Mixing
Number of solutions $X$ to $AX=XB$ in $mathbb F_2$
$begingroup$
It is a well-known theorem that in an arbitrary field $F$, if $A$ is an $mtimes m$ square matrix and $B$ is an $ntimes n$ square matrix, then there is a unique $mtimes n$ solution $X$ to the equation
$$AX=XB$$
if and only if $A$ and $B$ share no eigenvalues.
My question is this: what can be said about the number of solutions $Xin mathbb F_2$ to the above equation? Is the number of nonzero solutions equal to the number of common eigenvalues? If so, how can one prove this?
linear-algebra matrices matrix-equations
asked Jan 14 at 22:13
FrpzzdFrpzzd
23k841109
$endgroup$
|
show 3 more comments
$begingroup$
It is a well-known theorem that in an arbitrary field $F$, if $A$ is an $mtimes m$ square matrix and $B$ is an $ntimes n$ square matrix, then there is a unique $mtimes n$ solution $X$ to the equation
$$AX=XB$$
if and only if $A$ and $B$ share no eigenvalues.
My question is this: what can be said about the number of solutions $Xin mathbb F_2$ to the above equation? Is the number of nonzero solutions equal to the number of common eigenvalues? If so, how can one prove this?
linear-algebra matrices matrix-equations
asked Jan 14 at 22:13
FrpzzdFrpzzd
23k841109
$endgroup$
1
$begingroup$
You probably mean the number of linearly independent solutions. Otherwise, it's obvious that the answer is false. Because if $X$ is a solution, so is $lambda X$. Right? Does this theorem have a name? I'd like to see the proof.
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– stressed out
Jan 14 at 22:16
1
$begingroup$
You're right; I am actually working mostly in $mathbb F_2$ so that specification isn't important, and I should mention that in my question. As for the theorem: I don't know if it has a name, but the equation is called a "Sylvester Equation" and the Wikipedia page has a proof of said theorem.
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– Frpzzd
Jan 14 at 22:18
1
$begingroup$
@stressedout It is Sylvester equation.
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– A.Γ.
Jan 14 at 22:18
2
$begingroup$
@stressedout The unique solution, when it is unique, is $X = 0$.
$endgroup$
– Robert Israel
Jan 14 at 22:20
2
$begingroup$
Try Cecioni-Frobenius theorem..
$endgroup$
– user
Jan 14 at 22:31
|
show 3 more comments
$begingroup$
It is a well-known theorem that in an arbitrary field $F$, if $A$ is an $mtimes m$ square matrix and $B$ is an $ntimes n$ square matrix, then there is a unique $mtimes n$ solution $X$ to the equation
$$AX=XB$$
if and only if $A$ and $B$ share no eigenvalues.
My question is this: what can be said about the number of solutions $Xin mathbb F_2$ to the above equation? Is the number of nonzero solutions equal to the number of common eigenvalues? If so, how can one prove this?
linear-algebra matrices matrix-equations
asked Jan 14 at 22:13
FrpzzdFrpzzd
23k841109
$endgroup$
It is a well-known theorem that in an arbitrary field $F$, if $A$ is an $mtimes m$ square matrix and $B$ is an $ntimes n$ square matrix, then there is a unique $mtimes n$ solution $X$ to the equation
$$AX=XB$$
if and only if $A$ and $B$ share no eigenvalues.
My question is this: what can be said about the number of solutions $Xin mathbb F_2$ to the above equation? Is the number of nonzero solutions equal to the number of common eigenvalues? If so, how can one prove this?
linear-algebra matrices matrix-equations
linear-algebra matrices matrix-equations
asked Jan 14 at 22:13
FrpzzdFrpzzd
23k841109
asked Jan 14 at 22:13
FrpzzdFrpzzd
23k841109
edited Jan 14 at 22:19
Frpzzd
asked Jan 14 at 22:13
FrpzzdFrpzzd
23k841109
asked Jan 14 at 22:13
FrpzzdFrpzzd
23k841109
asked Jan 14 at 22:13
FrpzzdFrpzzd
23k841109
23k841109
1
$begingroup$
You probably mean the number of linearly independent solutions. Otherwise, it's obvious that the answer is false. Because if $X$ is a solution, so is $lambda X$. Right? Does this theorem have a name? I'd like to see the proof.
$endgroup$
– stressed out
Jan 14 at 22:16
1
$begingroup$
You're right; I am actually working mostly in $mathbb F_2$ so that specification isn't important, and I should mention that in my question. As for the theorem: I don't know if it has a name, but the equation is called a "Sylvester Equation" and the Wikipedia page has a proof of said theorem.
$endgroup$
– Frpzzd
Jan 14 at 22:18
1
$begingroup$
@stressedout It is Sylvester equation.
$endgroup$
– A.Γ.
Jan 14 at 22:18
2
$begingroup$
@stressedout The unique solution, when it is unique, is $X = 0$.
$endgroup$
– Robert Israel
Jan 14 at 22:20
2
$begingroup$
Try Cecioni-Frobenius theorem..
$endgroup$
– user
Jan 14 at 22:31
|
show 3 more comments
1
$begingroup$
You probably mean the number of linearly independent solutions. Otherwise, it's obvious that the answer is false. Because if $X$ is a solution, so is $lambda X$. Right? Does this theorem have a name? I'd like to see the proof.
$endgroup$
– stressed out
Jan 14 at 22:16
1
$begingroup$
You're right; I am actually working mostly in $mathbb F_2$ so that specification isn't important, and I should mention that in my question. As for the theorem: I don't know if it has a name, but the equation is called a "Sylvester Equation" and the Wikipedia page has a proof of said theorem.
$endgroup$
– Frpzzd
Jan 14 at 22:18
1
$begingroup$
@stressedout It is Sylvester equation.
$endgroup$
– A.Γ.
Jan 14 at 22:18
2
$begingroup$
@stressedout The unique solution, when it is unique, is $X = 0$.
