Mixing
If $(A+B)^n$ is binomial for some $n$, does that imply $AB = BA$?
$begingroup$
We say that a matrix power $(A+B)^n$ is binomial iff it satisfies the matrix equality $$(A+B)^n = sum limits_{j,=,0}^n binom{n}{j}A^jB^{n-j}.$$ If two matrices have a binomial power for some $n$, does that imply that $AB = BA$?
My approach was the most obvious (I think): I tried to find a non-commutative expression for $(A+B)^n$ to finally get all those ugly permutations with the same degree equal to something that adds up to the term with the respective degree on the other side, but they could "chaotically" add up to that without being commutative, so it's not interesting. I also tried to use a "inductive-like" thinking to get the simplest case of commutativity, i.e., the case where $n=2$, so, by the Euclidean algorithm, $n = 2q + r$. I didn't manage to advance much, though.
linear-algebra matrices algebra-precalculus binomial-theorem
edited Jan 30 at 15:01
J. W. Tanner
4,4411320
asked Jan 30 at 13:51
Lucas HenriqueLucas Henrique
1,031414
$endgroup$
add a comment |
$begingroup$
We say that a matrix power $(A+B)^n$ is binomial iff it satisfies the matrix equality $$(A+B)^n = sum limits_{j,=,0}^n binom{n}{j}A^jB^{n-j}.$$ If two matrices have a binomial power for some $n$, does that imply that $AB = BA$?
My approach was the most obvious (I think): I tried to find a non-commutative expression for $(A+B)^n$ to finally get all those ugly permutations with the same degree equal to something that adds up to the term with the respective degree on the other side, but they could "chaotically" add up to that without being commutative, so it's not interesting. I also tried to use a "inductive-like" thinking to get the simplest case of commutativity, i.e., the case where $n=2$, so, by the Euclidean algorithm, $n = 2q + r$. I didn't manage to advance much, though.
linear-algebra matrices algebra-precalculus binomial-theorem
edited Jan 30 at 15:01
J. W. Tanner
4,4411320
asked Jan 30 at 13:51
Lucas HenriqueLucas Henrique
1,031414
$endgroup$
2
$begingroup$
What if this "some $n$" is $0$ or $1$?
$endgroup$
– user587192
Jan 30 at 14:20
add a comment |
$begingroup$
We say that a matrix power $(A+B)^n$ is binomial iff it satisfies the matrix equality $$(A+B)^n = sum limits_{j,=,0}^n binom{n}{j}A^jB^{n-j}.$$ If two matrices have a binomial power for some $n$, does that imply that $AB = BA$?
My approach was the most obvious (I think): I tried to find a non-commutative expression for $(A+B)^n$ to finally get all those ugly permutations with the same degree equal to something that adds up to the term with the respective degree on the other side, but they could "chaotically" add up to that without being commutative, so it's not interesting. I also tried to use a "inductive-like" thinking to get the simplest case of commutativity, i.e., the case where $n=2$, so, by the Euclidean algorithm, $n = 2q + r$. I didn't manage to advance much, though.
linear-algebra matrices algebra-precalculus binomial-theorem
edited Jan 30 at 15:01
J. W. Tanner
4,4411320
asked Jan 30 at 13:51
Lucas HenriqueLucas Henrique
1,031414
$endgroup$
We say that a matrix power $(A+B)^n$ is binomial iff it satisfies the matrix equality $$(A+B)^n = sum limits_{j,=,0}^n binom{n}{j}A^jB^{n-j}.$$ If two matrices have a binomial power for some $n$, does that imply that $AB = BA$?
My approach was the most obvious (I think): I tried to find a non-commutative expression for $(A+B)^n$ to finally get all those ugly permutations with the same degree equal to something that adds up to the term with the respective degree on the other side, but they could "chaotically" add up to that without being commutative, so it's not interesting. I also tried to use a "inductive-like" thinking to get the simplest case of commutativity, i.e., the case where $n=2$, so, by the Euclidean algorithm, $n = 2q + r$. I didn't manage to advance much, though.
linear-algebra matrices algebra-precalculus binomial-theorem
linear-algebra matrices algebra-precalculus binomial-theorem
edited Jan 30 at 15:01
J. W. Tanner
4,4411320
asked Jan 30 at 13:51
Lucas HenriqueLucas Henrique
1,031414
edited Jan 30 at 15:01
J. W. Tanner
4,4411320
asked Jan 30 at 13:51
Lucas HenriqueLucas Henrique
1,031414
edited Jan 30 at 15:01
J. W. Tanner
4,4411320
edited Jan 30 at 15:01
J. W. Tanner
4,4411320
edited Jan 30 at 15:01
J. W. Tanner
4,4411320
4,4411320
asked Jan 30 at 13:51
Lucas HenriqueLucas Henrique
1,031414
asked Jan 30 at 13:51
Lucas HenriqueLucas Henrique
1,031414
asked Jan 30 at 13:51
Lucas HenriqueLucas Henrique
1,031414
1,031414
2
$begingroup$
What if this "some $n$" is $0$ or $1$?
