Mixing
$X^m=O_n$ implies $X^n=O_n$.
$begingroup$
Let $Xin M_n(mathbb C) $ and $m>n$ a positive integer. Does $X^m=O_n$ imply $X^n=O_n$?
My approach : I wrote the HC theorem and started multiplying it by $m-n,m... $ until I get to something like $alpha X^n=O_n$,but this doesn't imply $X^n=O_n$.
linear-algebra matrices
asked Jan 27 at 19:13
JustAnAmateurJustAnAmateur
1096
$endgroup$
add a comment |
$begingroup$
Let $Xin M_n(mathbb C) $ and $m>n$ a positive integer. Does $X^m=O_n$ imply $X^n=O_n$?
My approach : I wrote the HC theorem and started multiplying it by $m-n,m... $ until I get to something like $alpha X^n=O_n$,but this doesn't imply $X^n=O_n$.
linear-algebra matrices
asked Jan 27 at 19:13
JustAnAmateurJustAnAmateur
1096
$endgroup$
add a comment |
$begingroup$
Let $Xin M_n(mathbb C) $ and $m>n$ a positive integer. Does $X^m=O_n$ imply $X^n=O_n$?
My approach : I wrote the HC theorem and started multiplying it by $m-n,m... $ until I get to something like $alpha X^n=O_n$,but this doesn't imply $X^n=O_n$.
linear-algebra matrices
asked Jan 27 at 19:13
JustAnAmateurJustAnAmateur
1096
$endgroup$
Let $Xin M_n(mathbb C) $ and $m>n$ a positive integer. Does $X^m=O_n$ imply $X^n=O_n$?
My approach : I wrote the HC theorem and started multiplying it by $m-n,m... $ until I get to something like $alpha X^n=O_n$,but this doesn't imply $X^n=O_n$.
linear-algebra matrices
linear-algebra matrices
asked Jan 27 at 19:13
JustAnAmateurJustAnAmateur
1096
asked Jan 27 at 19:13
JustAnAmateurJustAnAmateur
1096
asked Jan 27 at 19:13
JustAnAmateurJustAnAmateur
1096
asked Jan 27 at 19:13
JustAnAmateurJustAnAmateur
1096
asked Jan 27 at 19:13
JustAnAmateurJustAnAmateur
1096
1096
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The answer is yes. One approach is as follows: if $X^m = 0$, then $X$ has only $0$ as its eigenvalue. Thus, its characteristic polynomial is $p(x) = x^n$. Now, apply the Cayley-Hamilton theorem.
answered Jan 27 at 19:15
OmnomnomnomOmnomnomnom
129k792185
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– JustAnAmateur
Jan 27 at 19:18
$begingroup$
@JustAnAmateur One can provide a proof involving kernels and induction only. See my answer.
$endgroup$
– Pedro Tamaroff♦
Jan 27 at 19:35
add a comment |
$begingroup$
Let $V$ be an $n$-dimensional space where $X$ acts, and consider the subspaces $ker(X^p) = V_p$. These are increasing, that is, $V_p subseteq V_{p+1}$. I claim that if $V_nneq V$, then all the inclusions in the chain of $n$ proper subspaces
$0=V_0 subseteq V_1subseteq cdotssubseteq V_n subsetneq V$
are strict. This is a contradiction, since as soon as an inclusion is proper, the dimension goes up by one, meaning the dimension of $V$ is at least $n+1$. Indeed, suppose that for some $p<n$ we have $V_{p-1} =V_p$, and let us show that $V_p = V_{p+1}$. So suppose that $X^{p+1}v=0$. Then $Xv$ is in the kernel of $X^p$. Since we assumed this is the same as the kernel of $X^{p-1}$, we see that $X^pv=0$, and we're done.
answered Jan 27 at 19:31
Pedro Tamaroff♦Pedro Tamaroff
97.5k10153297
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The answer is yes. One approach is as follows: if $X^m = 0$, then $X$ has only $0$ as its eigenvalue. Thus, its characteristic polynomial is $p(x) = x^n$. Now, apply the Cayley-Hamilton theorem.
answered Jan 27 at 19:15
OmnomnomnomOmnomnomnom
129k792185
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– JustAnAmateur
Jan 27 at 19:18
$begingroup$
@JustAnAmateur One can provide a proof involving kernels and induction only. See my answer.
$endgroup$
– Pedro Tamaroff♦
Jan 27 at 19:35
add a comment |
$begingroup$
The answer is yes. One approach is as follows: if $X^m = 0$, then $X$ has only $0$ as its eigenvalue. Thus, its characteristic polynomial is $p(x) = x^n$. Now, apply the Cayley-Hamilton theorem.
answered Jan 27 at 19:15
OmnomnomnomOmnomnomnom
129k792185
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– JustAnAmateur
Jan 27 at 19:18
$begingroup$
@JustAnAmateur One can provide a proof involving kernels and induction only. See my answer.
$endgroup$
– Pedro Tamaroff♦
Jan 27 at 19:35
add a comment |
$begingroup$
The answer is yes. One approach is as follows: if $X^m = 0$, then $X$ has only $0$ as its eigenvalue. Thus, its characteristic polynomial is $p(x) = x^n$. Now, apply the Cayley-Hamilton theorem.
answered Jan 27 at 19:15
OmnomnomnomOmnomnomnom
129k792185
$endgroup$
The answer is yes. One approach is as follows: if $X^m = 0$, then $X$ has only $0$ as its eigenvalue. Thus, its characteristic polynomial is $p(x) = x^n$. Now, apply the Cayley-Hamilton theorem.
answered Jan 27 at 19:15
OmnomnomnomOmnomnomnom
129k792185
answered Jan 27 at 19:15
OmnomnomnomOmnomnomnom
129k792185
answered Jan 27 at 19:15
OmnomnomnomOmnomnomnom
129k792185
answered Jan 27 at 19:15
OmnomnomnomOmnomnomnom
129k792185
129k792185
$begingroup$
Thank you very much!
