Mixing
Nilpotent commutative matrices $A, B$ $Longrightarrow$ $A+B$ nilpotent.
$begingroup$
How to prove that if $A, B$ are matrix of $ntimes n$ nilpotents so that $AB=BA$ then $A+B$ is nilpotent.
linear-algebra nilpotence
edited Jan 17 at 11:04
Widawensen
4,51321446
asked Aug 17 '13 at 3:36
Roiner Segura CuberoRoiner Segura Cubero
1,9051729
$endgroup$
add a comment |
$begingroup$
How to prove that if $A, B$ are matrix of $ntimes n$ nilpotents so that $AB=BA$ then $A+B$ is nilpotent.
linear-algebra nilpotence
edited Jan 17 at 11:04
Widawensen
4,51321446
asked Aug 17 '13 at 3:36
Roiner Segura CuberoRoiner Segura Cubero
1,9051729
$endgroup$
$begingroup$
binomial theorem
$endgroup$
– yoyo
Aug 17 '13 at 3:40
$begingroup$
yes but do not know how to prove
$endgroup$
– Roiner Segura Cubero
Aug 17 '13 at 3:41
add a comment |
$begingroup$
How to prove that if $A, B$ are matrix of $ntimes n$ nilpotents so that $AB=BA$ then $A+B$ is nilpotent.
linear-algebra nilpotence
edited Jan 17 at 11:04
Widawensen
4,51321446
asked Aug 17 '13 at 3:36
Roiner Segura CuberoRoiner Segura Cubero
1,9051729
$endgroup$
How to prove that if $A, B$ are matrix of $ntimes n$ nilpotents so that $AB=BA$ then $A+B$ is nilpotent.
linear-algebra nilpotence
linear-algebra nilpotence
edited Jan 17 at 11:04
Widawensen
4,51321446
asked Aug 17 '13 at 3:36
Roiner Segura CuberoRoiner Segura Cubero
1,9051729
edited Jan 17 at 11:04
Widawensen
4,51321446
asked Aug 17 '13 at 3:36
Roiner Segura CuberoRoiner Segura Cubero
1,9051729
edited Jan 17 at 11:04
Widawensen
4,51321446
edited Jan 17 at 11:04
Widawensen
4,51321446
edited Jan 17 at 11:04
Widawensen
4,51321446
4,51321446
asked Aug 17 '13 at 3:36
Roiner Segura CuberoRoiner Segura Cubero
1,9051729
asked Aug 17 '13 at 3:36
Roiner Segura CuberoRoiner Segura Cubero
1,9051729
asked Aug 17 '13 at 3:36
Roiner Segura CuberoRoiner Segura Cubero
1,9051729
1,9051729
$begingroup$
binomial theorem
$endgroup$
– yoyo
Aug 17 '13 at 3:40
$begingroup$
yes but do not know how to prove
$endgroup$
– Roiner Segura Cubero
Aug 17 '13 at 3:41
add a comment |
$begingroup$
binomial theorem
$endgroup$
– yoyo
Aug 17 '13 at 3:40
$begingroup$
yes but do not know how to prove
$endgroup$
– Roiner Segura Cubero
Aug 17 '13 at 3:41
$begingroup$
binomial theorem
$endgroup$
– yoyo
Aug 17 '13 at 3:40
$begingroup$
binomial theorem
$endgroup$
– yoyo
Aug 17 '13 at 3:40
$begingroup$
yes but do not know how to prove
$endgroup$
– Roiner Segura Cubero
Aug 17 '13 at 3:41
$begingroup$
yes but do not know how to prove
$endgroup$
– Roiner Segura Cubero
Aug 17 '13 at 3:41
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If $A^m = 0$ and $B^n = 0$ then what is $(A+B)^{m+n}$? Use the binomial theorem (uses commutativity!)
Edit: I'll give you one more step.
$$
(A+B)^{m+n} = sum_{i=0}^{m+n} {m+n choose i}A^{i}B^{m+n-i}
$$
can you show that either $A^i = 0$ or $B^{m+n-i}=0$?
answered Aug 17 '13 at 3:41
nullUsernullUser
16.7k442104
$endgroup$
$begingroup$
yes..thanks nullUser
$endgroup$
– Roiner Segura Cubero
Aug 17 '13 at 4:11
$begingroup$
...and note that it is enough to take $;m+n-1;$ ...
$endgroup$
– DonAntonio
Aug 17 '13 at 8:56
add a comment |
$begingroup$
Another approach to the problem can be based on the well known fact that commuting matrices are simultaneously triangularizable.