$endgroup$
– user587192
Jan 30 at 14:20
add a comment |
2
$begingroup$
What if this "some $n$" is $0$ or $1$?
$endgroup$
– user587192
Jan 30 at 14:20
2
2
$begingroup$
What if this "some $n$" is $0$ or $1$?
$endgroup$
– user587192
Jan 30 at 14:20
$begingroup$
What if this "some $n$" is $0$ or $1$?
$endgroup$
– user587192
Jan 30 at 14:20
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
When $n=2$, the answer is obviously yes.
When $nge3$, the answer is no in general. Here is a counterexample that works over any field. Let
$$
A=pmatrix{0&1&0\ 0&0&0\ 0&0&0}oplus0
text{ and } B=pmatrix{0&0&0\ 0&0&1\ 0&0&0}oplus0.
$$
Then $AB=E_{13}ne0=BA$, but $A+B$ is nilpotent and all trigrams generated by $A$ and $B$ (i.e. $AAA,AAB,ABA,BAA,ABB,BAB,BBA$ and $BBB$) are zero. Thus $(A+B)^n=sum_{k=0}^nbinom{n}{k}A^kB^{n-k}=0$ when $nge3$.
answered Feb 1 at 16:20
user1551user1551
73.9k566129
$endgroup$
$begingroup$
What does the "$oplus$" stand for?
$endgroup$
– Lucas Henrique
Feb 7 at 22:04
$begingroup$
@LucasHenrique It's the direct sum of two matrices. E.g. $Aoplus B=pmatrix{A&0\ 0&B}$. This is a standard notation in English linear algebra textbooks.
$endgroup$
– user1551
Feb 8 at 10:33
add a comment |
$begingroup$
BIG EDITION: Added the general case, and expanded the computation of the solution for $n=2$
The answer is yes for $n=2$ and no in general.
If $n=3$, you can rearrange
$$A^3+3A^2B+3AB^2+B^3=(A+B)^3=A^3+A^2B+ABA+BA^2+B^2A+BAB+AB^2$$ to get
$$(2A+B)[A,B]+[A,B](A+2B)=0 ;;;; (1),$$
where $[A,B]:=AB-BA$, by taking into account that $[A^2,B]=A[A,B]+[A,B]A$.
Observe that equation (1), via $B':=2A+B$, $A':=A+2B$, since $[A,B']=[A,2A+B]=[A,B]$, is equivalent to
$$B'[A,B']+[A,B']A'=0,$$
and since $[A',B']=[A+2B,2A+B]=[A,B]+4[B,A]=[A,B]-4[A,B]=-3[A,B]=-3[A,B']$, by multiplying by $-3$ we get the equivalent equation
$$B[A,B]+[A,B]A=0,$$
(I have changed $A',B'$ to $A,B$ by abuse of notation) which, after swap of $A,B$ and a change of sign amounts to
$$A[A,B]+[A,B]B=0 ;;;; (2).$$
Equation (2) has solutions with $[A,B]neq0$ already for $2×2$ matrices. To see this, take into account that $[$M$_2(K),$M$_2(K)]$ is the subspace of trace $0$ matrices. We can find $A,B$ such that
$$[A,B]=begin{pmatrix}0 & c \ 0 & 0end{pmatrix}neq0,$$
and in fact we can do so in such a way that $A[A,B]=0=[A,B]B$; it is enough to pick
$$A=begin{pmatrix}0 & 0 \ 0 & 1end{pmatrix}, B=begin{pmatrix}0 & 1 \ 0 & 0end{pmatrix},$$
which give $[A,B]=begin{pmatrix}0 & -1 \ 0 & 0end{pmatrix}.$
Going backwards we can find now an interesting solution for (1):
$$A=begin{pmatrix} 0 & 0 \ 2 & -1end{pmatrix}, B=begin{pmatrix} 0 & 0 \ -1& 2end{pmatrix}, [A,B]=begin{pmatrix} 0 & 0 \ -3& 0end{pmatrix}.$$
Observe that if $k>2$ then M$_k(K)$ contains M$_2(K)$ as a subring.
Now, if $ngeq4$:
Consider the free (noncommutative) algebra $Klangle X,Yrangle$ generated by two elements $X,Y$. Since when $A,B$ commute we have $(A+B)^n=sum_{i=0}^nbinom{n}i A^iB^{n-i}$ in any $K$-algebra, this means that the element $$p_n(X,Y)=(X+Y)^n-sum_{i=0}^nbinom{n}i X^iY^{n-i}$$
belongs to the ideal generated by $[X,Y]$ in $Klangle X,Yrangle$. Therefore for each $n$ we can write
$$p_n(X,Y)=a(X,Y)[X,Y]+[X,Y]b(X,Y)+sum_i c_i(X,Y)[X,Y]d_i(X,Y),$$
where $a,b,c_i,d_i$ are (noncommutative) polynomials in $X,Y$.
