Mixing
If $A^k$ commutes with $B$ then $A$ commutes with $B$.
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Let $A$ and $B$ are two $n times n$ Complex matrices. Assume that $(A-I)^n=0$ and $A^kB=BA^k$ for some $k in mathbb{N}$. Then I want to prove that $AB= BA$.
Clearly $1$ is the only eigen value of $A$ and also $A^k$ and $B$ are simultaneously triangulable. But how do I get down to $A$ to commute with $B$. Any help will be appreciated. Thanks.
linear-algebra abstract-algebra matrices field-theory
asked yesterday
Panja
286213
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up vote
10
down vote
favorite
Let $A$ and $B$ are two $n times n$ Complex matrices. Assume that $(A-I)^n=0$ and $A^kB=BA^k$ for some $k in mathbb{N}$. Then I want to prove that $AB= BA$.
Clearly $1$ is the only eigen value of $A$ and also $A^k$ and $B$ are simultaneously triangulable. But how do I get down to $A$ to commute with $B$. Any help will be appreciated. Thanks.
linear-algebra abstract-algebra matrices field-theory
asked yesterday
Panja
286213
2
Note that $mathbb{N}$ has to mean $left{1,2,3,ldotsright}$ for this to be correct.
– darij grinberg
19 hours ago
add a comment |
up vote
10
down vote
favorite
up vote
10
down vote
favorite
Let $A$ and $B$ are two $n times n$ Complex matrices. Assume that $(A-I)^n=0$ and $A^kB=BA^k$ for some $k in mathbb{N}$. Then I want to prove that $AB= BA$.
Clearly $1$ is the only eigen value of $A$ and also $A^k$ and $B$ are simultaneously triangulable. But how do I get down to $A$ to commute with $B$. Any help will be appreciated. Thanks.
linear-algebra abstract-algebra matrices field-theory
asked yesterday
Panja
286213
Let $A$ and $B$ are two $n times n$ Complex matrices. Assume that $(A-I)^n=0$ and $A^kB=BA^k$ for some $k in mathbb{N}$. Then I want to prove that $AB= BA$.
Clearly $1$ is the only eigen value of $A$ and also $A^k$ and $B$ are simultaneously triangulable. But how do I get down to $A$ to commute with $B$. Any help will be appreciated. Thanks.
linear-algebra abstract-algebra matrices field-theory
linear-algebra abstract-algebra matrices field-theory
asked yesterday
Panja
286213
asked yesterday
Panja
286213
asked yesterday
Panja
286213
asked yesterday
Panja
286213
asked yesterday
Panja
286213
286213
2
Note that $mathbb{N}$ has to mean $left{1,2,3,ldotsright}$ for this to be correct.
– darij grinberg
19 hours ago
add a comment |
2
Note that $mathbb{N}$ has to mean $left{1,2,3,ldotsright}$ for this to be correct.
– darij grinberg
19 hours ago
2
2
Note that $mathbb{N}$ has to mean $left{1,2,3,ldotsright}$ for this to be correct.
– darij grinberg
19 hours ago
Note that $mathbb{N}$ has to mean $left{1,2,3,ldotsright}$ for this to be correct.
– darij grinberg
19 hours ago
add a comment |
2 Answers
2
active
oldest
votes
up vote
8
down vote
accepted
Hint. Try to prove that $A$ is a polynomial in $A^k$.
Edit. To prove the hint, you may follow darij grinberg's comment below. I did essentially the same thing, but from a matrix analytic rather than linear algebraic perspective: I considered the primary matrix function $f(X)=(I+X)^{1/k}=sum_{i=0}^infty frac{f^{(i)}(0)}{k!}X^i$ for a nilpotent matrix $X$ associated with the scalar function $f(x)=(1+x)^{1/k}$. Put $X=A^k-I$ and we are done.
Alternatively, the hint can be proved using only Jordan forms, but the argument is much longer.
