear-algebra linear-transformations
edited Mar 26 at 16:56
J. W. Tanner
4,9371420
asked Mar 26 at 16:36
Akash PatalwanshiAkash Patalwanshi
1,0171820
$endgroup$
Consider the linear operator $f_A$ on $K^n×n$ defined by $f_A(X)=AX-XA$ where $K$ is field. Is $f_A$ is surjective? If no, then under what conditions $f_A$ is surjective?
My attempt: I can see easily that $f_A$ is not surjective if I take $K=mathbbR$. Hence in general $f_A$ is not surjective.
But I am not able to determine the condition under which $f_A$ is surjective.
It is given in a hint that, if $char(K)$ does not divide $n,$ then $f_A$ is not surjective. So does that mean, if I consider $K$ to be a finite field having characteristic which divides $n,$ then $f_A$ is surjective? How? Please help!
linear-algebra linear-transformations
linear-algebra linear-transformations
edited Mar 26 at 16:56
J. W. Tanner
4,9371420
asked Mar 26 at 16:36
Akash PatalwanshiAkash Patalwanshi
1,0171820
edited Mar 26 at 16:56
J. W. Tanner
4,9371420
asked Mar 26 at 16:36
Akash PatalwanshiAkash Patalwanshi
1,0171820
edited Mar 26 at 16:56
J. W. Tanner
4,9371420
edited Mar 26 at 16:56
J. W. Tanner
4,9371420
edited Mar 26 at 16:56
J. W. Tanner
4,9371420
4,9371420
asked Mar 26 at 16:36
Akash PatalwanshiAkash Patalwanshi
1,0171820
asked Mar 26 at 16:36
Akash PatalwanshiAkash Patalwanshi
1,0171820
asked Mar 26 at 16:36
Akash PatalwanshiAkash Patalwanshi
1,0171820
1,0171820
4
$begingroup$
It's surjective iff its injective, but $I$ is always in the kernel.
$endgroup$
– Lord Shark the Unknown
Mar 26 at 18:04
$begingroup$
Thanks to all of you.
$endgroup$
– Akash Patalwanshi
Mar 27 at 2:38
add a comment |
4
$begingroup$
It's surjective iff its injective, but $I$ is always in the kernel.
$endgroup$
– Lord Shark the Unknown
Mar 26 at 18:04
$begingroup$
Thanks to all of you.
$endgroup$
– Akash Patalwanshi
Mar 27 at 2:38
4
4
$begingroup$
It's surjective iff its injective, but $I$ is always in the kernel.
$endgroup$
– Lord Shark the Unknown
Mar 26 at 18:04
$begingroup$
It's surjective iff its injective, but $I$ is always in the kernel.
$endgroup$
– Lord Shark the Unknown
Mar 26 at 18:04
$begingroup$
Thanks to all of you.
$endgroup$
– Akash Patalwanshi
Mar 27 at 2:38
$begingroup$
Thanks to all of you.
$endgroup$
– Akash Patalwanshi
Mar 27 at 2:38
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: Examine the trace of $f_A(X)$.
answered Mar 26 at 16:47
lhflhf
168k11172405
$endgroup$
$begingroup$
Sir trace of $f_A(X)=0$
$endgroup$
– Akash Patalwanshi
Mar 26 at 16:50
$begingroup$
@AkashPatalwanshi, right. And do all matrices have zero trace?
$endgroup$
– lhf
Mar 26 at 16:52
$begingroup$
Sir, yes identify matrix does not have zero trace and hence $f_A$ is not surjective. But I didn't understand what and why in hint it is given that "if charK does not divides $n$ then $f_A$ is not surjective?
$endgroup$
– Akash Patalwanshi
Mar 26 at 16:55
2
$begingroup$
If $mboxcharK$ does not divide $n$, then $I_n$ is not trace zero.
$endgroup$
– chhro
Mar 26 at 17:06
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%2f3163440%2fis-this-linear-operator-surjective%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: Examine the trace of $f_A(X)$.
answered Mar 26 at 16:47
lhflhf
168k11172405
$endgroup$
$begingroup$
Sir trace of $f_A(X)=0$
$endgroup$
– Akash Patalwanshi
Mar 26 at 16:50
$begingroup$
@AkashPatalwanshi, right. And do all matrices have zero trace?
$endgroup$
– lhf
Mar 26 at 16:52
$begingroup$
Sir, yes identify matrix does not have zero trace and hence $f_A$ is not surjective. But I didn't understand what and why in hint it is given that "if charK does not divides $n$ then $f_A$ is not surjective?
$endgroup$
– Akash Patalwanshi
Mar 26 at 16:55
2
$begingroup$
If $mboxcharK$ does not divide $n$, then $I_n$ is not trace zero.