E.g., if I did not get it wrong, for $n=4$ we have
$$p_4(A,B)=(3A^2+2AB+B^2)[A,B]+[A,B](4B^2+BA+A^2)+A[A,B](A+2B)+(A+2B)[A,B]B.$$
I have found this by pairing monomials of the same degrees in $A$ and $B$. For example: $$A^3B-A^2BA=A^2[A,B], A^3B-ABA^2=A[A^2,B]=A^2[A,B]+A[A,B]A,$$
and the difficult one,
$$A^2B^2-BABA=A^2B^2-B[A,B]A-B^2A^2=[A^2,B^2]-B[A,B]A=AB[A,B]+A[A,B]B+[A,B]BA.$$
Now observe that since $p_n$ is homogeneous and $ngeq3$, by a degree argument we must have $a,b,c_i,d_i$ either $0$ or of degree at least $1$, i.e., we cannot have a term of the form $alpha[X,Y]$ with $0neqalphain K$. This implies that touching the $[X,Y]$ factor of each term there is either an $X$ or an $Y$ either at its left or at its right.
Therefore, to get $p_n(A,B)=0$ it is enough to have
$$A[A,B]=[A,B]A=B[A,B]=[A,B]B=0.$$
This can be done with $[A,B]neq0$ for matrices of size $3$ or more (just pick sparse matrices with the nonzero elements not in the diagonal, in order to get $[A,B]$ with just one nonzero element and $0$ trace, as in the explanation above for $2times 2$ matrices).
(Note that the general argument of this section is also valid for $n=3$; the previous section shows that in this particular case we can easily find examples also of size $2$.)
answered Jan 30 at 14:33
Jose BroxJose Brox
3,44711129
$endgroup$
$begingroup$
Thank you for your answer. how did you deduce that the answer is no in general? In fact there's evidence but I'd be glad if you could construct such matrices.
$endgroup$
– Lucas Henrique
Jan 31 at 18:18
$begingroup$
@LucasHenrique Please, see my answer again. I have added a lot of information.
$endgroup$
– Jose Brox
Feb 1 at 13:32
add a comment |
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2 Answers
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2 Answers
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$begingroup$
When $n=2$, the answer is obviously yes.
When $nge3$, the answer is no in general. Here is a counterexample that works over any field. Let
$$
A=pmatrix{0&1&0\ 0&0&0\ 0&0&0}oplus0
text{ and } B=pmatrix{0&0&0\ 0&0&1\ 0&0&0}oplus0.
$$
Then $AB=E_{13}ne0=BA$, but $A+B$ is nilpotent and all trigrams generated by $A$ and $B$ (i.e. $AAA,AAB,ABA,BAA,ABB,BAB,BBA$ and $BBB$) are zero. Thus $(A+B)^n=sum_{k=0}^nbinom{n}{k}A^kB^{n-k}=0$ when $nge3$.
answered Feb 1 at 16:20
user1551user1551
73.9k566129
$endgroup$
$begingroup$
What does the "$oplus$" stand for?
$endgroup$
– Lucas Henrique
Feb 7 at 22:04
$begingroup$
@LucasHenrique It's the direct sum of two matrices. E.g. $Aoplus B=pmatrix{A&0\ 0&B}$. This is a standard notation in English linear algebra textbooks.
$endgroup$
– user1551
Feb 8 at 10:33
add a comment |
$begingroup$
When $n=2$, the answer is obviously yes.
When $nge3$, the answer is no in general. Here is a counterexample that works over any field. Let
$$
A=pmatrix{0&1&0\ 0&0&0\ 0&0&0}oplus0
text{ and } B=pmatrix{0&0&0\ 0&0&1\ 0&0&0}oplus0.
$$
Then $AB=E_{13}ne0=BA$, but $A+B$ is nilpotent and all trigrams generated by $A$ and $B$ (i.e. $AAA,AAB,ABA,BAA,ABB,BAB,BBA$ and $BBB$) are zero. Thus $(A+B)^n=sum_{k=0}^nbinom{n}{k}A^kB^{n-k}=0$ when $nge3$.
answered Feb 1 at 16:20
user1551user1551
73.9k566129
$endgroup$
$begingroup$
What does the "$oplus$" stand for?
$endgroup$
– Lucas Henrique
Feb 7 at 22:04
$begingroup$
@LucasHenrique It's the direct sum of two matrices. E.g. $Aoplus B=pmatrix{A&0\ 0&B}$. This is a standard notation in English linear algebra textbooks.
$endgroup$
– user1551
Feb 8 at 10:33
add a comment |
$begingroup$
When $n=2$, the answer is obviously yes.
When $nge3$, the answer is no in general. Here is a counterexample that works over any field. Let
$$
A=pmatrix{0&1&0\ 0&0&0\ 0&0&0}oplus0
text{ and } B=pmatrix{0&0&0\ 0&0&1\ 0&0&0}oplus0.
$$
Then $AB=E_{13}ne0=BA$, but $A+B$ is nilpotent and all trigrams generated by $A$ and $B$ (i.e. $AAA,AAB,ABA,BAA,ABB,BAB,BBA$ and $BBB$) are zero. Thus $(A+B)^n=sum_{k=0}^nbinom{n}{k}A^kB^{n-k}=0$ when $nge3$.
answered Feb 1 at 16:20
user1551user1551
73.9k566129
$endgroup$
When $n=2$, the answer is obviously yes.
