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 $T$
How widely used is the term Treppenwitz? Is it something that most Germans know?
How to deal with a team lead who never gives me credit?
Seeking colloquialism for “just because”
How to answer "Have you ever been terminated?"
Why is "Consequences inflicted." not a sentence?
English words in a non-english sci-fi novel
Check which numbers satisfy the condition [A*B*C = A! + B! + C!]
How does the particle を relate to the verb 行く in the structure「A を + B に行く」?
What does the "x" in "x86" represent?
Why did the IBM 650 use bi-quinary?
Single word antonym of "flightless"
How does debian/ubuntu knows a package has a updated version
Generate an RGB colour grid
What would be the ideal power source for a cybernetic eye?
Extract all GPU name, model and GPU ram
Output the ŋarâþ crîþ alphabet song without using (m)any letters
Error "illegal generic type for instanceof" when using local classes
When a candle burns, why does the top of wick glow if bottom of flame is hottest?
Why was the term "discrete" used in discrete logarithm?
If a contract sometimes uses the wrong name, is it still valid?
Apollo command module space walk?
When do you get frequent flier miles - when you buy, or when you fly?
Sci-Fi book where patients in a coma ward all live in a subconscious world linked together
Short Story with Cinderella as a Voo-doo Witch
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