$endgroup$
– Robert Israel
Jan 14 at 22:20
2
$begingroup$
Try Cecioni-Frobenius theorem..
$endgroup$
– user
Jan 14 at 22:31
1
1
$begingroup$
You probably mean the number of linearly independent solutions. Otherwise, it's obvious that the answer is false. Because if $X$ is a solution, so is $lambda X$. Right? Does this theorem have a name? I'd like to see the proof.
$endgroup$
– stressed out
Jan 14 at 22:16
$begingroup$
You probably mean the number of linearly independent solutions. Otherwise, it's obvious that the answer is false. Because if $X$ is a solution, so is $lambda X$. Right? Does this theorem have a name? I'd like to see the proof.
$endgroup$
– stressed out
Jan 14 at 22:16
1
1
$begingroup$
You're right; I am actually working mostly in $mathbb F_2$ so that specification isn't important, and I should mention that in my question. As for the theorem: I don't know if it has a name, but the equation is called a "Sylvester Equation" and the Wikipedia page has a proof of said theorem.
$endgroup$
– Frpzzd
Jan 14 at 22:18
$begingroup$
You're right; I am actually working mostly in $mathbb F_2$ so that specification isn't important, and I should mention that in my question. As for the theorem: I don't know if it has a name, but the equation is called a "Sylvester Equation" and the Wikipedia page has a proof of said theorem.
$endgroup$
– Frpzzd
Jan 14 at 22:18
1
1
$begingroup$
@stressedout It is Sylvester equation.
$endgroup$
– A.Γ.
Jan 14 at 22:18
$begingroup$
@stressedout It is Sylvester equation.
$endgroup$
– A.Γ.
Jan 14 at 22:18
2
2
$begingroup$
@stressedout The unique solution, when it is unique, is $X = 0$.
$endgroup$
– Robert Israel
Jan 14 at 22:20
$begingroup$
@stressedout The unique solution, when it is unique, is $X = 0$.
$endgroup$
– Robert Israel
Jan 14 at 22:20
2
2
$begingroup$
Try Cecioni-Frobenius theorem..
$endgroup$
– user
Jan 14 at 22:31
$begingroup$
Try Cecioni-Frobenius theorem..
$endgroup$
– user
Jan 14 at 22:31
|
show 3 more comments
5 Answers
5
active
oldest
votes
$begingroup$
$X$ maps each eigenvector of $B$ to either $0$ or an eigenvector of the same eigenvalue of $A$, since if $Bv = lambda v$, $AX v = X B v = lambda X v$.
Let's suppose for simplicity that $B$ is diagonalizable, with a basis $b_j$ of eigenvectors corresponding to eigenvalues $lambda_j$ that may or may not be eigenvalues of $A$.
For each $j$, we can choose $X b_j$ to be any member of the eigenspace of $A$ for $lambda_j$, and once we have done that we have specified a solution $X$. Thus the number of solutions for $X$ is the product over $j$ of the cardinalities of the eigenspaces of $A$ for $lambda_j$ (note that this is $1$ if $lambda_j$ is not an eigenvalue of $A$).
answered Jan 14 at 22:34
Robert IsraelRobert Israel
323k23213467
$endgroup$
add a comment |
$begingroup$
Your statement of well-known theorem requires saying that the eigenvalues are in the algebraic closure of the given field $F$. This is an easy version of Cecioni-Frobenius theorem suggested in the comments by 'user'.
Method 1 : Module Theory
Under your assumption for matrices $A$, $B$, $X$, define
$$
C_{A,B}={Xin M_{mtimes n}(F) mid AX - XB = 0 }.$$
Let $nu_{A,B}$ be the dimension of $C_{A,B}$ over $F$. The actual version of Cecioni-Frobenius theorem gives a formula
$$
nu_{A,B}=sum_p(deg p)sum_{i,j} min{lambda_{p,i}, mu_{p,j}}
$$
Here the first sum is over all irreducible polynomials $p$ which are common in the primary decompositions of $F[x]$-modules $M^A$ and $M^B$, and the indices $i, j$ of second double sum is from the partition $lambda_p=sum_i lambda_{p,i}$ that indicates the powers of $p$ in $p$-primary part of $M^A$, $mu_p=sum_j mu_{p,j}$ that of powers of $p$ in $p$-primary part of $M^B$.
Then the number of solutions to $AX=XB$ is $2^{nu_{A,B}}$.
Method 2 : Kronecker Product
Alternatively, there is a module-theory-free way. You will have an advantage of not having to solve for eigenvalues of $A$, $B$. But, a disadvantage is that there is almost no hope for writing down an explicit formula for $nu_{A,B}$. This is by Kronecker product, the method is mainly this: Solution of a Sylvester equation?
In our case, the equation $AX-XB=0$ can be written as
$$
(I_n otimes A - B^T otimes I_m) textrm{vec}(X)=0
$$
where $textrm{vec}(X)$ is a stack of columns of $X$ in order.
Then perform elementary row operations on the $mntimes mn$ matrix $I_n otimes A - B^T otimes I_m$. Find the number of free-variables, which is $nu_{A,B}$. Again the number of solutions to $AX=XB$ is $2^{nu_{A,B}}$.
answered Jan 15 at 2:20
i707107i707107
12.4k21547
$endgroup$
add a comment |
$begingroup$
To ask for the number of solutions of $f(X)=AX-XB=0$ is equivalent to ask for $dim(ker(f))$. Moreover, the required dimension is the same over the underlying field $K$ or over its algebraic closure $overline{K}$.
If we stack the considered matrices row by row, then $f=Aotimes I-Iotimes B^T$. Let $(lambda_i),(mu_i)$ be the spectra of $A,B$ in $overline{K}$. Since the functions $Xmapsto AX,Xmapsto XB$ commute, $spectrum(f)=(lambda_i-mu_j)_{i,j}$. Unfortunately, that does not give the required dimension; we can only say that $ker(f)not= 0$ iff $degree(gcd(chi_A,chi_B))geq 1$, where $chi_.$ is the characteristic polynomial.