- Let $J$ be the Jordan form of $A$. Since all eigenvalues of $A$ are ones, we may write $J=J_{m_1}oplus J_{m_2}opluscdotsoplus J_{m_b}$, where $1le m_1le m_2lecdotsle m_b$ and $J_m$ denotes a Jordan block of size $m$ for the eigenvalue $1$.
- Note that if $rge0$ and $m<n$, then $J_m^r$ coincides with the leading principal $mtimes m$ submatrix of $J_n^r$. Hence $p(J_m^k)$ coincides with a leading principal submatrix of $p(J_n^k)$ for any polynomial $p$.
- It follows that if $p(J_n^k)=J_n$, then $p(J_m^k)=J_m$ for every $m<n$. This is true in particular when $min{m_1,m_2,ldots,m_b}$. Consequently, $p(A^k)=A$.
- Hence the problem boils down to finding a polynomial $p$ such that $p(J_n^k)=J_n$. This should be straightforward and I will leave it to you.
answered yesterday
user1551
70.1k566125
1
I was trying to prove your hint. I was thinking that if I can show that $A^k$ has a cyclic vector then since $A^k$ commutes with $A$, $A$ would have been a polynomial in $A^k$. But it may not be the case that $A^k$ has a cyclic vector. Can you kindly give me one more hint to prove your hint ?
– Panja
yesterday
5
My favorite way of proving the hint is to recall Newton's binomial formula $left(1+xright)^r = sumlimits_{i=0}^{infty} dbinom{r}{i} x^i$ for all $r in mathbb{Q}$. This is, per se, an equality between formal power series in the indeterminate $x$ over $mathbb{Q}$. But since the matrix $A^k - I$ is nilpotent (make sure you understand why!), we can substitute $A^k - I$ for $x$ in this formula (make sure you understand why!), and apply it to $r = 1/k$, thus obtaining $A$ on the left hand side (make sure you understand how!), and on the right hand side a polynomial in $A^k - I$.
– darij grinberg
19 hours ago
1
I never knew you could put non-integers into binomial coefficients. Thanks for sharing!
– user25959
14 hours ago
1
@Song There is no need to do this. Since $X$ is nilpotent, the infinite sum is actually a finite sum (or more specifically, a sum of no more than $n$ terms).
– user1551
12 hours ago
1
@Song $A-I$ is nilpotent. Therefore $1$ is an eigenvalue fo $A$ of multiplicity $n$. Hence $1$ is an eigenvalue of $A^k$ of multiplicity $n$, meaning that $A^k$ is nilpotent. You may also triangularise $A$ first to see this.
– user1551
12 hours ago
|
show 4 more comments
up vote
0
down vote
If $A$ has only $1$ as eigenvalue then it is invertible (and of the form $I+N$ where $N$ is nilpotent with $N^n=0$).
$A $ is satisfying also any equation $(A^i-I)^n=0$ so we can write
$(A-I)^n=0, (A^2-I)^n=0, dots (A^k-I)^n=0$.
Also we can write ${A^k} B(A^k)^{-1}=B$.
In similar fashion
${A^k} B(A^k)^{-1}=({A^k})^{-1} B(A^k)$
${A^{2k}} B = B(A^{2k})$, etc..
For any natural $m$:
${A^{mk}} B = B(A^{mk})$
If powers $(A^k)^m$ could be a basis for expression of $A$ then commutativity would follow...
answered yesterday
Widawensen
4,34921444
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
Hint. Try to prove that $A$ is a polynomial in $A^k$.
Edit. To prove the hint, you may follow darij grinberg's comment below. I did essentially the same thing, but from a matrix analytic rather than linear algebraic perspective: I considered the primary matrix function $f(X)=(I+X)^{1/k}=sum_{i=0}^infty frac{f^{(i)}(0)}{k!}X^i$ for a nilpotent matrix $X$ associated with the scalar function $f(x)=(1+x)^{1/k}$. Put $X=A^k-I$ and we are done.