When $nge3$, the answer is no in general. Here is a counterexample that works over any field. Let
$$
A=pmatrix{0&1&0\ 0&0&0\ 0&0&0}oplus0
text{ and } B=pmatrix{0&0&0\ 0&0&1\ 0&0&0}oplus0.
$$
Then $AB=E_{13}ne0=BA$, but $A+B$ is nilpotent and all trigrams generated by $A$ and $B$ (i.e. $AAA,AAB,ABA,BAA,ABB,BAB,BBA$ and $BBB$) are zero. Thus $(A+B)^n=sum_{k=0}^nbinom{n}{k}A^kB^{n-k}=0$ when $nge3$.
answered Feb 1 at 16:20
user1551user1551
73.9k566129
answered Feb 1 at 16:20
user1551user1551
73.9k566129
answered Feb 1 at 16:20
user1551user1551
73.9k566129
answered Feb 1 at 16:20
user1551user1551
73.9k566129
73.9k566129
$begingroup$
What does the "$oplus$" stand for?
$endgroup$
– Lucas Henrique
Feb 7 at 22:04
$begingroup$
@LucasHenrique It's the direct sum of two matrices. E.g. $Aoplus B=pmatrix{A&0\ 0&B}$. This is a standard notation in English linear algebra textbooks.
$endgroup$
– user1551
Feb 8 at 10:33
add a comment |
$begingroup$
What does the "$oplus$" stand for?
$endgroup$
– Lucas Henrique
Feb 7 at 22:04
$begingroup$
@LucasHenrique It's the direct sum of two matrices. E.g. $Aoplus B=pmatrix{A&0\ 0&B}$. This is a standard notation in English linear algebra textbooks.
$endgroup$
– user1551
Feb 8 at 10:33
$begingroup$
What does the "$oplus$" stand for?
$endgroup$
– Lucas Henrique
Feb 7 at 22:04
$begingroup$
What does the "$oplus$" stand for?
$endgroup$
– Lucas Henrique
Feb 7 at 22:04
$begingroup$
@LucasHenrique It's the direct sum of two matrices. E.g. $Aoplus B=pmatrix{A&0\ 0&B}$. This is a standard notation in English linear algebra textbooks.
$endgroup$
– user1551
Feb 8 at 10:33
$begingroup$
@LucasHenrique It's the direct sum of two matrices. E.g. $Aoplus B=pmatrix{A&0\ 0&B}$. This is a standard notation in English linear algebra textbooks.
$endgroup$
– user1551
Feb 8 at 10:33
add a comment |
$begingroup$
BIG EDITION: Added the general case, and expanded the computation of the solution for $n=2$
The answer is yes for $n=2$ and no in general.
If $n=3$, you can rearrange
$$A^3+3A^2B+3AB^2+B^3=(A+B)^3=A^3+A^2B+ABA+BA^2+B^2A+BAB+AB^2$$ to get
$$(2A+B)[A,B]+[A,B](A+2B)=0 ;;;; (1),$$
where $[A,B]:=AB-BA$, by taking into account that $[A^2,B]=A[A,B]+[A,B]A$.
Observe that equation (1), via $B':=2A+B$, $A':=A+2B$, since $[A,B']=[A,2A+B]=[A,B]$, is equivalent to
$$B'[A,B']+[A,B']A'=0,$$
and since $[A',B']=[A+2B,2A+B]=[A,B]+4[B,A]=[A,B]-4[A,B]=-3[A,B]=-3[A,B']$, by multiplying by $-3$ we get the equivalent equation
$$B[A,B]+[A,B]A=0,$$
(I have changed $A',B'$ to $A,B$ by abuse of notation) which, after swap of $A,B$ and a change of sign amounts to
$$A[A,B]+[A,B]B=0 ;;;; (2).$$
Equation (2) has solutions with $[A,B]neq0$ already for $2×2$ matrices. To see this, take into account that $[$M$_2(K),$M$_2(K)]$ is the subspace of trace $0$ matrices. We can find $A,B$ such that
$$[A,B]=begin{pmatrix}0 & c \ 0 & 0end{pmatrix}neq0,$$
and in fact we can do so in such a way that $A[A,B]=0=[A,B]B$; it is enough to pick
$$A=begin{pmatrix}0 & 0 \ 0 & 1end{pmatrix}, B=begin{pmatrix}0 & 1 \ 0 & 0end{pmatrix},$$
which give $[A,B]=begin{pmatrix}0 & -1 \ 0 & 0end{pmatrix}.$
Going backwards we can find now an interesting solution for (1):
$$A=begin{pmatrix} 0 & 0 \ 2 & -1end{pmatrix}, B=begin{pmatrix} 0 & 0 \ -1& 2end{pmatrix}, [A,B]=begin{pmatrix} 0 & 0 \ -3& 0end{pmatrix}.$$
Observe that if $k>2$ then M$_k(K)$ contains M$_2(K)$ as a subring.