The Cecioni-Frobenius (CF) theorem (1910) provides a solution if we can calculate the invariant factors of $A,B$.
$dim(ker(f))=nu_{A,B}=sum_{i,j}degree(gcd(d_i(A),d_j(B))$ where $d_i$ is the $i^{th}$ invariant factor.
I had heard about this theorem in 2009 and then I had completely forgotten about it. It was used by Byrnes-Gauger (1979) to show that
$A,B$ are similar iff $2nu_{A,B}=nu_{A,A}+nu_{B,B}$.
Note that, in great dimension, the invariant factor algorithm (the calculation of the Smith normal form of $xI-A$) is not effective. Fortunately, Byrnes-Gauger before proved (1977) this extraordinary result
$A,B$ are similar iff $det(xI-A)=det(xI-B),rank(Aotimes I-Iotimes A)=rank(Botimes I-Iotimes B)=rank(Aotimes I-Iotimes B)$.
Here is an example of using the CF theorem
let $J_k$ be the nilpotent Jordan block of dimension $k$, $A=diag(J_3,J_2),B=diag(J_4,J_1)$. Then the Smith forms are
$S(A)=diag(1,1,1,x^2,x^3),S(B)=diag(1,1,1,x,x^4)$.
Then $nu_{A,B}=deg(gcd(x^2,x))+deg(gcd(x^2,x^4))+deg(gcd(x^3,x))+deg(gcd(x^3,x^4))=7$.
answered Feb 5 at 9:50
loup blancloup blanc
23.3k21851
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$begingroup$
(+1) for mentioning Byrnes-Gauger.
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– i707107
Feb 8 at 16:33
add a comment |
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This is just a special case of Sylvester equation.
Let $A$ and $B$ be operators with spectra $sigma(A)$ and $sigma(B)$, respectively. If $sigma(A)$ and $sigma(B)$ are disjoint, then the equation
$AX-XB=Y$ has a unique solution $X$ for every $Y$. Moreover, the solution is given by $X=sum_{n=0}^infty A^{-n-1}YB$. Hence, for $Y=0$ unique solution is $X=0$ as Robert Israel claimed.
The proof of the theorem above can be found in Bhatia's Matrix Analysis.
answered Jan 14 at 22:52
Mustafa SaidMustafa Said
2,9911913
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add a comment |
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Suppose $X_1$ and $X_2$ are both solutions. Then $A(X_1+X_2)=AX_1+AX_2=X_1B+X_2B=(X_1+X_2)B$. Thus, $X_1+X_2$ is a solution. So the number of solutions can be $0$, $1$, or infinity. This is a common situation in linear algebra: you either have no solutions, a unique solution, or infinite solutions.
answered Jan 14 at 22:20
AcccumulationAcccumulation
6,9892618
$endgroup$
$begingroup$
@Accumulation Not true in $mathbb F_2$, since $X+X=0$. Thanks for the answer, though.
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– Frpzzd
Jan 14 at 22:21
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Nor in any finite field, since there are only finitely many $m times n$ matrices there.
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– Robert Israel
Jan 14 at 22:22
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I wrote my answer before the question was edited to say that it's over $mathbb F_2$. If the downvote is due to me assuming that it's over the reals, I don't think that's an unreasonable assumption.
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– Acccumulation
Jan 14 at 22:25
add a comment |
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5 Answers
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5 Answers
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$begingroup$
$X$ maps each eigenvector of $B$ to either $0$ or an eigenvector of the same eigenvalue of $A$, since if $Bv = lambda v$, $AX v = X B v = lambda X v$.
Let's suppose for simplicity that $B$ is diagonalizable, with a basis $b_j$ of eigenvectors corresponding to eigenvalues $lambda_j$ that may or may not be eigenvalues of $A$.
For each $j$, we can choose $X b_j$ to be any member of the eigenspace of $A$ for $lambda_j$, and once we have done that we have specified a solution $X$. Thus the number of solutions for $X$ is the product over $j$ of the cardinalities of the eigenspaces of $A$ for $lambda_j$ (note that this is $1$ if $lambda_j$ is not an eigenvalue of $A$).
answered Jan 14 at 22:34
Robert IsraelRobert Israel
323k23213467
$endgroup$
add a comment |
$begingroup$
$X$ maps each eigenvector of $B$ to either $0$ or an eigenvector of the same eigenvalue of $A$, since if $Bv = lambda v$, $AX v = X B v = lambda X v$.
Let's suppose for simplicity that $B$ is diagonalizable, with a basis $b_j$ of eigenvectors corresponding to eigenvalues $lambda_j$ that may or may not be eigenvalues of $A$.
For each $j$, we can choose $X b_j$ to be any member of the eigenspace of $A$ for $lambda_j$, and once we have done that we have specified a solution $X$. Thus the number of solutions for $X$ is the product over $j$ of the cardinalities of the eigenspaces of $A$ for $lambda_j$ (note that this is $1$ if $lambda_j$ is not an eigenvalue of $A$).
answered Jan 14 at 22:34
Robert IsraelRobert Israel
323k23213467
$endgroup$
add a comment |
$begingroup$
$X$ maps each eigenvector of $B$ to either $0$ or an eigenvector of the same eigenvalue of $A$, since if $Bv = lambda v$, $AX v = X B v = lambda X v$.
Let's suppose for simplicity that $B$ is diagonalizable, with a basis $b_j$ of eigenvectors corresponding to eigenvalues $lambda_j$ that may or may not be eigenvalues of $A$.
For each $j$, we can choose $X b_j$ to be any member of the eigenspace of $A$ for $lambda_j$, and once we have done that we have specified a solution $X$. Thus the number of solutions for $X$ is the product over $j$ of the cardinalities of the eigenspaces of $A$ for $lambda_j$ (note that this is $1$ if $lambda_j$ is not an eigenvalue of $A$).
answered Jan 14 at 22:34
Robert IsraelRobert Israel
323k23213467
$endgroup$
$X$ maps each eigenvector of $B$ to either $0$ or an eigenvector of the same eigenvalue of $A$, since if $Bv = lambda v$, $AX v = X B v = lambda X v$.