Alternatively, the hint can be proved using only Jordan forms, but the argument is much longer.
- Let $J$ be the Jordan form of $A$. Since all eigenvalues of $A$ are ones, we may write $J=J_{m_1}oplus J_{m_2}opluscdotsoplus J_{m_b}$, where $1le m_1le m_2lecdotsle m_b$ and $J_m$ denotes a Jordan block of size $m$ for the eigenvalue $1$.
- Note that if $rge0$ and $m<n$, then $J_m^r$ coincides with the leading principal $mtimes m$ submatrix of $J_n^r$. Hence $p(J_m^k)$ coincides with a leading principal submatrix of $p(J_n^k)$ for any polynomial $p$.
- It follows that if $p(J_n^k)=J_n$, then $p(J_m^k)=J_m$ for every $m<n$. This is true in particular when $min{m_1,m_2,ldots,m_b}$. Consequently, $p(A^k)=A$.
- Hence the problem boils down to finding a polynomial $p$ such that $p(J_n^k)=J_n$. This should be straightforward and I will leave it to you.
answered yesterday
user1551
70.1k566125
1
I was trying to prove your hint. I was thinking that if I can show that $A^k$ has a cyclic vector then since $A^k$ commutes with $A$, $A$ would have been a polynomial in $A^k$. But it may not be the case that $A^k$ has a cyclic vector. Can you kindly give me one more hint to prove your hint ?
– Panja
yesterday
5
My favorite way of proving the hint is to recall Newton's binomial formula $left(1+xright)^r = sumlimits_{i=0}^{infty} dbinom{r}{i} x^i$ for all $r in mathbb{Q}$. This is, per se, an equality between formal power series in the indeterminate $x$ over $mathbb{Q}$. But since the matrix $A^k - I$ is nilpotent (make sure you understand why!), we can substitute $A^k - I$ for $x$ in this formula (make sure you understand why!), and apply it to $r = 1/k$, thus obtaining $A$ on the left hand side (make sure you understand how!), and on the right hand side a polynomial in $A^k - I$.
– darij grinberg
19 hours ago
1
I never knew you could put non-integers into binomial coefficients. Thanks for sharing!
– user25959
14 hours ago
1
@Song There is no need to do this. Since $X$ is nilpotent, the infinite sum is actually a finite sum (or more specifically, a sum of no more than $n$ terms).
– user1551
12 hours ago
1
@Song $A-I$ is nilpotent. Therefore $1$ is an eigenvalue fo $A$ of multiplicity $n$. Hence $1$ is an eigenvalue of $A^k$ of multiplicity $n$, meaning that $A^k$ is nilpotent. You may also triangularise $A$ first to see this.
– user1551
12 hours ago
|
show 4 more comments
up vote
8
down vote
accepted
Hint. Try to prove that $A$ is a polynomial in $A^k$.
Edit. To prove the hint, you may follow darij grinberg's comment below. I did essentially the same thing, but from a matrix analytic rather than linear algebraic perspective: I considered the primary matrix function $f(X)=(I+X)^{1/k}=sum_{i=0}^infty frac{f^{(i)}(0)}{k!}X^i$ for a nilpotent matrix $X$ associated with the scalar function $f(x)=(1+x)^{1/k}$. Put $X=A^k-I$ and we are done.
Alternatively, the hint can be proved using only Jordan forms, but the argument is much longer.
- Let $J$ be the Jordan form of $A$. Since all eigenvalues of $A$ are ones, we may write $J=J_{m_1}oplus J_{m_2}opluscdotsoplus J_{m_b}$, where $1le m_1le m_2lecdotsle m_b$ and $J_m$ denotes a Jordan block of size $m$ for the eigenvalue $1$.
- Note that if $rge0$ and $m<n$, then $J_m^r$ coincides with the leading principal $mtimes m$ submatrix of $J_n^r$. Hence $p(J_m^k)$ coincides with a leading principal submatrix of $p(J_n^k)$ for any polynomial $p$.
