Minimal polynomial for any power of Jordan block is same as the minimal polynomial of the Jordan block.
1
Let $J$ be the $n times n$ Jordan block corresponding to the eigen value $1$. For any natural number $r$ is it true that the minimal polynomial for $J^r$ is $(X-1)^n$ ?
Another way to think about it to produce a cyclic vector of $J^r$. I can’t prove it. I need some help. Thanks.
linear-algebra abstract-algebra matrices jordan-normal-form canonical-transformation
asked Dec 6 at 10:12
user371231
579511
add a comment |
1
Let $J$ be the $n times n$ Jordan block corresponding to the eigen value $1$. For any natural number $r$ is it true that the minimal polynomial for $J^r$ is $(X-1)^n$ ?
Another way to think about it to produce a cyclic vector of $J^r$. I can’t prove it. I need some help. Thanks.
linear-algebra abstract-algebra matrices jordan-normal-form canonical-transformation
asked Dec 6 at 10:12
user371231
579511
This is true for any non-zero eigenvalue, not just eigenvalue $1$.
– Zvi
Dec 6 at 11:19
2
Write $J=I+N$ where $I$ is the identity and $N$ is 1 on the super diagonal and zero otherwise (hence nilpotent of order $n$). Then $J^r=I+sum_{k=1}^r {{r}choose{k}} N^k$. You can show that $sum_{k=1}^r {{r}choose{k}} N^k$ is nilpotent of order $n$.
– Eric
Dec 6 at 15:40
1
The result is false without some restriction on $n,r$ and the characteristic of the field in question. For example, in characteristic $p$ if $n=p$ we have that $(I+N)^p=I$ which has minimal polynomial $X-1$.
– ancientmathematician
Dec 6 at 16:34
My interest is in 0 characteristic. Thanks for your counter example in positive characteristic.
– user371231
Dec 6 at 18:46
add a comment |
1
1
1
2
Let $J$ be the $n times n$ Jordan block corresponding to the eigen value $1$. For any natural number $r$ is it true that the minimal polynomial for $J^r$ is $(X-1)^n$ ?
Another way to think about it to produce a cyclic vector of $J^r$. I can’t prove it. I need some help. Thanks.
linear-algebra abstract-algebra matrices jordan-normal-form canonical-transformation
asked Dec 6 at 10:12
user371231
579511
Let $J$ be the $n times n$ Jordan block corresponding to the eigen value $1$. For any natural number $r$ is it true that the minimal polynomial for $J^r$ is $(X-1)^n$ ?
Another way to think about it to produce a cyclic vector of $J^r$. I can’t prove it. I need some help. Thanks.
linear-algebra abstract-algebra matrices jordan-normal-form canonical-transformation
linear-algebra abstract-algebra matrices jordan-normal-form canonical-transformation
asked Dec 6 at 10:12
user371231
579511
asked Dec 6 at 10:12
user371231
579511
asked Dec 6 at 10:12
user371231
579511
asked Dec 6 at 10:12
user371231
579511
asked Dec 6 at 10:12
user371231
579511
579511
This is true for any non-zero eigenvalue, not just eigenvalue $1$.
– Zvi
Dec 6 at 11:19
2
Write $J=I+N$ where $I$ is the identity and $N$ is 1 on the super diagonal and zero otherwise (hence nilpotent of order $n$). Then $J^r=I+sum_{k=1}^r {{r}choose{k}} N^k$. You can show that $sum_{k=1}^r {{r}choose{k}} N^k$ is nilpotent of order $n$.
– Eric
Dec 6 at 15:40
1
The result is false without some restriction on $n,r$ and the characteristic of the field in question. For example, in characteristic $p$ if $n=p$ we have that $(I+N)^p=I$ which has minimal polynomial $X-1$.
– ancientmathematician
Dec 6 at 16:34
My interest is in 0 characteristic. Thanks for your counter example in positive characteristic.
– user371231
Dec 6 at 18:46
add a comment |
This is true for any non-zero eigenvalue, not just eigenvalue $1$.
– Zvi
Dec 6 at 11:19
2
Write $J=I+N$ where $I$ is the identity and $N$ is 1 on the super diagonal and zero otherwise (hence nilpotent of order $n$). Then $J^r=I+sum_{k=1}^r {{r}choose{k}} N^k$. You can show that $sum_{k=1}^r {{r}choose{k}} N^k$ is nilpotent of order $n$.