Let's suppose for simplicity that $B$ is diagonalizable, with a basis $b_j$ of eigenvectors corresponding to eigenvalues $lambda_j$ that may or may not be eigenvalues of $A$.
For each $j$, we can choose $X b_j$ to be any member of the eigenspace of $A$ for $lambda_j$, and once we have done that we have specified a solution $X$. Thus the number of solutions for $X$ is the product over $j$ of the cardinalities of the eigenspaces of $A$ for $lambda_j$ (note that this is $1$ if $lambda_j$ is not an eigenvalue of $A$).
answered Jan 14 at 22:34
Robert IsraelRobert Israel
323k23213467
answered Jan 14 at 22:34
Robert IsraelRobert Israel
323k23213467
answered Jan 14 at 22:34
Robert IsraelRobert Israel
323k23213467
answered Jan 14 at 22:34
Robert IsraelRobert Israel
323k23213467
323k23213467
add a comment |
add a comment |
$begingroup$
Your statement of well-known theorem requires saying that the eigenvalues are in the algebraic closure of the given field $F$. This is an easy version of Cecioni-Frobenius theorem suggested in the comments by 'user'.
Method 1 : Module Theory
Under your assumption for matrices $A$, $B$, $X$, define
$$
C_{A,B}={Xin M_{mtimes n}(F) mid AX - XB = 0 }.$$
Let $nu_{A,B}$ be the dimension of $C_{A,B}$ over $F$. The actual version of Cecioni-Frobenius theorem gives a formula
$$
nu_{A,B}=sum_p(deg p)sum_{i,j} min{lambda_{p,i}, mu_{p,j}}
$$
Here the first sum is over all irreducible polynomials $p$ which are common in the primary decompositions of $F[x]$-modules $M^A$ and $M^B$, and the indices $i, j$ of second double sum is from the partition $lambda_p=sum_i lambda_{p,i}$ that indicates the powers of $p$ in $p$-primary part of $M^A$, $mu_p=sum_j mu_{p,j}$ that of powers of $p$ in $p$-primary part of $M^B$.
Then the number of solutions to $AX=XB$ is $2^{nu_{A,B}}$.
Method 2 : Kronecker Product
Alternatively, there is a module-theory-free way. You will have an advantage of not having to solve for eigenvalues of $A$, $B$. But, a disadvantage is that there is almost no hope for writing down an explicit formula for $nu_{A,B}$. This is by Kronecker product, the method is mainly this: Solution of a Sylvester equation?
In our case, the equation $AX-XB=0$ can be written as
$$
(I_n otimes A - B^T otimes I_m) textrm{vec}(X)=0
$$
where $textrm{vec}(X)$ is a stack of columns of $X$ in order.
Then perform elementary row operations on the $mntimes mn$ matrix $I_n otimes A - B^T otimes I_m$. Find the number of free-variables, which is $nu_{A,B}$. Again the number of solutions to $AX=XB$ is $2^{nu_{A,B}}$.
answered Jan 15 at 2:20
i707107i707107
12.4k21547
$endgroup$
add a comment |
$begingroup$
Your statement of well-known theorem requires saying that the eigenvalues are in the algebraic closure of the given field $F$. This is an easy version of Cecioni-Frobenius theorem suggested in the comments by 'user'.
Method 1 : Module Theory
Under your assumption for matrices $A$, $B$, $X$, define
$$
C_{A,B}={Xin M_{mtimes n}(F) mid AX - XB = 0 }.$$
Let $nu_{A,B}$ be the dimension of $C_{A,B}$ over $F$. The actual version of Cecioni-Frobenius theorem gives a formula
$$
nu_{A,B}=sum_p(deg p)sum_{i,j} min{lambda_{p,i}, mu_{p,j}}
$$
Here the first sum is over all irreducible polynomials $p$ which are common in the primary decompositions of $F[x]$-modules $M^A$ and $M^B$, and the indices $i, j$ of second double sum is from the partition $lambda_p=sum_i lambda_{p,i}$ that indicates the powers of $p$ in $p$-primary part of $M^A$, $mu_p=sum_j mu_{p,j}$ that of powers of $p$ in $p$-primary part of $M^B$.
Then the number of solutions to $AX=XB$ is $2^{nu_{A,B}}$.
Method 2 : Kronecker Product
Alternatively, there is a module-theory-free way. You will have an advantage of not having to solve for eigenvalues of $A$, $B$. But, a disadvantage is that there is almost no hope for writing down an explicit formula for $nu_{A,B}$. This is by Kronecker product, the method is mainly this: Solution of a Sylvester equation?
In our case, the equation $AX-XB=0$ can be written as
$$
(I_n otimes A - B^T otimes I_m) textrm{vec}(X)=0
$$
where $textrm{vec}(X)$ is a stack of columns of $X$ in order.
Then perform elementary row operations on the $mntimes mn$ matrix $I_n otimes A - B^T otimes I_m$. Find the number of free-variables, which is $nu_{A,B}$. Again the number of solutions to $AX=XB$ is $2^{nu_{A,B}}$.
answered Jan 15 at 2:20
i707107i707107
12.4k21547
$endgroup$
add a comment |
$begingroup$
Your statement of well-known theorem requires saying that the eigenvalues are in the algebraic closure of the given field $F$. This is an easy version of Cecioni-Frobenius theorem suggested in the comments by 'user'.
