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Suppose $BA$ is nilpotent, is it always true that $AB$ is nilpotent?

Nilpotent matrices with same minimal polynomial and nullityIs the zero matrix the only symmetric, nilpotent matrix with real values?Smoothness of a MatrixSuppose $A^2B+BA^2=2ABA$.Prove that there exists a positive integer $k$ such that $(AB-BA)^k=0$.Product of nilpotent matrices is nilpotentProduct of nilpotent matrices invertibleExample of non-nilpotent product of nilpotent matricesUnitriangular group $UT_n(Bbb Z)$ is nilpotent with class $n$Proving an operator is nilpotentNilpotent matrix sets

2

$begingroup$

Given matrices $A$ and $B$ such that $BA$ is nilpotent. Is it true that $AB$ is nilpotent?

My thoughts so far:

$BA$ nilpotent $Rightarrow (BA)^k = 0$ for some positive integer k.

$(BA)^k = (BA)(BA)....(BA) = BABA...BA = B(AB)(AB)....(AB)A = B(AB)^k-1A = 0$

But I have no idea how to proceed from here. If its not true, is there any counterexample?

linear-algebra

share|cite|improve this question

edited 2 days ago

Asaf Karagila♦

307k33440773

asked 2 days ago

aaaaaa

374

$endgroup$

add a comment | 

2

$begingroup$

Given matrices $A$ and $B$ such that $BA$ is nilpotent. Is it true that $AB$ is nilpotent?

My thoughts so far:

$BA$ nilpotent $Rightarrow (BA)^k = 0$ for some positive integer k.

$(BA)^k = (BA)(BA)....(BA) = BABA...BA = B(AB)(AB)....(AB)A = B(AB)^k-1A = 0$

But I have no idea how to proceed from here. If its not true, is there any counterexample?

linear-algebra

share|cite|improve this question

edited 2 days ago

Asaf Karagila♦

307k33440773

asked 2 days ago

aaaaaa

374

$endgroup$

add a comment | 

$begingroup$

Given matrices $A$ and $B$ such that $BA$ is nilpotent. Is it true that $AB$ is nilpotent?

My thoughts so far:

$BA$ nilpotent $Rightarrow (BA)^k = 0$ for some positive integer k.

$(BA)^k = (BA)(BA)....(BA) = BABA...BA = B(AB)(AB)....(AB)A = B(AB)^k-1A = 0$

But I have no idea how to proceed from here. If its not true, is there any counterexample?

linear-algebra

share|cite|improve this question

edited 2 days ago

Asaf Karagila♦

307k33440773

asked 2 days ago

aaaaaa

374

$endgroup$

share|cite|improve this question

edited 2 days ago

Asaf Karagila♦

307k33440773

asked 2 days ago

aaaaaa

374

share|cite|improve this question

edited 2 days ago

Asaf Karagila♦

307k33440773

asked 2 days ago

aaaaaa

374

edited 2 days ago

Asaf Karagila♦

307k33440773

edited 2 days ago

Asaf Karagila♦

307k33440773

asked 2 days ago

aaaaaa

374

asked 2 days ago

aaaaaa

374

3 Answers
3

active

oldest

votes

4

$begingroup$

$(BA)^k = (BA)(BA)....(BA) = BABA...BA =0$

Multiply from the left by $A$ and from the right by $B$, you'll get $$ A.BABA...BA.B= (AB)(AB)...(AB)=(AB)^k+1=0$$

share|cite|improve this answer

answered 2 days ago

Fareed AFFareed AF

692112

$endgroup$

add a comment | 

5

$begingroup$

You just have to replace $A$ by $B$ in your computation. Then $(AB)^k +1= A(BA)^k B$. I assume you can finish from here.

share|cite|improve this answer

answered 2 days ago

Severin SchravenSeverin Schraven

6,7802936

$endgroup$

add a comment | 

2

$begingroup$

You can multiply by A on the left side and B on the right side, and you get $(AB)^k+1 = 0$

share|cite|improve this answer

answered 2 days ago

GatgatGatgat

1775

$endgroup$

add a comment | 

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4

$begingroup$

$(BA)^k = (BA)(BA)....(BA) = BABA...BA =0$

Multiply from the left by $A$ and from the right by $B$, you'll get $$ A.BABA...BA.B= (AB)(AB)...(AB)=(AB)^k+1=0$$

share|cite|improve this answer

answered 2 days ago

Fareed AFFareed AF

692112

$endgroup$

add a comment | 

4

$begingroup$

$(BA)^k = (BA)(BA)....(BA) = BABA...BA =0$

Multiply from the left by $A$ and from the right by $B$, you'll get $$ A.BABA...BA.B= (AB)(AB)...(AB)=(AB)^k+1=0$$

share|cite|improve this answer

answered 2 days ago

Fareed AFFareed AF

692112

$endgroup$

add a comment | 

$begingroup$

$(BA)^k = (BA)(BA)....(BA) = BABA...BA =0$

Multiply from the left by $A$ and from the right by $B$, you'll get $$ A.BABA...BA.B= (AB)(AB)...(AB)=(AB)^k+1=0$$

share|cite|improve this answer

answered 2 days ago

Fareed AFFareed AF

692112

$endgroup$

share|cite|improve this answer

answered 2 days ago

Fareed AFFareed AF

692112

answered 2 days ago

Fareed AFFareed AF

692112

answered 2 days ago

Fareed AFFareed AF

692112

5

$begingroup$

You just have to replace $A$ by $B$ in your computation. Then $(AB)^k +1= A(BA)^k B$. I assume you can finish from here.

share|cite|improve this answer

answered 2 days ago

Severin SchravenSeverin Schraven

6,7802936

$endgroup$

add a comment | 

5

$begingroup$

You just have to replace $A$ by $B$ in your computation. Then $(AB)^k +1= A(BA)^k B$. I assume you can finish from here.

share|cite|improve this answer

answered 2 days ago

Severin SchravenSeverin Schraven

6,7802936

$endgroup$

add a comment | 

$begingroup$

You just have to replace $A$ by $B$ in your computation. Then $(AB)^k +1= A(BA)^k B$. I assume you can finish from here.

share|cite|improve this answer

answered 2 days ago

Severin SchravenSeverin Schraven

6,7802936

$endgroup$

share|cite|improve this answer

answered 2 days ago

Severin SchravenSeverin Schraven

6,7802936

answered 2 days ago

Severin SchravenSeverin Schraven

6,7802936

answered 2 days ago

Severin SchravenSeverin Schraven

6,7802936

2

$begingroup$

You can multiply by A on the left side and B on the right side, and you get $(AB)^k+1 = 0$

share|cite|improve this answer

answered 2 days ago

GatgatGatgat

1775

$endgroup$

add a comment | 

2

$begingroup$

You can multiply by A on the left side and B on the right side, and you get $(AB)^k+1 = 0$

share|cite|improve this answer

answered 2 days ago

GatgatGatgat

1775

$endgroup$

add a comment | 

$begingroup$

You can multiply by A on the left side and B on the right side, and you get $(AB)^k+1 = 0$

share|cite|improve this answer

answered 2 days ago

GatgatGatgat

1775

$endgroup$

share|cite|improve this answer

answered 2 days ago

GatgatGatgat

1775

answered 2 days ago

GatgatGatgat

1775

answered 2 days ago

GatgatGatgat

1775