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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 $T$










1












$begingroup$


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!










share|cite|improve this question











$endgroup$







  • 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















1












$begingroup$


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!










share|cite|improve this question











$endgroup$







  • 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













1












1








1


0



$begingroup$


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!










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 26 at 16:56









J. W. Tanner

4,9371420




4,9371420










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












  • 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










1 Answer
1






active

oldest

votes


















1












$begingroup$

Hint: Examine the trace of $f_A(X)$.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    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












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Hint: Examine the trace of $f_A(X)$.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    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
















1












$begingroup$

Hint: Examine the trace of $f_A(X)$.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    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














1












1








1





$begingroup$

Hint: Examine the trace of $f_A(X)$.






share|cite|improve this answer









$endgroup$



Hint: Examine the trace of $f_A(X)$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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

















  • $begingroup$
    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


















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