– Eric
Dec 6 at 15:40
1
The result is false without some restriction on $n,r$ and the characteristic of the field in question. For example, in characteristic $p$ if $n=p$ we have that $(I+N)^p=I$ which has minimal polynomial $X-1$.
– ancientmathematician
Dec 6 at 16:34
My interest is in 0 characteristic. Thanks for your counter example in positive characteristic.
– user371231
Dec 6 at 18:46
This is true for any non-zero eigenvalue, not just eigenvalue $1$.
– Zvi
Dec 6 at 11:19
This is true for any non-zero eigenvalue, not just eigenvalue $1$.
– Zvi
Dec 6 at 11:19
2
2
Write $J=I+N$ where $I$ is the identity and $N$ is 1 on the super diagonal and zero otherwise (hence nilpotent of order $n$). Then $J^r=I+sum_{k=1}^r {{r}choose{k}} N^k$. You can show that $sum_{k=1}^r {{r}choose{k}} N^k$ is nilpotent of order $n$.
– Eric
Dec 6 at 15:40
Write $J=I+N$ where $I$ is the identity and $N$ is 1 on the super diagonal and zero otherwise (hence nilpotent of order $n$). Then $J^r=I+sum_{k=1}^r {{r}choose{k}} N^k$. You can show that $sum_{k=1}^r {{r}choose{k}} N^k$ is nilpotent of order $n$.
– Eric
Dec 6 at 15:40
1
1
The result is false without some restriction on $n,r$ and the characteristic of the field in question. For example, in characteristic $p$ if $n=p$ we have that $(I+N)^p=I$ which has minimal polynomial $X-1$.
– ancientmathematician
Dec 6 at 16:34
The result is false without some restriction on $n,r$ and the characteristic of the field in question. For example, in characteristic $p$ if $n=p$ we have that $(I+N)^p=I$ which has minimal polynomial $X-1$.
– ancientmathematician
Dec 6 at 16:34
My interest is in 0 characteristic. Thanks for your counter example in positive characteristic.
– user371231
Dec 6 at 18:46
My interest is in 0 characteristic. Thanks for your counter example in positive characteristic.
– user371231
Dec 6 at 18:46
add a comment |
2 Answers
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1
Hint: write $J=I+N$ where $N$ is the shift matrix. $N$ is nilpotent with index $n$. Now expand $J^r=(I+N)^r=...$ and find out what is the smallest $m$ we need in order to $(J^r-I)^m=0$.
answered Dec 6 at 11:17
A.Γ.
21.7k22455
add a comment |
1
As $r(J-I)=r(J^r-I)$, so geometric multiplicity is $1$in both case are same and hence same minimal polynomial. Here $r$ means rank of matrix.
answered Dec 6 at 12:20
neelkanth
2,0461927
It is not true. Take $3times 3$ case: $$J-I=begin{bmatrix}0 & 1& 0\0 & 0 & 1\0 & 0 & 0end{bmatrix},quad J^2-I=begin{bmatrix}0 & 2& color{red}{1}\0 & 0 & 2\0 & 0 & 0end{bmatrix}.$$ In what sense are they equal?
– A.Γ.
Dec 6 at 12:50
Both has same rank as $2.$
– neelkanth
Dec 6 at 13:03
@neelkanth If by "r" you mean rank then you should write that out clearly. Your equation communicates that the constant multiple r times J-I is equal to the constant multiple r times (J^r-I) which is not true as pointed out by A.Γ.
– Eric
Dec 6 at 15:47
1
It’s standard notation if we work in matrices
– neelkanth
Dec 6 at 16:15
Can you explain why the ranks are same ?
– user371231
Dec 6 at 16:18
|
show 2 more comments
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2 Answers
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2 Answers
2
active
oldest
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oldest
votes
1
Hint: write $J=I+N$ where $N$ is the shift matrix. $N$ is nilpotent with index $n$. Now expand $J^r=(I+N)^r=...$ and find out what is the smallest $m$ we need in order to $(J^r-I)^m=0$.
answered Dec 6 at 11:17
A.Γ.