- It follows that if $p(J_n^k)=J_n$, then $p(J_m^k)=J_m$ for every $m<n$. This is true in particular when $min{m_1,m_2,ldots,m_b}$. Consequently, $p(A^k)=A$.
- Hence the problem boils down to finding a polynomial $p$ such that $p(J_n^k)=J_n$. This should be straightforward and I will leave it to you.
answered yesterday
user1551
70.1k566125
1
I was trying to prove your hint. I was thinking that if I can show that $A^k$ has a cyclic vector then since $A^k$ commutes with $A$, $A$ would have been a polynomial in $A^k$. But it may not be the case that $A^k$ has a cyclic vector. Can you kindly give me one more hint to prove your hint ?
– Panja
yesterday
5
My favorite way of proving the hint is to recall Newton's binomial formula $left(1+xright)^r = sumlimits_{i=0}^{infty} dbinom{r}{i} x^i$ for all $r in mathbb{Q}$. This is, per se, an equality between formal power series in the indeterminate $x$ over $mathbb{Q}$. But since the matrix $A^k - I$ is nilpotent (make sure you understand why!), we can substitute $A^k - I$ for $x$ in this formula (make sure you understand why!), and apply it to $r = 1/k$, thus obtaining $A$ on the left hand side (make sure you understand how!), and on the right hand side a polynomial in $A^k - I$.
– darij grinberg
19 hours ago
1
I never knew you could put non-integers into binomial coefficients. Thanks for sharing!
– user25959
14 hours ago
1
@Song There is no need to do this. Since $X$ is nilpotent, the infinite sum is actually a finite sum (or more specifically, a sum of no more than $n$ terms).
– user1551
12 hours ago
1
@Song $A-I$ is nilpotent. Therefore $1$ is an eigenvalue fo $A$ of multiplicity $n$. Hence $1$ is an eigenvalue of $A^k$ of multiplicity $n$, meaning that $A^k$ is nilpotent. You may also triangularise $A$ first to see this.
– user1551
12 hours ago
|
show 4 more comments
up vote
8
down vote
accepted
up vote
8
down vote
accepted
Hint. Try to prove that $A$ is a polynomial in $A^k$.
Edit. To prove the hint, you may follow darij grinberg's comment below. I did essentially the same thing, but from a matrix analytic rather than linear algebraic perspective: I considered the primary matrix function $f(X)=(I+X)^{1/k}=sum_{i=0}^infty frac{f^{(i)}(0)}{k!}X^i$ for a nilpotent matrix $X$ associated with the scalar function $f(x)=(1+x)^{1/k}$. Put $X=A^k-I$ and we are done.
Alternatively, the hint can be proved using only Jordan forms, but the argument is much longer.
- Let $J$ be the Jordan form of $A$. Since all eigenvalues of $A$ are ones, we may write $J=J_{m_1}oplus J_{m_2}opluscdotsoplus J_{m_b}$, where $1le m_1le m_2lecdotsle m_b$ and $J_m$ denotes a Jordan block of size $m$ for the eigenvalue $1$.
- Note that if $rge0$ and $m<n$, then $J_m^r$ coincides with the leading principal $mtimes m$ submatrix of $J_n^r$. Hence $p(J_m^k)$ coincides with a leading principal submatrix of $p(J_n^k)$ for any polynomial $p$.
- It follows that if $p(J_n^k)=J_n$, then $p(J_m^k)=J_m$ for every $m<n$. This is true in particular when $min{m_1,m_2,ldots,m_b}$. Consequently, $p(A^k)=A$.
- Hence the problem boils down to finding a polynomial $p$ such that $p(J_n^k)=J_n$. This should be straightforward and I will leave it to you.
answered yesterday
user1551
70.1k566125
Hint. Try to prove that $A$ is a polynomial in $A^k$.