Method 1 : Module Theory
Under your assumption for matrices $A$, $B$, $X$, define
$$
C_{A,B}={Xin M_{mtimes n}(F) mid AX - XB = 0 }.$$
Let $nu_{A,B}$ be the dimension of $C_{A,B}$ over $F$. The actual version of Cecioni-Frobenius theorem gives a formula
$$
nu_{A,B}=sum_p(deg p)sum_{i,j} min{lambda_{p,i}, mu_{p,j}}
$$
Here the first sum is over all irreducible polynomials $p$ which are common in the primary decompositions of $F[x]$-modules $M^A$ and $M^B$, and the indices $i, j$ of second double sum is from the partition $lambda_p=sum_i lambda_{p,i}$ that indicates the powers of $p$ in $p$-primary part of $M^A$, $mu_p=sum_j mu_{p,j}$ that of powers of $p$ in $p$-primary part of $M^B$.
Then the number of solutions to $AX=XB$ is $2^{nu_{A,B}}$.
Method 2 : Kronecker Product
Alternatively, there is a module-theory-free way. You will have an advantage of not having to solve for eigenvalues of $A$, $B$. But, a disadvantage is that there is almost no hope for writing down an explicit formula for $nu_{A,B}$. This is by Kronecker product, the method is mainly this: Solution of a Sylvester equation?
In our case, the equation $AX-XB=0$ can be written as
$$
(I_n otimes A - B^T otimes I_m) textrm{vec}(X)=0
$$
where $textrm{vec}(X)$ is a stack of columns of $X$ in order.
Then perform elementary row operations on the $mntimes mn$ matrix $I_n otimes A - B^T otimes I_m$. Find the number of free-variables, which is $nu_{A,B}$. Again the number of solutions to $AX=XB$ is $2^{nu_{A,B}}$.
answered Jan 15 at 2:20
i707107i707107
12.4k21547
$endgroup$
Your statement of well-known theorem requires saying that the eigenvalues are in the algebraic closure of the given field $F$. This is an easy version of Cecioni-Frobenius theorem suggested in the comments by 'user'.
Method 1 : Module Theory
Under your assumption for matrices $A$, $B$, $X$, define
$$
C_{A,B}={Xin M_{mtimes n}(F) mid AX - XB = 0 }.$$
Let $nu_{A,B}$ be the dimension of $C_{A,B}$ over $F$. The actual version of Cecioni-Frobenius theorem gives a formula
$$
nu_{A,B}=sum_p(deg p)sum_{i,j} min{lambda_{p,i}, mu_{p,j}}
$$
Here the first sum is over all irreducible polynomials $p$ which are common in the primary decompositions of $F[x]$-modules $M^A$ and $M^B$, and the indices $i, j$ of second double sum is from the partition $lambda_p=sum_i lambda_{p,i}$ that indicates the powers of $p$ in $p$-primary part of $M^A$, $mu_p=sum_j mu_{p,j}$ that of powers of $p$ in $p$-primary part of $M^B$.
Then the number of solutions to $AX=XB$ is $2^{nu_{A,B}}$.
Method 2 : Kronecker Product
Alternatively, there is a module-theory-free way. You will have an advantage of not having to solve for eigenvalues of $A$, $B$. But, a disadvantage is that there is almost no hope for writing down an explicit formula for $nu_{A,B}$. This is by Kronecker product, the method is mainly this: Solution of a Sylvester equation?
In our case, the equation $AX-XB=0$ can be written as
$$
(I_n otimes A - B^T otimes I_m) textrm{vec}(X)=0
$$
where $textrm{vec}(X)$ is a stack of columns of $X$ in order.
Then perform elementary row operations on the $mntimes mn$ matrix $I_n otimes A - B^T otimes I_m$. Find the number of free-variables, which is $nu_{A,B}$. Again the number of solutions to $AX=XB$ is $2^{nu_{A,B}}$.
answered Jan 15 at 2:20
i707107i707107
12.4k21547
answered Jan 15 at 2:20
i707107i707107
12.4k21547
answered Jan 15 at 2:20
i707107i707107
12.4k21547
answered Jan 15 at 2:20
i707107i707107
12.4k21547
12.4k21547
add a comment |
add a comment |
$begingroup$
To ask for the number of solutions of $f(X)=AX-XB=0$ is equivalent to ask for $dim(ker(f))$. Moreover, the required dimension is the same over the underlying field $K$ or over its algebraic closure $overline{K}$.
If we stack the considered matrices row by row, then $f=Aotimes I-Iotimes B^T$. Let $(lambda_i),(mu_i)$ be the spectra of $A,B$ in $overline{K}$. Since the functions $Xmapsto AX,Xmapsto XB$ commute, $spectrum(f)=(lambda_i-mu_j)_{i,j}$. Unfortunately, that does not give the required dimension; we can only say that $ker(f)not= 0$ iff $degree(gcd(chi_A,chi_B))geq 1$, where $chi_.$ is the characteristic polynomial.
The Cecioni-Frobenius (CF) theorem (1910) provides a solution if we can calculate the invariant factors of $A,B$.
$dim(ker(f))=nu_{A,B}=sum_{i,j}degree(gcd(d_i(A),d_j(B))$ where $d_i$ is the $i^{th}$ invariant factor.
I had heard about this theorem in 2009 and then I had completely forgotten about it. It was used by Byrnes-Gauger (1979) to show that
$A,B$ are similar iff $2nu_{A,B}=nu_{A,A}+nu_{B,B}$.
Note that, in great dimension, the invariant factor algorithm (the calculation of the Smith normal form of $xI-A$) is not effective. Fortunately, Byrnes-Gauger before proved (1977) this extraordinary result
$A,B$ are similar iff $det(xI-A)=det(xI-B),rank(Aotimes I-Iotimes A)=rank(Botimes I-Iotimes B)=rank(Aotimes I-Iotimes B)$.
Here is an example of using the CF theorem
let $J_k$ be the nilpotent Jordan block of dimension $k$, $A=diag(J_3,J_2),B=diag(J_4,J_1)$. Then the Smith forms are
$S(A)=diag(1,1,1,x^2,x^3),S(B)=diag(1,1,1,x,x^4)$.