Now, if $ngeq4$:
Consider the free (noncommutative) algebra $Klangle X,Yrangle$ generated by two elements $X,Y$. Since when $A,B$ commute we have $(A+B)^n=sum_{i=0}^nbinom{n}i A^iB^{n-i}$ in any $K$-algebra, this means that the element $$p_n(X,Y)=(X+Y)^n-sum_{i=0}^nbinom{n}i X^iY^{n-i}$$
belongs to the ideal generated by $[X,Y]$ in $Klangle X,Yrangle$. Therefore for each $n$ we can write
$$p_n(X,Y)=a(X,Y)[X,Y]+[X,Y]b(X,Y)+sum_i c_i(X,Y)[X,Y]d_i(X,Y),$$
where $a,b,c_i,d_i$ are (noncommutative) polynomials in $X,Y$.
E.g., if I did not get it wrong, for $n=4$ we have
$$p_4(A,B)=(3A^2+2AB+B^2)[A,B]+[A,B](4B^2+BA+A^2)+A[A,B](A+2B)+(A+2B)[A,B]B.$$
I have found this by pairing monomials of the same degrees in $A$ and $B$. For example: $$A^3B-A^2BA=A^2[A,B], A^3B-ABA^2=A[A^2,B]=A^2[A,B]+A[A,B]A,$$
and the difficult one,
$$A^2B^2-BABA=A^2B^2-B[A,B]A-B^2A^2=[A^2,B^2]-B[A,B]A=AB[A,B]+A[A,B]B+[A,B]BA.$$
Now observe that since $p_n$ is homogeneous and $ngeq3$, by a degree argument we must have $a,b,c_i,d_i$ either $0$ or of degree at least $1$, i.e., we cannot have a term of the form $alpha[X,Y]$ with $0neqalphain K$. This implies that touching the $[X,Y]$ factor of each term there is either an $X$ or an $Y$ either at its left or at its right.
Therefore, to get $p_n(A,B)=0$ it is enough to have
$$A[A,B]=[A,B]A=B[A,B]=[A,B]B=0.$$
This can be done with $[A,B]neq0$ for matrices of size $3$ or more (just pick sparse matrices with the nonzero elements not in the diagonal, in order to get $[A,B]$ with just one nonzero element and $0$ trace, as in the explanation above for $2times 2$ matrices).
(Note that the general argument of this section is also valid for $n=3$; the previous section shows that in this particular case we can easily find examples also of size $2$.)
answered Jan 30 at 14:33
Jose BroxJose Brox
3,44711129
$endgroup$
$begingroup$
Thank you for your answer. how did you deduce that the answer is no in general? In fact there's evidence but I'd be glad if you could construct such matrices.
$endgroup$
– Lucas Henrique
Jan 31 at 18:18
$begingroup$
@LucasHenrique Please, see my answer again. I have added a lot of information.
$endgroup$
– Jose Brox
Feb 1 at 13:32
add a comment |
$begingroup$
BIG EDITION: Added the general case, and expanded the computation of the solution for $n=2$
The answer is yes for $n=2$ and no in general.
If $n=3$, you can rearrange
$$A^3+3A^2B+3AB^2+B^3=(A+B)^3=A^3+A^2B+ABA+BA^2+B^2A+BAB+AB^2$$ to get
$$(2A+B)[A,B]+[A,B](A+2B)=0 ;;;; (1),$$
where $[A,B]:=AB-BA$, by taking into account that $[A^2,B]=A[A,B]+[A,B]A$.
Observe that equation (1), via $B':=2A+B$, $A':=A+2B$, since $[A,B']=[A,2A+B]=[A,B]$, is equivalent to
$$B'[A,B']+[A,B']A'=0,$$
and since $[A',B']=[A+2B,2A+B]=[A,B]+4[B,A]=[A,B]-4[A,B]=-3[A,B]=-3[A,B']$, by multiplying by $-3$ we get the equivalent equation
$$B[A,B]+[A,B]A=0,$$
(I have changed $A',B'$ to $A,B$ by abuse of notation) which, after swap of $A,B$ and a change of sign amounts to
$$A[A,B]+[A,B]B=0 ;;;; (2).$$
Equation (2) has solutions with $[A,B]neq0$ already for $2×2$ matrices. To see this, take into account that $[$M$_2(K),$M$_2(K)]$ is the subspace of trace $0$ matrices. We can find $A,B$ such that
$$[A,B]=begin{pmatrix}0 & c \ 0 & 0end{pmatrix}neq0,$$
and in fact we can do so in such a way that $A[A,B]=0=[A,B]B$; it is enough to pick
$$A=begin{pmatrix}0 & 0 \ 0 & 1end{pmatrix}, B=begin{pmatrix}0 & 1 \ 0 & 0end{pmatrix},$$
which give $[A,B]=begin{pmatrix}0 & -1 \ 0 & 0end{pmatrix}.$
Going backwards we can find now an interesting solution for (1):
$$A=begin{pmatrix} 0 & 0 \ 2 & -1end{pmatrix}, B=begin{pmatrix} 0 & 0 \ -1& 2end{pmatrix}, [A,B]=begin{pmatrix} 0 & 0 \ -3& 0end{pmatrix}.$$
Observe that if $k>2$ then M$_k(K)$ contains M$_2(K)$ as a subring.