Edit. To prove the hint, you may follow darij grinberg's comment below. I did essentially the same thing, but from a matrix analytic rather than linear algebraic perspective: I considered the primary matrix function $f(X)=(I+X)^{1/k}=sum_{i=0}^infty frac{f^{(i)}(0)}{k!}X^i$ for a nilpotent matrix $X$ associated with the scalar function $f(x)=(1+x)^{1/k}$. Put $X=A^k-I$ and we are done.
Alternatively, the hint can be proved using only Jordan forms, but the argument is much longer.
- Let $J$ be the Jordan form of $A$. Since all eigenvalues of $A$ are ones, we may write $J=J_{m_1}oplus J_{m_2}opluscdotsoplus J_{m_b}$, where $1le m_1le m_2lecdotsle m_b$ and $J_m$ denotes a Jordan block of size $m$ for the eigenvalue $1$.
- Note that if $rge0$ and $m<n$, then $J_m^r$ coincides with the leading principal $mtimes m$ submatrix of $J_n^r$. Hence $p(J_m^k)$ coincides with a leading principal submatrix of $p(J_n^k)$ for any polynomial $p$.
- It follows that if $p(J_n^k)=J_n$, then $p(J_m^k)=J_m$ for every $m<n$. This is true in particular when $min{m_1,m_2,ldots,m_b}$. Consequently, $p(A^k)=A$.
- Hence the problem boils down to finding a polynomial $p$ such that $p(J_n^k)=J_n$. This should be straightforward and I will leave it to you.
answered yesterday
user1551
70.1k566125
edited 12 hours ago
answered yesterday
user1551
70.1k566125
answered yesterday
user1551
70.1k566125
answered yesterday
user1551
70.1k566125
70.1k566125
1
I was trying to prove your hint. I was thinking that if I can show that $A^k$ has a cyclic vector then since $A^k$ commutes with $A$, $A$ would have been a polynomial in $A^k$. But it may not be the case that $A^k$ has a cyclic vector. Can you kindly give me one more hint to prove your hint ?
– Panja
yesterday
5
My favorite way of proving the hint is to recall Newton's binomial formula $left(1+xright)^r = sumlimits_{i=0}^{infty} dbinom{r}{i} x^i$ for all $r in mathbb{Q}$. This is, per se, an equality between formal power series in the indeterminate $x$ over $mathbb{Q}$. But since the matrix $A^k - I$ is nilpotent (make sure you understand why!), we can substitute $A^k - I$ for $x$ in this formula (make sure you understand why!), and apply it to $r = 1/k$, thus obtaining $A$ on the left hand side (make sure you understand how!), and on the right hand side a polynomial in $A^k - I$.
– darij grinberg
19 hours ago
1
I never knew you could put non-integers into binomial coefficients. Thanks for sharing!
– user25959
14 hours ago
1
@Song There is no need to do this. Since $X$ is nilpotent, the infinite sum is actually a finite sum (or more specifically, a sum of no more than $n$ terms).
– user1551
12 hours ago
1
@Song $A-I$ is nilpotent. Therefore $1$ is an eigenvalue fo $A$ of multiplicity $n$. Hence $1$ is an eigenvalue of $A^k$ of multiplicity $n$, meaning that $A^k$ is nilpotent. You may also triangularise $A$ first to see this.
– user1551
12 hours ago
|
show 4 more comments
1
I was trying to prove your hint. I was thinking that if I can show that $A^k$ has a cyclic vector then since $A^k$ commutes with $A$, $A$ would have been a polynomial in $A^k$. But it may not be the case that $A^k$ has a cyclic vector. Can you kindly give me one more hint to prove your hint ?