Then $nu_{A,B}=deg(gcd(x^2,x))+deg(gcd(x^2,x^4))+deg(gcd(x^3,x))+deg(gcd(x^3,x^4))=7$.
answered Feb 5 at 9:50
loup blancloup blanc
23.3k21851
$endgroup$
$begingroup$
(+1) for mentioning Byrnes-Gauger.
$endgroup$
– i707107
Feb 8 at 16:33
add a comment |
$begingroup$
To ask for the number of solutions of $f(X)=AX-XB=0$ is equivalent to ask for $dim(ker(f))$. Moreover, the required dimension is the same over the underlying field $K$ or over its algebraic closure $overline{K}$.
If we stack the considered matrices row by row, then $f=Aotimes I-Iotimes B^T$. Let $(lambda_i),(mu_i)$ be the spectra of $A,B$ in $overline{K}$. Since the functions $Xmapsto AX,Xmapsto XB$ commute, $spectrum(f)=(lambda_i-mu_j)_{i,j}$. Unfortunately, that does not give the required dimension; we can only say that $ker(f)not= 0$ iff $degree(gcd(chi_A,chi_B))geq 1$, where $chi_.$ is the characteristic polynomial.
The Cecioni-Frobenius (CF) theorem (1910) provides a solution if we can calculate the invariant factors of $A,B$.
$dim(ker(f))=nu_{A,B}=sum_{i,j}degree(gcd(d_i(A),d_j(B))$ where $d_i$ is the $i^{th}$ invariant factor.
I had heard about this theorem in 2009 and then I had completely forgotten about it. It was used by Byrnes-Gauger (1979) to show that
$A,B$ are similar iff $2nu_{A,B}=nu_{A,A}+nu_{B,B}$.
Note that, in great dimension, the invariant factor algorithm (the calculation of the Smith normal form of $xI-A$) is not effective. Fortunately, Byrnes-Gauger before proved (1977) this extraordinary result
$A,B$ are similar iff $det(xI-A)=det(xI-B),rank(Aotimes I-Iotimes A)=rank(Botimes I-Iotimes B)=rank(Aotimes I-Iotimes B)$.
Here is an example of using the CF theorem
let $J_k$ be the nilpotent Jordan block of dimension $k$, $A=diag(J_3,J_2),B=diag(J_4,J_1)$. Then the Smith forms are
$S(A)=diag(1,1,1,x^2,x^3),S(B)=diag(1,1,1,x,x^4)$.
Then $nu_{A,B}=deg(gcd(x^2,x))+deg(gcd(x^2,x^4))+deg(gcd(x^3,x))+deg(gcd(x^3,x^4))=7$.
answered Feb 5 at 9:50
loup blancloup blanc
23.3k21851
$endgroup$
$begingroup$
(+1) for mentioning Byrnes-Gauger.
$endgroup$
– i707107
Feb 8 at 16:33
add a comment |
$begingroup$
To ask for the number of solutions of $f(X)=AX-XB=0$ is equivalent to ask for $dim(ker(f))$. Moreover, the required dimension is the same over the underlying field $K$ or over its algebraic closure $overline{K}$.
If we stack the considered matrices row by row, then $f=Aotimes I-Iotimes B^T$. Let $(lambda_i),(mu_i)$ be the spectra of $A,B$ in $overline{K}$. Since the functions $Xmapsto AX,Xmapsto XB$ commute, $spectrum(f)=(lambda_i-mu_j)_{i,j}$. Unfortunately, that does not give the required dimension; we can only say that $ker(f)not= 0$ iff $degree(gcd(chi_A,chi_B))geq 1$, where $chi_.$ is the characteristic polynomial.
The Cecioni-Frobenius (CF) theorem (1910) provides a solution if we can calculate the invariant factors of $A,B$.
$dim(ker(f))=nu_{A,B}=sum_{i,j}degree(gcd(d_i(A),d_j(B))$ where $d_i$ is the $i^{th}$ invariant factor.
I had heard about this theorem in 2009 and then I had completely forgotten about it. It was used by Byrnes-Gauger (1979) to show that
$A,B$ are similar iff $2nu_{A,B}=nu_{A,A}+nu_{B,B}$.
Note that, in great dimension, the invariant factor algorithm (the calculation of the Smith normal form of $xI-A$) is not effective. Fortunately, Byrnes-Gauger before proved (1977) this extraordinary result
$A,B$ are similar iff $det(xI-A)=det(xI-B),rank(Aotimes I-Iotimes A)=rank(Botimes I-Iotimes B)=rank(Aotimes I-Iotimes B)$.
Here is an example of using the CF theorem
let $J_k$ be the nilpotent Jordan block of dimension $k$, $A=diag(J_3,J_2),B=diag(J_4,J_1)$. Then the Smith forms are
$S(A)=diag(1,1,1,x^2,x^3),S(B)=diag(1,1,1,x,x^4)$.
Then $nu_{A,B}=deg(gcd(x^2,x))+deg(gcd(x^2,x^4))+deg(gcd(x^3,x))+deg(gcd(x^3,x^4))=7$.
answered Feb 5 at 9:50
loup blancloup blanc
23.3k21851
$endgroup$
To ask for the number of solutions of $f(X)=AX-XB=0$ is equivalent to ask for $dim(ker(f))$. Moreover, the required dimension is the same over the underlying field $K$ or over its algebraic closure $overline{K}$.
If we stack the considered matrices row by row, then $f=Aotimes I-Iotimes B^T$. Let $(lambda_i),(mu_i)$ be the spectra of $A,B$ in $overline{K}$. Since the functions $Xmapsto AX,Xmapsto XB$ commute, $spectrum(f)=(lambda_i-mu_j)_{i,j}$. Unfortunately, that does not give the required dimension; we can only say that $ker(f)not= 0$ iff $degree(gcd(chi_A,chi_B))geq 1$, where $chi_.$ is the characteristic polynomial.
The Cecioni-Frobenius (CF) theorem (1910) provides a solution if we can calculate the invariant factors of $A,B$.
$dim(ker(f))=nu_{A,B}=sum_{i,j}degree(gcd(d_i(A),d_j(B))$ where $d_i$ is the $i^{th}$ invariant factor.
