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
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Finitely generated matrix groups whose eigenvalues are all algebraic
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
$begingroup$
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
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
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
edited 2 days ago
Asaf Karagila♦
307k33440773
asked 2 days ago
aaaaaa
374
$endgroup$
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
linear-algebra
edited 2 days ago
Asaf Karagila♦
307k33440773
asked 2 days ago
aaaaaa
374
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
edited 2 days ago
Asaf Karagila♦
307k33440773
307k33440773
asked 2 days ago
aaaaaa
374
asked 2 days ago
aaaaaa
374
asked 2 days ago
aaaaaa
374
374
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
$(BA)^k = (BA)(BA)....(BA) = BABA...BA =0$
Multiply from the left by $A$ and from the right by $B$, you'll get $$ A.BABA...BA.B= (AB)(AB)...(AB)=(AB)^k+1=0$$
answered 2 days ago
Fareed AFFareed AF
692112
$endgroup$
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.
answered 2 days ago
Severin SchravenSeverin Schraven
6,7802936
$endgroup$
add a comment |
$begingroup$
You can multiply by A on the left side and B on the right side, and you get $(AB)^k+1 = 0$
answered 2 days ago
GatgatGatgat
1775
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$(BA)^k = (BA)(BA)....(BA) = BABA...BA =0$
Multiply from the left by $A$ and from the right by $B$, you'll get $$ A.BABA...BA.B= (AB)(AB)...(AB)=(AB)^k+1=0$$
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 $$ A.BABA...BA.B= (AB)(AB)...(AB)=(AB)^k+1=0$$
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 $$ A.BABA...BA.B= (AB)(AB)...(AB)=(AB)^k+1=0$$
answered 2 days ago
Fareed AFFareed AF
692112
$endgroup$
$(BA)^k = (BA)(BA)....(BA) = BABA...BA =0$
Multiply from the left by $A$ and from the right by $B$, you'll get $$ A.BABA...BA.B= (AB)(AB)...(AB)=(AB)^k+1=0$$
answered 2 days ago
Fareed AFFareed AF
692112
answered 2 days ago
Fareed AFFareed AF
692112
answered 2 days ago
Fareed AFFareed AF
692112
answered 2 days ago
Fareed AFFareed AF
692112
692112
add a comment |
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.
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.
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.
answered 2 days ago
Severin SchravenSeverin Schraven
6,7802936
$endgroup$
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.
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
answered 2 days ago
Severin SchravenSeverin Schraven
6,7802936
6,7802936
add a comment |
add a comment |
$begingroup$
You can multiply by A on the left side and B on the right side, and you get $(AB)^k+1 = 0$
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 $(AB)^k+1 = 0$
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 $(AB)^k+1 = 0$
answered 2 days ago
GatgatGatgat
1775
$endgroup$
You can multiply by A on the left side and B on the right side, and you get $(AB)^k+1 = 0$
answered 2 days ago
GatgatGatgat
1775
answered 2 days ago
GatgatGatgat
1775
answered 2 days ago
GatgatGatgat
1775
answered 2 days ago
GatgatGatgat
1775
1775
add a comment |
add a comment |
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