Now, if $ngeq4$:
Consider the free (noncommutative) algebra $Klangle X,Yrangle$ generated by two elements $X,Y$. Since when $A,B$ commute we have $(A+B)^n=sum_{i=0}^nbinom{n}i A^iB^{n-i}$ in any $K$-algebra, this means that the element $$p_n(X,Y)=(X+Y)^n-sum_{i=0}^nbinom{n}i X^iY^{n-i}$$
belongs to the ideal generated by $[X,Y]$ in $Klangle X,Yrangle$. Therefore for each $n$ we can write
$$p_n(X,Y)=a(X,Y)[X,Y]+[X,Y]b(X,Y)+sum_i c_i(X,Y)[X,Y]d_i(X,Y),$$
where $a,b,c_i,d_i$ are (noncommutative) polynomials in $X,Y$.
E.g., if I did not get it wrong, for $n=4$ we have
$$p_4(A,B)=(3A^2+2AB+B^2)[A,B]+[A,B](4B^2+BA+A^2)+A[A,B](A+2B)+(A+2B)[A,B]B.$$
I have found this by pairing monomials of the same degrees in $A$ and $B$. For example: $$A^3B-A^2BA=A^2[A,B], A^3B-ABA^2=A[A^2,B]=A^2[A,B]+A[A,B]A,$$
and the difficult one,
$$A^2B^2-BABA=A^2B^2-B[A,B]A-B^2A^2=[A^2,B^2]-B[A,B]A=AB[A,B]+A[A,B]B+[A,B]BA.$$
Now observe that since $p_n$ is homogeneous and $ngeq3$, by a degree argument we must have $a,b,c_i,d_i$ either $0$ or of degree at least $1$, i.e., we cannot have a term of the form $alpha[X,Y]$ with $0neqalphain K$. This implies that touching the $[X,Y]$ factor of each term there is either an $X$ or an $Y$ either at its left or at its right.
Therefore, to get $p_n(A,B)=0$ it is enough to have
$$A[A,B]=[A,B]A=B[A,B]=[A,B]B=0.$$
This can be done with $[A,B]neq0$ for matrices of size $3$ or more (just pick sparse matrices with the nonzero elements not in the diagonal, in order to get $[A,B]$ with just one nonzero element and $0$ trace, as in the explanation above for $2times 2$ matrices).
(Note that the general argument of this section is also valid for $n=3$; the previous section shows that in this particular case we can easily find examples also of size $2$.)
answered Jan 30 at 14:33
Jose BroxJose Brox
3,44711129
$endgroup$
$begingroup$
Thank you for your answer. how did you deduce that the answer is no in general? In fact there's evidence but I'd be glad if you could construct such matrices.
$endgroup$
– Lucas Henrique
Jan 31 at 18:18
$begingroup$
@LucasHenrique Please, see my answer again. I have added a lot of information.
$endgroup$
– Jose Brox
Feb 1 at 13:32
add a comment |
$begingroup$
BIG EDITION: Added the general case, and expanded the computation of the solution for $n=2$
The answer is yes for $n=2$ and no in general.
If $n=3$, you can rearrange
$$A^3+3A^2B+3AB^2+B^3=(A+B)^3=A^3+A^2B+ABA+BA^2+B^2A+BAB+AB^2$$ to get
$$(2A+B)[A,B]+[A,B](A+2B)=0 ;;;; (1),$$
where $[A,B]:=AB-BA$, by taking into account that $[A^2,B]=A[A,B]+[A,B]A$.
Observe that equation (1), via $B':=2A+B$, $A':=A+2B$, since $[A,B']=[A,2A+B]=[A,B]$, is equivalent to
$$B'[A,B']+[A,B']A'=0,$$
and since $[A',B']=[A+2B,2A+B]=[A,B]+4[B,A]=[A,B]-4[A,B]=-3[A,B]=-3[A,B']$, by multiplying by $-3$ we get the equivalent equation
$$B[A,B]+[A,B]A=0,$$
(I have changed $A',B'$ to $A,B$ by abuse of notation) which, after swap of $A,B$ and a change of sign amounts to
$$A[A,B]+[A,B]B=0 ;;;; (2).$$
Equation (2) has solutions with $[A,B]neq0$ already for $2×2$ matrices. To see this, take into account that $[$M$_2(K),$M$_2(K)]$ is the subspace of trace $0$ matrices. We can find $A,B$ such that
$$[A,B]=begin{pmatrix}0 & c \ 0 & 0end{pmatrix}neq0,$$
and in fact we can do so in such a way that $A[A,B]=0=[A,B]B$; it is enough to pick
$$A=begin{pmatrix}0 & 0 \ 0 & 1end{pmatrix}, B=begin{pmatrix}0 & 1 \ 0 & 0end{pmatrix},$$
which give $[A,B]=begin{pmatrix}0 & -1 \ 0 & 0end{pmatrix}.$
Going backwards we can find now an interesting solution for (1):
$$A=begin{pmatrix} 0 & 0 \ 2 & -1end{pmatrix}, B=begin{pmatrix} 0 & 0 \ -1& 2end{pmatrix}, [A,B]=begin{pmatrix} 0 & 0 \ -3& 0end{pmatrix}.$$
Observe that if $k>2$ then M$_k(K)$ contains M$_2(K)$ as a subring.