I had heard about this theorem in 2009 and then I had completely forgotten about it. It was used by Byrnes-Gauger (1979) to show that
$A,B$ are similar iff $2nu_{A,B}=nu_{A,A}+nu_{B,B}$.
Note that, in great dimension, the invariant factor algorithm (the calculation of the Smith normal form of $xI-A$) is not effective. Fortunately, Byrnes-Gauger before proved (1977) this extraordinary result
$A,B$ are similar iff $det(xI-A)=det(xI-B),rank(Aotimes I-Iotimes A)=rank(Botimes I-Iotimes B)=rank(Aotimes I-Iotimes B)$.
Here is an example of using the CF theorem
let $J_k$ be the nilpotent Jordan block of dimension $k$, $A=diag(J_3,J_2),B=diag(J_4,J_1)$. Then the Smith forms are
$S(A)=diag(1,1,1,x^2,x^3),S(B)=diag(1,1,1,x,x^4)$.
Then $nu_{A,B}=deg(gcd(x^2,x))+deg(gcd(x^2,x^4))+deg(gcd(x^3,x))+deg(gcd(x^3,x^4))=7$.
answered Feb 5 at 9:50
loup blancloup blanc
23.3k21851
edited Feb 5 at 10:09
answered Feb 5 at 9:50
loup blancloup blanc
23.3k21851
answered Feb 5 at 9:50
loup blancloup blanc
23.3k21851
answered Feb 5 at 9:50
loup blancloup blanc
23.3k21851
23.3k21851
$begingroup$
(+1) for mentioning Byrnes-Gauger.
$endgroup$
– i707107
Feb 8 at 16:33
add a comment |
$begingroup$
(+1) for mentioning Byrnes-Gauger.
$endgroup$
– i707107
Feb 8 at 16:33
$begingroup$
(+1) for mentioning Byrnes-Gauger.
$endgroup$
– i707107
Feb 8 at 16:33
$begingroup$
(+1) for mentioning Byrnes-Gauger.
$endgroup$
– i707107
Feb 8 at 16:33
add a comment |
$begingroup$
This is just a special case of Sylvester equation.
Let $A$ and $B$ be operators with spectra $sigma(A)$ and $sigma(B)$, respectively. If $sigma(A)$ and $sigma(B)$ are disjoint, then the equation
$AX-XB=Y$ has a unique solution $X$ for every $Y$. Moreover, the solution is given by $X=sum_{n=0}^infty A^{-n-1}YB$. Hence, for $Y=0$ unique solution is $X=0$ as Robert Israel claimed.
The proof of the theorem above can be found in Bhatia's Matrix Analysis.
answered Jan 14 at 22:52
Mustafa SaidMustafa Said
2,9911913
$endgroup$
add a comment |
$begingroup$
This is just a special case of Sylvester equation.
Let $A$ and $B$ be operators with spectra $sigma(A)$ and $sigma(B)$, respectively. If $sigma(A)$ and $sigma(B)$ are disjoint, then the equation
$AX-XB=Y$ has a unique solution $X$ for every $Y$. Moreover, the solution is given by $X=sum_{n=0}^infty A^{-n-1}YB$. Hence, for $Y=0$ unique solution is $X=0$ as Robert Israel claimed.
The proof of the theorem above can be found in Bhatia's Matrix Analysis.
answered Jan 14 at 22:52
Mustafa SaidMustafa Said
2,9911913
$endgroup$
add a comment |
$begingroup$
This is just a special case of Sylvester equation.
Let $A$ and $B$ be operators with spectra $sigma(A)$ and $sigma(B)$, respectively. If $sigma(A)$ and $sigma(B)$ are disjoint, then the equation
$AX-XB=Y$ has a unique solution $X$ for every $Y$. Moreover, the solution is given by $X=sum_{n=0}^infty A^{-n-1}YB$. Hence, for $Y=0$ unique solution is $X=0$ as Robert Israel claimed.
The proof of the theorem above can be found in Bhatia's Matrix Analysis.
answered Jan 14 at 22:52
Mustafa SaidMustafa Said
2,9911913
$endgroup$
This is just a special case of Sylvester equation.
Let $A$ and $B$ be operators with spectra $sigma(A)$ and $sigma(B)$, respectively. If $sigma(A)$ and $sigma(B)$ are disjoint, then the equation
$AX-XB=Y$ has a unique solution $X$ for every $Y$. Moreover, the solution is given by $X=sum_{n=0}^infty A^{-n-1}YB$. Hence, for $Y=0$ unique solution is $X=0$ as Robert Israel claimed.
The proof of the theorem above can be found in Bhatia's Matrix Analysis.
answered Jan 14 at 22:52
Mustafa SaidMustafa Said
2,9911913
answered Jan 14 at 22:52
Mustafa SaidMustafa Said
2,9911913
answered Jan 14 at 22:52
Mustafa SaidMustafa Said
2,9911913
answered Jan 14 at 22:52
Mustafa SaidMustafa Said
2,9911913
2,9911913
add a comment |
add a comment |
$begingroup$
Suppose $X_1$ and $X_2$ are both solutions. Then $A(X_1+X_2)=AX_1+AX_2=X_1B+X_2B=(X_1+X_2)B$. Thus, $X_1+X_2$ is a solution. So the number of solutions can be $0$, $1$, or infinity. This is a common situation in linear algebra: you either have no solutions, a unique solution, or infinite solutions.
answered Jan 14 at 22:20
AcccumulationAcccumulation
6,9892618
$endgroup$
$begingroup$
@Accumulation Not true in $mathbb F_2$, since $X+X=0$. Thanks for the answer, though.
$endgroup$
– Frpzzd
Jan 14 at 22:21
$begingroup$
Nor in any finite field, since there are only finitely many $m times n$ matrices there.