21.7k22455
add a comment |
1
Hint: write $J=I+N$ where $N$ is the shift matrix. $N$ is nilpotent with index $n$. Now expand $J^r=(I+N)^r=...$ and find out what is the smallest $m$ we need in order to $(J^r-I)^m=0$.
answered Dec 6 at 11:17
A.Γ.
21.7k22455
add a comment |
1
1
1
Hint: write $J=I+N$ where $N$ is the shift matrix. $N$ is nilpotent with index $n$. Now expand $J^r=(I+N)^r=...$ and find out what is the smallest $m$ we need in order to $(J^r-I)^m=0$.
answered Dec 6 at 11:17
A.Γ.
21.7k22455
Hint: write $J=I+N$ where $N$ is the shift matrix. $N$ is nilpotent with index $n$. Now expand $J^r=(I+N)^r=...$ and find out what is the smallest $m$ we need in order to $(J^r-I)^m=0$.
answered Dec 6 at 11:17
A.Γ.
21.7k22455
answered Dec 6 at 11:17
A.Γ.
21.7k22455
answered Dec 6 at 11:17
A.Γ.
21.7k22455
answered Dec 6 at 11:17
A.Γ.
21.7k22455
21.7k22455
add a comment |
add a comment |
1
As $r(J-I)=r(J^r-I)$, so geometric multiplicity is $1$in both case are same and hence same minimal polynomial. Here $r$ means rank of matrix.
answered Dec 6 at 12:20
neelkanth
2,0461927
It is not true. Take $3times 3$ case: $$J-I=begin{bmatrix}0 & 1& 0\0 & 0 & 1\0 & 0 & 0end{bmatrix},quad J^2-I=begin{bmatrix}0 & 2& color{red}{1}\0 & 0 & 2\0 & 0 & 0end{bmatrix}.$$ In what sense are they equal?
– A.Γ.
Dec 6 at 12:50
Both has same rank as $2.$
– neelkanth
Dec 6 at 13:03
@neelkanth If by "r" you mean rank then you should write that out clearly. Your equation communicates that the constant multiple r times J-I is equal to the constant multiple r times (J^r-I) which is not true as pointed out by A.Γ.
– Eric
Dec 6 at 15:47
1
It’s standard notation if we work in matrices
– neelkanth
Dec 6 at 16:15
Can you explain why the ranks are same ?
– user371231
Dec 6 at 16:18
|
show 2 more comments
1
As $r(J-I)=r(J^r-I)$, so geometric multiplicity is $1$in both case are same and hence same minimal polynomial. Here $r$ means rank of matrix.
answered Dec 6 at 12:20
neelkanth
2,0461927
It is not true. Take $3times 3$ case: $$J-I=begin{bmatrix}0 & 1& 0\0 & 0 & 1\0 & 0 & 0end{bmatrix},quad J^2-I=begin{bmatrix}0 & 2& color{red}{1}\0 & 0 & 2\0 & 0 & 0end{bmatrix}.$$ In what sense are they equal?
– A.Γ.
Dec 6 at 12:50
Both has same rank as $2.$
– neelkanth
Dec 6 at 13:03
@neelkanth If by "r" you mean rank then you should write that out clearly. Your equation communicates that the constant multiple r times J-I is equal to the constant multiple r times (J^r-I) which is not true as pointed out by A.Γ.
– Eric
Dec 6 at 15:47
1
It’s standard notation if we work in matrices
– neelkanth
Dec 6 at 16:15
Can you explain why the ranks are same ?
– user371231
Dec 6 at 16:18
|
show 2 more comments
1
1
1
As $r(J-I)=r(J^r-I)$, so geometric multiplicity is $1$in both case are same and hence same minimal polynomial. Here $r$ means rank of matrix.
answered Dec 6 at 12:20
neelkanth
2,0461927
As $r(J-I)=r(J^r-I)$, so geometric multiplicity is $1$in both case are same and hence same minimal polynomial. Here $r$ means rank of matrix.
answered Dec 6 at 12:20
neelkanth
2,0461927
edited Dec 6 at 16:21
answered Dec 6 at 12:20
neelkanth
2,0461927
answered Dec 6 at 12:20
neelkanth
2,0461927
answered Dec 6 at 12:20
neelkanth
2,0461927
2,0461927
It is not true. Take $3times 3$ case: $$J-I=begin{bmatrix}0 & 1& 0\0 & 0 & 1\0 & 0 & 0end{bmatrix},quad J^2-I=begin{bmatrix}0 & 2& color{red}{1}\0 & 0 & 2\0 & 0 & 0end{bmatrix}.$$ In what sense are they equal?