$endgroup$
– JustAnAmateur
Jan 27 at 19:18
$begingroup$
@JustAnAmateur One can provide a proof involving kernels and induction only. See my answer.
$endgroup$
– Pedro Tamaroff♦
Jan 27 at 19:35
add a comment |
$begingroup$
Thank you very much!
$endgroup$
– JustAnAmateur
Jan 27 at 19:18
$begingroup$
@JustAnAmateur One can provide a proof involving kernels and induction only. See my answer.
$endgroup$
– Pedro Tamaroff♦
Jan 27 at 19:35
$begingroup$
Thank you very much!
$endgroup$
– JustAnAmateur
Jan 27 at 19:18
$begingroup$
Thank you very much!
$endgroup$
– JustAnAmateur
Jan 27 at 19:18
$begingroup$
@JustAnAmateur One can provide a proof involving kernels and induction only. See my answer.
$endgroup$
– Pedro Tamaroff♦
Jan 27 at 19:35
$begingroup$
@JustAnAmateur One can provide a proof involving kernels and induction only. See my answer.
$endgroup$
– Pedro Tamaroff♦
Jan 27 at 19:35
add a comment |
$begingroup$
Let $V$ be an $n$-dimensional space where $X$ acts, and consider the subspaces $ker(X^p) = V_p$. These are increasing, that is, $V_p subseteq V_{p+1}$. I claim that if $V_nneq V$, then all the inclusions in the chain of $n$ proper subspaces
$0=V_0 subseteq V_1subseteq cdotssubseteq V_n subsetneq V$
are strict. This is a contradiction, since as soon as an inclusion is proper, the dimension goes up by one, meaning the dimension of $V$ is at least $n+1$. Indeed, suppose that for some $p<n$ we have $V_{p-1} =V_p$, and let us show that $V_p = V_{p+1}$. So suppose that $X^{p+1}v=0$. Then $Xv$ is in the kernel of $X^p$. Since we assumed this is the same as the kernel of $X^{p-1}$, we see that $X^pv=0$, and we're done.
answered Jan 27 at 19:31
Pedro Tamaroff♦Pedro Tamaroff
97.5k10153297
$endgroup$
add a comment |
$begingroup$
Let $V$ be an $n$-dimensional space where $X$ acts, and consider the subspaces $ker(X^p) = V_p$. These are increasing, that is, $V_p subseteq V_{p+1}$. I claim that if $V_nneq V$, then all the inclusions in the chain of $n$ proper subspaces
$0=V_0 subseteq V_1subseteq cdotssubseteq V_n subsetneq V$
are strict. This is a contradiction, since as soon as an inclusion is proper, the dimension goes up by one, meaning the dimension of $V$ is at least $n+1$. Indeed, suppose that for some $p<n$ we have $V_{p-1} =V_p$, and let us show that $V_p = V_{p+1}$. So suppose that $X^{p+1}v=0$. Then $Xv$ is in the kernel of $X^p$. Since we assumed this is the same as the kernel of $X^{p-1}$, we see that $X^pv=0$, and we're done.
answered Jan 27 at 19:31
Pedro Tamaroff♦Pedro Tamaroff
97.5k10153297
$endgroup$
add a comment |
$begingroup$
Let $V$ be an $n$-dimensional space where $X$ acts, and consider the subspaces $ker(X^p) = V_p$. These are increasing, that is, $V_p subseteq V_{p+1}$. I claim that if $V_nneq V$, then all the inclusions in the chain of $n$ proper subspaces
$0=V_0 subseteq V_1subseteq cdotssubseteq V_n subsetneq V$
are strict. This is a contradiction, since as soon as an inclusion is proper, the dimension goes up by one, meaning the dimension of $V$ is at least $n+1$. Indeed, suppose that for some $p<n$ we have $V_{p-1} =V_p$, and let us show that $V_p = V_{p+1}$. So suppose that $X^{p+1}v=0$. Then $Xv$ is in the kernel of $X^p$. Since we assumed this is the same as the kernel of $X^{p-1}$, we see that $X^pv=0$, and we're done.
answered Jan 27 at 19:31
Pedro Tamaroff♦Pedro Tamaroff
97.5k10153297
$endgroup$
Let $V$ be an $n$-dimensional space where $X$ acts, and consider the subspaces $ker(X^p) = V_p$. These are increasing, that is, $V_p subseteq V_{p+1}$. I claim that if $V_nneq V$, then all the inclusions in the chain of $n$ proper subspaces
$0=V_0 subseteq V_1subseteq cdotssubseteq V_n subsetneq V$
are strict. This is a contradiction, since as soon as an inclusion is proper, the dimension goes up by one, meaning the dimension of $V$ is at least $n+1$. Indeed, suppose that for some $p<n$ we have $V_{p-1} =V_p$, and let us show that $V_p = V_{p+1}$. So suppose that $X^{p+1}v=0$. Then $Xv$ is in the kernel of $X^p$. Since we assumed this is the same as the kernel of $X^{p-1}$, we see that $X^pv=0$, and we're done.
answered Jan 27 at 19:31
Pedro Tamaroff♦Pedro Tamaroff
97.5k10153297
answered Jan 27 at 19:31
Pedro Tamaroff♦Pedro Tamaroff
97.5k10153297
answered Jan 27 at 19:31
Pedro Tamaroff♦Pedro Tamaroff
97.5k10153297
answered Jan 27 at 19:31
Pedro Tamaroff♦Pedro Tamaroff
97.5k10153297
97.5k10153297
add a comment |
add a comment |
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