$endgroup$
– chhro
Mar 26 at 17:06
add a comment |
$begingroup$
Hint: Examine the trace of $f_A(X)$.
answered Mar 26 at 16:47
lhflhf
168k11172405
$endgroup$
$begingroup$
Sir trace of $f_A(X)=0$
$endgroup$
– Akash Patalwanshi
Mar 26 at 16:50
$begingroup$
@AkashPatalwanshi, right. And do all matrices have zero trace?
$endgroup$
– lhf
Mar 26 at 16:52
$begingroup$
Sir, yes identify matrix does not have zero trace and hence $f_A$ is not surjective. But I didn't understand what and why in hint it is given that "if charK does not divides $n$ then $f_A$ is not surjective?
$endgroup$
– Akash Patalwanshi
Mar 26 at 16:55
2
$begingroup$
If $mboxcharK$ does not divide $n$, then $I_n$ is not trace zero.
$endgroup$
– chhro
Mar 26 at 17:06
add a comment |
$begingroup$
Hint: Examine the trace of $f_A(X)$.
answered Mar 26 at 16:47
lhflhf
168k11172405
$endgroup$
Hint: Examine the trace of $f_A(X)$.
answered Mar 26 at 16:47
lhflhf
168k11172405
answered Mar 26 at 16:47
lhflhf
168k11172405
answered Mar 26 at 16:47
lhflhf
168k11172405
answered Mar 26 at 16:47
lhflhf
168k11172405
168k11172405
$begingroup$
Sir trace of $f_A(X)=0$
$endgroup$
– Akash Patalwanshi
Mar 26 at 16:50
$begingroup$
@AkashPatalwanshi, right. And do all matrices have zero trace?
$endgroup$
– lhf
Mar 26 at 16:52
$begingroup$
Sir, yes identify matrix does not have zero trace and hence $f_A$ is not surjective. But I didn't understand what and why in hint it is given that "if charK does not divides $n$ then $f_A$ is not surjective?
$endgroup$
– Akash Patalwanshi
Mar 26 at 16:55
2
$begingroup$
If $mboxcharK$ does not divide $n$, then $I_n$ is not trace zero.
$endgroup$
– chhro
Mar 26 at 17:06
add a comment |
$begingroup$
Sir trace of $f_A(X)=0$
$endgroup$
– Akash Patalwanshi
Mar 26 at 16:50
$begingroup$
@AkashPatalwanshi, right. And do all matrices have zero trace?
$endgroup$
– lhf
Mar 26 at 16:52
$begingroup$
Sir, yes identify matrix does not have zero trace and hence $f_A$ is not surjective. But I didn't understand what and why in hint it is given that "if charK does not divides $n$ then $f_A$ is not surjective?
$endgroup$
– Akash Patalwanshi
Mar 26 at 16:55
2
$begingroup$
If $mboxcharK$ does not divide $n$, then $I_n$ is not trace zero.
$endgroup$
– chhro
Mar 26 at 17:06
$begingroup$
Sir trace of $f_A(X)=0$
$endgroup$
– Akash Patalwanshi
Mar 26 at 16:50
$begingroup$
Sir trace of $f_A(X)=0$
$endgroup$
– Akash Patalwanshi
Mar 26 at 16:50
$begingroup$
@AkashPatalwanshi, right. And do all matrices have zero trace?
$endgroup$
– lhf
Mar 26 at 16:52
$begingroup$
@AkashPatalwanshi, right. And do all matrices have zero trace?
$endgroup$
– lhf
Mar 26 at 16:52
$begingroup$
Sir, yes identify matrix does not have zero trace and hence $f_A$ is not surjective. But I didn't understand what and why in hint it is given that "if charK does not divides $n$ then $f_A$ is not surjective?
$endgroup$
– Akash Patalwanshi
Mar 26 at 16:55
$begingroup$
Sir, yes identify matrix does not have zero trace and hence $f_A$ is not surjective. But I didn't understand what and why in hint it is given that "if charK does not divides $n$ then $f_A$ is not surjective?
$endgroup$
– Akash Patalwanshi
Mar 26 at 16:55
2
2
$begingroup$
If $mboxcharK$ does not divide $n$, then $I_n$ is not trace zero.
$endgroup$
– chhro
Mar 26 at 17:06
$begingroup$
If $mboxcharK$ does not divide $n$, then $I_n$ is not trace zero.
$endgroup$
– chhro
Mar 26 at 17:06
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%2f3163440%2fis-this-linear-operator-surjective%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
4
$begingroup$
It's surjective iff its injective, but $I$ is always in the kernel.
$endgroup$
– Lord Shark the Unknown
Mar 26 at 18:04
$begingroup$
Thanks to all of you.
$endgroup$
– Akash Patalwanshi
Mar 27 at 2:38