If so then exists such matrix $P$ that $A=PT_AP^{-1}$ and $B=PT_BP^{-1}$ where $T_A,T_B$ are, assume, upper triangular matrices.
Additionally they have on diagonal only $0$ values as the eigenvalues of nilpotent matrices are $0$.
Hence it's obvious that $T_A+T_B$ has also only eigenvalues equal to $0$ and consequently also $A+B= PT_AP^{-1} +PT_BP^{-1}=P(T_A+T_B)P^{-1}$
This kind of argumentation allows additionally to extend the claim for other types of expressions with commuting nilpotent $A,B$ so we have for example that
any sums of powers $A^k, B^i$ $(k,i>0)$ (and their products) are also nilpotent.
answered Jan 18 at 9:09
WidawensenWidawensen
4,51321446
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $A^m = 0$ and $B^n = 0$ then what is $(A+B)^{m+n}$? Use the binomial theorem (uses commutativity!)
Edit: I'll give you one more step.
$$
(A+B)^{m+n} = sum_{i=0}^{m+n} {m+n choose i}A^{i}B^{m+n-i}
$$
can you show that either $A^i = 0$ or $B^{m+n-i}=0$?
answered Aug 17 '13 at 3:41
nullUsernullUser
16.7k442104
$endgroup$
$begingroup$
yes..thanks nullUser
$endgroup$
– Roiner Segura Cubero
Aug 17 '13 at 4:11
$begingroup$
...and note that it is enough to take $;m+n-1;$ ...
$endgroup$
– DonAntonio
Aug 17 '13 at 8:56
add a comment |
$begingroup$
If $A^m = 0$ and $B^n = 0$ then what is $(A+B)^{m+n}$? Use the binomial theorem (uses commutativity!)
Edit: I'll give you one more step.
$$
(A+B)^{m+n} = sum_{i=0}^{m+n} {m+n choose i}A^{i}B^{m+n-i}
$$
can you show that either $A^i = 0$ or $B^{m+n-i}=0$?
answered Aug 17 '13 at 3:41
nullUsernullUser
16.7k442104
$endgroup$
$begingroup$
yes..thanks nullUser
$endgroup$
– Roiner Segura Cubero
Aug 17 '13 at 4:11
$begingroup$
...and note that it is enough to take $;m+n-1;$ ...
$endgroup$
– DonAntonio
Aug 17 '13 at 8:56
add a comment |
$begingroup$
If $A^m = 0$ and $B^n = 0$ then what is $(A+B)^{m+n}$? Use the binomial theorem (uses commutativity!)
Edit: I'll give you one more step.
$$
(A+B)^{m+n} = sum_{i=0}^{m+n} {m+n choose i}A^{i}B^{m+n-i}
$$
can you show that either $A^i = 0$ or $B^{m+n-i}=0$?
answered Aug 17 '13 at 3:41
nullUsernullUser
16.7k442104
$endgroup$
If $A^m = 0$ and $B^n = 0$ then what is $(A+B)^{m+n}$? Use the binomial theorem (uses commutativity!)
Edit: I'll give you one more step.
$$
(A+B)^{m+n} = sum_{i=0}^{m+n} {m+n choose i}A^{i}B^{m+n-i}
$$
can you show that either $A^i = 0$ or $B^{m+n-i}=0$?
answered Aug 17 '13 at 3:41
nullUsernullUser
16.7k442104
answered Aug 17 '13 at 3:41
nullUsernullUser
16.7k442104
answered Aug 17 '13 at 3:41
nullUsernullUser
16.7k442104
answered Aug 17 '13 at 3:41
nullUsernullUser
16.7k442104
16.7k442104
$begingroup$
yes..thanks nullUser
$endgroup$
– Roiner Segura Cubero
Aug 17 '13 at 4:11
$begingroup$
...and note that it is enough to take $;m+n-1;$ ...
$endgroup$
– DonAntonio
Aug 17 '13 at 8:56
add a comment |
$begingroup$
yes..thanks nullUser
$endgroup$
– Roiner Segura Cubero
Aug 17 '13 at 4:11
$begingroup$
...and note that it is enough to take $;m+n-1;$ ...
$endgroup$
– DonAntonio
Aug 17 '13 at 8:56
$begingroup$
yes..thanks nullUser
$endgroup$
– Roiner Segura Cubero
Aug 17 '13 at 4:11
$begingroup$
yes..thanks nullUser
$endgroup$
– Roiner Segura Cubero
Aug 17 '13 at 4:11
$begingroup$
...and note that it is enough to take $;m+n-1;$ ...