$endgroup$
– Robert Israel
Jan 14 at 22:22
$begingroup$
I wrote my answer before the question was edited to say that it's over $mathbb F_2$. If the downvote is due to me assuming that it's over the reals, I don't think that's an unreasonable assumption.
$endgroup$
– Acccumulation
Jan 14 at 22:25
add a comment |
$begingroup$
Suppose $X_1$ and $X_2$ are both solutions. Then $A(X_1+X_2)=AX_1+AX_2=X_1B+X_2B=(X_1+X_2)B$. Thus, $X_1+X_2$ is a solution. So the number of solutions can be $0$, $1$, or infinity. This is a common situation in linear algebra: you either have no solutions, a unique solution, or infinite solutions.
answered Jan 14 at 22:20
AcccumulationAcccumulation
6,9892618
$endgroup$
$begingroup$
@Accumulation Not true in $mathbb F_2$, since $X+X=0$. Thanks for the answer, though.
$endgroup$
– Frpzzd
Jan 14 at 22:21
$begingroup$
Nor in any finite field, since there are only finitely many $m times n$ matrices there.
$endgroup$
– Robert Israel
Jan 14 at 22:22
$begingroup$
I wrote my answer before the question was edited to say that it's over $mathbb F_2$. If the downvote is due to me assuming that it's over the reals, I don't think that's an unreasonable assumption.
$endgroup$
– Acccumulation
Jan 14 at 22:25
add a comment |
$begingroup$
Suppose $X_1$ and $X_2$ are both solutions. Then $A(X_1+X_2)=AX_1+AX_2=X_1B+X_2B=(X_1+X_2)B$. Thus, $X_1+X_2$ is a solution. So the number of solutions can be $0$, $1$, or infinity. This is a common situation in linear algebra: you either have no solutions, a unique solution, or infinite solutions.
answered Jan 14 at 22:20
AcccumulationAcccumulation
6,9892618
$endgroup$
Suppose $X_1$ and $X_2$ are both solutions. Then $A(X_1+X_2)=AX_1+AX_2=X_1B+X_2B=(X_1+X_2)B$. Thus, $X_1+X_2$ is a solution. So the number of solutions can be $0$, $1$, or infinity. This is a common situation in linear algebra: you either have no solutions, a unique solution, or infinite solutions.
answered Jan 14 at 22:20
AcccumulationAcccumulation
6,9892618
answered Jan 14 at 22:20
AcccumulationAcccumulation
6,9892618
answered Jan 14 at 22:20
AcccumulationAcccumulation
6,9892618
answered Jan 14 at 22:20
AcccumulationAcccumulation
6,9892618
6,9892618
$begingroup$
@Accumulation Not true in $mathbb F_2$, since $X+X=0$. Thanks for the answer, though.
$endgroup$
– Frpzzd
Jan 14 at 22:21
$begingroup$
Nor in any finite field, since there are only finitely many $m times n$ matrices there.
$endgroup$
– Robert Israel
Jan 14 at 22:22
$begingroup$
I wrote my answer before the question was edited to say that it's over $mathbb F_2$. If the downvote is due to me assuming that it's over the reals, I don't think that's an unreasonable assumption.
$endgroup$
– Acccumulation
Jan 14 at 22:25
add a comment |
$begingroup$
@Accumulation Not true in $mathbb F_2$, since $X+X=0$. Thanks for the answer, though.
$endgroup$
– Frpzzd
Jan 14 at 22:21
$begingroup$
Nor in any finite field, since there are only finitely many $m times n$ matrices there.
$endgroup$
– Robert Israel
Jan 14 at 22:22
$begingroup$
I wrote my answer before the question was edited to say that it's over $mathbb F_2$. If the downvote is due to me assuming that it's over the reals, I don't think that's an unreasonable assumption.
$endgroup$
– Acccumulation
Jan 14 at 22:25
$begingroup$
@Accumulation Not true in $mathbb F_2$, since $X+X=0$. Thanks for the answer, though.
$endgroup$
– Frpzzd
Jan 14 at 22:21
$begingroup$
@Accumulation Not true in $mathbb F_2$, since $X+X=0$. Thanks for the answer, though.
$endgroup$
– Frpzzd
Jan 14 at 22:21
$begingroup$
Nor in any finite field, since there are only finitely many $m times n$ matrices there.
$endgroup$
– Robert Israel
Jan 14 at 22:22
$begingroup$
Nor in any finite field, since there are only finitely many $m times n$ matrices there.
$endgroup$
– Robert Israel
Jan 14 at 22:22
$begingroup$
I wrote my answer before the question was edited to say that it's over $mathbb F_2$. If the downvote is due to me assuming that it's over the reals, I don't think that's an unreasonable assumption.
$endgroup$
– Acccumulation
Jan 14 at 22:25
$begingroup$
I wrote my answer before the question was edited to say that it's over $mathbb F_2$. If the downvote is due to me assuming that it's over the reals, I don't think that's an unreasonable assumption.
$endgroup$
– Acccumulation
Jan 14 at 22:25
add a comment |
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$begingroup$
You probably mean the number of linearly independent solutions. Otherwise, it's obvious that the answer is false. Because if $X$ is a solution, so is $lambda X$. Right? Does this theorem have a name? I'd like to see the proof.
$endgroup$
– stressed out
Jan 14 at 22:16
1
$begingroup$
You're right; I am actually working mostly in $mathbb F_2$ so that specification isn't important, and I should mention that in my question. As for the theorem: I don't know if it has a name, but the equation is called a "Sylvester Equation" and the Wikipedia page has a proof of said theorem.
$endgroup$
– Frpzzd
Jan 14 at 22:18
1
$begingroup$
@stressedout It is Sylvester equation.
$endgroup$
– A.Γ.
Jan 14 at 22:18
2
$begingroup$
@stressedout The unique solution, when it is unique, is $X = 0$.
$endgroup$
– Robert Israel
Jan 14 at 22:20
2
$begingroup$
Try Cecioni-Frobenius theorem..
$endgroup$
– user
Jan 14 at 22:31