Now, if $ngeq4$:
Consider the free (noncommutative) algebra $Klangle X,Yrangle$ generated by two elements $X,Y$. Since when $A,B$ commute we have $(A+B)^n=sum_{i=0}^nbinom{n}i A^iB^{n-i}$ in any $K$-algebra, this means that the element $$p_n(X,Y)=(X+Y)^n-sum_{i=0}^nbinom{n}i X^iY^{n-i}$$
belongs to the ideal generated by $[X,Y]$ in $Klangle X,Yrangle$. Therefore for each $n$ we can write
$$p_n(X,Y)=a(X,Y)[X,Y]+[X,Y]b(X,Y)+sum_i c_i(X,Y)[X,Y]d_i(X,Y),$$
where $a,b,c_i,d_i$ are (noncommutative) polynomials in $X,Y$.
E.g., if I did not get it wrong, for $n=4$ we have
$$p_4(A,B)=(3A^2+2AB+B^2)[A,B]+[A,B](4B^2+BA+A^2)+A[A,B](A+2B)+(A+2B)[A,B]B.$$
I have found this by pairing monomials of the same degrees in $A$ and $B$. For example: $$A^3B-A^2BA=A^2[A,B], A^3B-ABA^2=A[A^2,B]=A^2[A,B]+A[A,B]A,$$
and the difficult one,
$$A^2B^2-BABA=A^2B^2-B[A,B]A-B^2A^2=[A^2,B^2]-B[A,B]A=AB[A,B]+A[A,B]B+[A,B]BA.$$
Now observe that since $p_n$ is homogeneous and $ngeq3$, by a degree argument we must have $a,b,c_i,d_i$ either $0$ or of degree at least $1$, i.e., we cannot have a term of the form $alpha[X,Y]$ with $0neqalphain K$. This implies that touching the $[X,Y]$ factor of each term there is either an $X$ or an $Y$ either at its left or at its right.
Therefore, to get $p_n(A,B)=0$ it is enough to have
$$A[A,B]=[A,B]A=B[A,B]=[A,B]B=0.$$
This can be done with $[A,B]neq0$ for matrices of size $3$ or more (just pick sparse matrices with the nonzero elements not in the diagonal, in order to get $[A,B]$ with just one nonzero element and $0$ trace, as in the explanation above for $2times 2$ matrices).
(Note that the general argument of this section is also valid for $n=3$; the previous section shows that in this particular case we can easily find examples also of size $2$.)
answered Jan 30 at 14:33
Jose BroxJose Brox
3,44711129
$endgroup$
BIG EDITION: Added the general case, and expanded the computation of the solution for $n=2$
The answer is yes for $n=2$ and no in general.
If $n=3$, you can rearrange
$$A^3+3A^2B+3AB^2+B^3=(A+B)^3=A^3+A^2B+ABA+BA^2+B^2A+BAB+AB^2$$ to get
$$(2A+B)[A,B]+[A,B](A+2B)=0 ;;;; (1),$$
where $[A,B]:=AB-BA$, by taking into account that $[A^2,B]=A[A,B]+[A,B]A$.
Observe that equation (1), via $B':=2A+B$, $A':=A+2B$, since $[A,B']=[A,2A+B]=[A,B]$, is equivalent to
$$B'[A,B']+[A,B']A'=0,$$
and since $[A',B']=[A+2B,2A+B]=[A,B]+4[B,A]=[A,B]-4[A,B]=-3[A,B]=-3[A,B']$, by multiplying by $-3$ we get the equivalent equation
$$B[A,B]+[A,B]A=0,$$
(I have changed $A',B'$ to $A,B$ by abuse of notation) which, after swap of $A,B$ and a change of sign amounts to
$$A[A,B]+[A,B]B=0 ;;;; (2).$$
Equation (2) has solutions with $[A,B]neq0$ already for $2×2$ matrices. To see this, take into account that $[$M$_2(K),$M$_2(K)]$ is the subspace of trace $0$ matrices. We can find $A,B$ such that
$$[A,B]=begin{pmatrix}0 & c \ 0 & 0end{pmatrix}neq0,$$
and in fact we can do so in such a way that $A[A,B]=0=[A,B]B$; it is enough to pick
$$A=begin{pmatrix}0 & 0 \ 0 & 1end{pmatrix}, B=begin{pmatrix}0 & 1 \ 0 & 0end{pmatrix},$$
which give $[A,B]=begin{pmatrix}0 & -1 \ 0 & 0end{pmatrix}.$
Going backwards we can find now an interesting solution for (1):
$$A=begin{pmatrix} 0 & 0 \ 2 & -1end{pmatrix}, B=begin{pmatrix} 0 & 0 \ -1& 2end{pmatrix}, [A,B]=begin{pmatrix} 0 & 0 \ -3& 0end{pmatrix}.$$
Observe that if $k>2$ then M$_k(K)$ contains M$_2(K)$ as a subring.