– A.Γ.
Dec 6 at 12:50
Both has same rank as $2.$
– neelkanth
Dec 6 at 13:03
@neelkanth If by "r" you mean rank then you should write that out clearly. Your equation communicates that the constant multiple r times J-I is equal to the constant multiple r times (J^r-I) which is not true as pointed out by A.Γ.
– Eric
Dec 6 at 15:47
1
It’s standard notation if we work in matrices
– neelkanth
Dec 6 at 16:15
Can you explain why the ranks are same ?
– user371231
Dec 6 at 16:18
|
show 2 more comments
It is not true. Take $3times 3$ case: $$J-I=begin{bmatrix}0 & 1& 0\0 & 0 & 1\0 & 0 & 0end{bmatrix},quad J^2-I=begin{bmatrix}0 & 2& color{red}{1}\0 & 0 & 2\0 & 0 & 0end{bmatrix}.$$ In what sense are they equal?
– A.Γ.
Dec 6 at 12:50
Both has same rank as $2.$
– neelkanth
Dec 6 at 13:03
@neelkanth If by "r" you mean rank then you should write that out clearly. Your equation communicates that the constant multiple r times J-I is equal to the constant multiple r times (J^r-I) which is not true as pointed out by A.Γ.
– Eric
Dec 6 at 15:47
1
It’s standard notation if we work in matrices
– neelkanth
Dec 6 at 16:15
Can you explain why the ranks are same ?
– user371231
Dec 6 at 16:18
It is not true. Take $3times 3$ case: $$J-I=begin{bmatrix}0 & 1& 0\0 & 0 & 1\0 & 0 & 0end{bmatrix},quad J^2-I=begin{bmatrix}0 & 2& color{red}{1}\0 & 0 & 2\0 & 0 & 0end{bmatrix}.$$ In what sense are they equal?
– A.Γ.
Dec 6 at 12:50
It is not true. Take $3times 3$ case: $$J-I=begin{bmatrix}0 & 1& 0\0 & 0 & 1\0 & 0 & 0end{bmatrix},quad J^2-I=begin{bmatrix}0 & 2& color{red}{1}\0 & 0 & 2\0 & 0 & 0end{bmatrix}.$$ In what sense are they equal?
– A.Γ.
Dec 6 at 12:50
Both has same rank as $2.$
– neelkanth
Dec 6 at 13:03
Both has same rank as $2.$
– neelkanth
Dec 6 at 13:03
@neelkanth If by "r" you mean rank then you should write that out clearly. Your equation communicates that the constant multiple r times J-I is equal to the constant multiple r times (J^r-I) which is not true as pointed out by A.Γ.
– Eric
Dec 6 at 15:47
@neelkanth If by "r" you mean rank then you should write that out clearly. Your equation communicates that the constant multiple r times J-I is equal to the constant multiple r times (J^r-I) which is not true as pointed out by A.Γ.
– Eric
Dec 6 at 15:47
1
1
It’s standard notation if we work in matrices
– neelkanth
Dec 6 at 16:15
It’s standard notation if we work in matrices
– neelkanth
Dec 6 at 16:15
Can you explain why the ranks are same ?
– user371231
Dec 6 at 16:18
Can you explain why the ranks are same ?
– user371231
Dec 6 at 16:18
|
show 2 more comments
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This is true for any non-zero eigenvalue, not just eigenvalue $1$.
– Zvi
Dec 6 at 11:19
2
Write $J=I+N$ where $I$ is the identity and $N$ is 1 on the super diagonal and zero otherwise (hence nilpotent of order $n$). Then $J^r=I+sum_{k=1}^r {{r}choose{k}} N^k$. You can show that $sum_{k=1}^r {{r}choose{k}} N^k$ is nilpotent of order $n$.
– Eric
Dec 6 at 15:40
1
The result is false without some restriction on $n,r$ and the characteristic of the field in question. For example, in characteristic $p$ if $n=p$ we have that $(I+N)^p=I$ which has minimal polynomial $X-1$.
– ancientmathematician
Dec 6 at 16:34
My interest is in 0 characteristic. Thanks for your counter example in positive characteristic.
– user371231
Dec 6 at 18:46