– Panja
yesterday
5
My favorite way of proving the hint is to recall Newton's binomial formula $left(1+xright)^r = sumlimits_{i=0}^{infty} dbinom{r}{i} x^i$ for all $r in mathbb{Q}$. This is, per se, an equality between formal power series in the indeterminate $x$ over $mathbb{Q}$. But since the matrix $A^k - I$ is nilpotent (make sure you understand why!), we can substitute $A^k - I$ for $x$ in this formula (make sure you understand why!), and apply it to $r = 1/k$, thus obtaining $A$ on the left hand side (make sure you understand how!), and on the right hand side a polynomial in $A^k - I$.
– darij grinberg
19 hours ago
1
I never knew you could put non-integers into binomial coefficients. Thanks for sharing!
– user25959
14 hours ago
1
@Song There is no need to do this. Since $X$ is nilpotent, the infinite sum is actually a finite sum (or more specifically, a sum of no more than $n$ terms).
– user1551
12 hours ago
1
@Song $A-I$ is nilpotent. Therefore $1$ is an eigenvalue fo $A$ of multiplicity $n$. Hence $1$ is an eigenvalue of $A^k$ of multiplicity $n$, meaning that $A^k$ is nilpotent. You may also triangularise $A$ first to see this.
– user1551
12 hours ago
1
1
I was trying to prove your hint. I was thinking that if I can show that $A^k$ has a cyclic vector then since $A^k$ commutes with $A$, $A$ would have been a polynomial in $A^k$. But it may not be the case that $A^k$ has a cyclic vector. Can you kindly give me one more hint to prove your hint ?
– Panja
yesterday
I was trying to prove your hint. I was thinking that if I can show that $A^k$ has a cyclic vector then since $A^k$ commutes with $A$, $A$ would have been a polynomial in $A^k$. But it may not be the case that $A^k$ has a cyclic vector. Can you kindly give me one more hint to prove your hint ?
– Panja
yesterday
5
5
My favorite way of proving the hint is to recall Newton's binomial formula $left(1+xright)^r = sumlimits_{i=0}^{infty} dbinom{r}{i} x^i$ for all $r in mathbb{Q}$. This is, per se, an equality between formal power series in the indeterminate $x$ over $mathbb{Q}$. But since the matrix $A^k - I$ is nilpotent (make sure you understand why!), we can substitute $A^k - I$ for $x$ in this formula (make sure you understand why!), and apply it to $r = 1/k$, thus obtaining $A$ on the left hand side (make sure you understand how!), and on the right hand side a polynomial in $A^k - I$.
– darij grinberg
19 hours ago
My favorite way of proving the hint is to recall Newton's binomial formula $left(1+xright)^r = sumlimits_{i=0}^{infty} dbinom{r}{i} x^i$ for all $r in mathbb{Q}$. This is, per se, an equality between formal power series in the indeterminate $x$ over $mathbb{Q}$. But since the matrix $A^k - I$ is nilpotent (make sure you understand why!), we can substitute $A^k - I$ for $x$ in this formula (make sure you understand why!), and apply it to $r = 1/k$, thus obtaining $A$ on the left hand side (make sure you understand how!), and on the right hand side a polynomial in $A^k - I$.
– darij grinberg
19 hours ago
1
1
I never knew you could put non-integers into binomial coefficients. Thanks for sharing!
– user25959
14 hours ago
I never knew you could put non-integers into binomial coefficients. Thanks for sharing!
– user25959
14 hours ago
1
1
@Song There is no need to do this. Since $X$ is nilpotent, the infinite sum is actually a finite sum (or more specifically, a sum of no more than $n$ terms).
– user1551
12 hours ago
@Song There is no need to do this. Since $X$ is nilpotent, the infinite sum is actually a finite sum (or more specifically, a sum of no more than $n$ terms).
– user1551
12 hours ago
1
1
@Song $A-I$ is nilpotent. Therefore $1$ is an eigenvalue fo $A$ of multiplicity $n$. Hence $1$ is an eigenvalue of $A^k$ of multiplicity $n$, meaning that $A^k$ is nilpotent. You may also triangularise $A$ first to see this.