Now, if $ngeq4$:
Consider the free (noncommutative) algebra $Klangle X,Yrangle$ generated by two elements $X,Y$. Since when $A,B$ commute we have $(A+B)^n=sum_{i=0}^nbinom{n}i A^iB^{n-i}$ in any $K$-algebra, this means that the element $$p_n(X,Y)=(X+Y)^n-sum_{i=0}^nbinom{n}i X^iY^{n-i}$$
belongs to the ideal generated by $[X,Y]$ in $Klangle X,Yrangle$. Therefore for each $n$ we can write
$$p_n(X,Y)=a(X,Y)[X,Y]+[X,Y]b(X,Y)+sum_i c_i(X,Y)[X,Y]d_i(X,Y),$$
where $a,b,c_i,d_i$ are (noncommutative) polynomials in $X,Y$.
E.g., if I did not get it wrong, for $n=4$ we have
$$p_4(A,B)=(3A^2+2AB+B^2)[A,B]+[A,B](4B^2+BA+A^2)+A[A,B](A+2B)+(A+2B)[A,B]B.$$
I have found this by pairing monomials of the same degrees in $A$ and $B$. For example: $$A^3B-A^2BA=A^2[A,B], A^3B-ABA^2=A[A^2,B]=A^2[A,B]+A[A,B]A,$$
and the difficult one,
$$A^2B^2-BABA=A^2B^2-B[A,B]A-B^2A^2=[A^2,B^2]-B[A,B]A=AB[A,B]+A[A,B]B+[A,B]BA.$$
Now observe that since $p_n$ is homogeneous and $ngeq3$, by a degree argument we must have $a,b,c_i,d_i$ either $0$ or of degree at least $1$, i.e., we cannot have a term of the form $alpha[X,Y]$ with $0neqalphain K$. This implies that touching the $[X,Y]$ factor of each term there is either an $X$ or an $Y$ either at its left or at its right.
Therefore, to get $p_n(A,B)=0$ it is enough to have
$$A[A,B]=[A,B]A=B[A,B]=[A,B]B=0.$$
This can be done with $[A,B]neq0$ for matrices of size $3$ or more (just pick sparse matrices with the nonzero elements not in the diagonal, in order to get $[A,B]$ with just one nonzero element and $0$ trace, as in the explanation above for $2times 2$ matrices).
(Note that the general argument of this section is also valid for $n=3$; the previous section shows that in this particular case we can easily find examples also of size $2$.)
answered Jan 30 at 14:33
Jose BroxJose Brox
3,44711129
edited Feb 1 at 13:37
answered Jan 30 at 14:33
Jose BroxJose Brox
3,44711129
answered Jan 30 at 14:33
Jose BroxJose Brox
3,44711129
answered Jan 30 at 14:33
Jose BroxJose Brox
3,44711129
3,44711129
$begingroup$
Thank you for your answer. how did you deduce that the answer is no in general? In fact there's evidence but I'd be glad if you could construct such matrices.
$endgroup$
– Lucas Henrique
Jan 31 at 18:18
$begingroup$
@LucasHenrique Please, see my answer again. I have added a lot of information.
$endgroup$
– Jose Brox
Feb 1 at 13:32
add a comment |
$begingroup$
Thank you for your answer. how did you deduce that the answer is no in general? In fact there's evidence but I'd be glad if you could construct such matrices.
$endgroup$
– Lucas Henrique
Jan 31 at 18:18
$begingroup$
@LucasHenrique Please, see my answer again. I have added a lot of information.
$endgroup$
– Jose Brox
Feb 1 at 13:32
$begingroup$
Thank you for your answer. how did you deduce that the answer is no in general? In fact there's evidence but I'd be glad if you could construct such matrices.
$endgroup$
– Lucas Henrique
Jan 31 at 18:18
$begingroup$
Thank you for your answer. how did you deduce that the answer is no in general? In fact there's evidence but I'd be glad if you could construct such matrices.
$endgroup$
– Lucas Henrique
Jan 31 at 18:18
$begingroup$
@LucasHenrique Please, see my answer again. I have added a lot of information.
$endgroup$
– Jose Brox
Feb 1 at 13:32
$begingroup$
@LucasHenrique Please, see my answer again. I have added a lot of information.
$endgroup$
– Jose Brox
Feb 1 at 13:32
add a comment |
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$begingroup$
What if this "some $n$" is $0$ or $1$?
$endgroup$
– user587192
Jan 30 at 14:20