$endgroup$
– DonAntonio
Aug 17 '13 at 8:56
$begingroup$
...and note that it is enough to take $;m+n-1;$ ...
$endgroup$
– DonAntonio
Aug 17 '13 at 8:56
add a comment |
$begingroup$
Another approach to the problem can be based on the well known fact that commuting matrices are simultaneously triangularizable.
If so then exists such matrix $P$ that $A=PT_AP^{-1}$ and $B=PT_BP^{-1}$ where $T_A,T_B$ are, assume, upper triangular matrices.
Additionally they have on diagonal only $0$ values as the eigenvalues of nilpotent matrices are $0$.
Hence it's obvious that $T_A+T_B$ has also only eigenvalues equal to $0$ and consequently also $A+B= PT_AP^{-1} +PT_BP^{-1}=P(T_A+T_B)P^{-1}$
This kind of argumentation allows additionally to extend the claim for other types of expressions with commuting nilpotent $A,B$ so we have for example that
any sums of powers $A^k, B^i$ $(k,i>0)$ (and their products) are also nilpotent.
answered Jan 18 at 9:09
WidawensenWidawensen
4,51321446
$endgroup$
add a comment |
$begingroup$
Another approach to the problem can be based on the well known fact that commuting matrices are simultaneously triangularizable.
If so then exists such matrix $P$ that $A=PT_AP^{-1}$ and $B=PT_BP^{-1}$ where $T_A,T_B$ are, assume, upper triangular matrices.
Additionally they have on diagonal only $0$ values as the eigenvalues of nilpotent matrices are $0$.
Hence it's obvious that $T_A+T_B$ has also only eigenvalues equal to $0$ and consequently also $A+B= PT_AP^{-1} +PT_BP^{-1}=P(T_A+T_B)P^{-1}$
This kind of argumentation allows additionally to extend the claim for other types of expressions with commuting nilpotent $A,B$ so we have for example that
any sums of powers $A^k, B^i$ $(k,i>0)$ (and their products) are also nilpotent.
answered Jan 18 at 9:09
WidawensenWidawensen
4,51321446
$endgroup$
add a comment |
$begingroup$
Another approach to the problem can be based on the well known fact that commuting matrices are simultaneously triangularizable.
If so then exists such matrix $P$ that $A=PT_AP^{-1}$ and $B=PT_BP^{-1}$ where $T_A,T_B$ are, assume, upper triangular matrices.
Additionally they have on diagonal only $0$ values as the eigenvalues of nilpotent matrices are $0$.
Hence it's obvious that $T_A+T_B$ has also only eigenvalues equal to $0$ and consequently also $A+B= PT_AP^{-1} +PT_BP^{-1}=P(T_A+T_B)P^{-1}$
This kind of argumentation allows additionally to extend the claim for other types of expressions with commuting nilpotent $A,B$ so we have for example that
any sums of powers $A^k, B^i$ $(k,i>0)$ (and their products) are also nilpotent.
answered Jan 18 at 9:09
WidawensenWidawensen
4,51321446
$endgroup$
Another approach to the problem can be based on the well known fact that commuting matrices are simultaneously triangularizable.
If so then exists such matrix $P$ that $A=PT_AP^{-1}$ and $B=PT_BP^{-1}$ where $T_A,T_B$ are, assume, upper triangular matrices.
Additionally they have on diagonal only $0$ values as the eigenvalues of nilpotent matrices are $0$.
Hence it's obvious that $T_A+T_B$ has also only eigenvalues equal to $0$ and consequently also $A+B= PT_AP^{-1} +PT_BP^{-1}=P(T_A+T_B)P^{-1}$
This kind of argumentation allows additionally to extend the claim for other types of expressions with commuting nilpotent $A,B$ so we have for example that
any sums of powers $A^k, B^i$ $(k,i>0)$ (and their products) are also nilpotent.
answered Jan 18 at 9:09
WidawensenWidawensen
4,51321446
edited Jan 18 at 9:30
answered Jan 18 at 9:09
WidawensenWidawensen
4,51321446
answered Jan 18 at 9:09
WidawensenWidawensen
4,51321446
answered Jan 18 at 9:09
WidawensenWidawensen
4,51321446
4,51321446
add a comment |
add a comment |
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$begingroup$
binomial theorem
$endgroup$
– yoyo
Aug 17 '13 at 3:40
$begingroup$
yes but do not know how to prove
$endgroup$
– Roiner Segura Cubero
Aug 17 '13 at 3:41