– user1551
12 hours ago
@Song $A-I$ is nilpotent. Therefore $1$ is an eigenvalue fo $A$ of multiplicity $n$. Hence $1$ is an eigenvalue of $A^k$ of multiplicity $n$, meaning that $A^k$ is nilpotent. You may also triangularise $A$ first to see this.
– user1551
12 hours ago
|
show 4 more comments
up vote
0
down vote
If $A$ has only $1$ as eigenvalue then it is invertible (and of the form $I+N$ where $N$ is nilpotent with $N^n=0$).
$A $ is satisfying also any equation $(A^i-I)^n=0$ so we can write
$(A-I)^n=0, (A^2-I)^n=0, dots (A^k-I)^n=0$.
Also we can write ${A^k} B(A^k)^{-1}=B$.
In similar fashion
${A^k} B(A^k)^{-1}=({A^k})^{-1} B(A^k)$
${A^{2k}} B = B(A^{2k})$, etc..
For any natural $m$:
${A^{mk}} B = B(A^{mk})$
If powers $(A^k)^m$ could be a basis for expression of $A$ then commutativity would follow...
answered yesterday
Widawensen
4,34921444
add a comment |
up vote
0
down vote
If $A$ has only $1$ as eigenvalue then it is invertible (and of the form $I+N$ where $N$ is nilpotent with $N^n=0$).
$A $ is satisfying also any equation $(A^i-I)^n=0$ so we can write
$(A-I)^n=0, (A^2-I)^n=0, dots (A^k-I)^n=0$.
Also we can write ${A^k} B(A^k)^{-1}=B$.
In similar fashion
${A^k} B(A^k)^{-1}=({A^k})^{-1} B(A^k)$
${A^{2k}} B = B(A^{2k})$, etc..
For any natural $m$:
${A^{mk}} B = B(A^{mk})$
If powers $(A^k)^m$ could be a basis for expression of $A$ then commutativity would follow...
answered yesterday
Widawensen
4,34921444
add a comment |
up vote
0
down vote
up vote
0
down vote
If $A$ has only $1$ as eigenvalue then it is invertible (and of the form $I+N$ where $N$ is nilpotent with $N^n=0$).
$A $ is satisfying also any equation $(A^i-I)^n=0$ so we can write
$(A-I)^n=0, (A^2-I)^n=0, dots (A^k-I)^n=0$.
Also we can write ${A^k} B(A^k)^{-1}=B$.
In similar fashion
${A^k} B(A^k)^{-1}=({A^k})^{-1} B(A^k)$
${A^{2k}} B = B(A^{2k})$, etc..
For any natural $m$:
${A^{mk}} B = B(A^{mk})$
If powers $(A^k)^m$ could be a basis for expression of $A$ then commutativity would follow...
answered yesterday
Widawensen
4,34921444
If $A$ has only $1$ as eigenvalue then it is invertible (and of the form $I+N$ where $N$ is nilpotent with $N^n=0$).
$A $ is satisfying also any equation $(A^i-I)^n=0$ so we can write
$(A-I)^n=0, (A^2-I)^n=0, dots (A^k-I)^n=0$.
Also we can write ${A^k} B(A^k)^{-1}=B$.
In similar fashion
${A^k} B(A^k)^{-1}=({A^k})^{-1} B(A^k)$
${A^{2k}} B = B(A^{2k})$, etc..
For any natural $m$:
${A^{mk}} B = B(A^{mk})$
If powers $(A^k)^m$ could be a basis for expression of $A$ then commutativity would follow...
answered yesterday
Widawensen
4,34921444
edited 23 hours ago
answered yesterday
Widawensen
4,34921444
answered yesterday
Widawensen
4,34921444
answered yesterday
Widawensen
4,34921444
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Note that $mathbb{N}$ has to mean $left{1,2,3,ldotsright}$ for this to be correct.
– darij grinberg
19 hours ago