Dtyjlui

I have two symmetric $3times 3$ matrices $A, B$. I am interested in solving the system

$$AX= XB$$

Is there a way this is usually done? The matrices are not necessarily non singular.

You can use vectorization and Kronecker products.

Assume we have the equation
$$bf AXB=C$$

$$({bf B}^T otimes {bf A})text{vec}({bf X}) = text{vec}({bf AXB}) = text{vec}({bf C})$$

Now how can we modify it to get what we want?

This is a partial answer, but for a generalized setting where $A,Bin M_n(Bbb{K})$ for some field $Bbb{K}$. Suppose that $A$ and $B$ are diagonalizable over $Bbb K$. Let $a_1,a_2,ldots,a_n$ be a complete set of eigenvectors of $A$ with eigenvalues $lambda_1,lambda_2,ldots,lambda_n$, and $b_1,b_2,ldots,b_n$ a complete set of eigenvectors of $B^t$ with eigenvealues $mu_1,mu_2,ldots,mu_n$. Then, the matrices $C_{i,j}=a_ib_j^t$ form a basis of $M_n(Bbb{K})$.

Consider the linear map $T:M_n(Bbb{K})to M_n(Bbb{K})$ given by $T(X)=AX-XB$. If $X=sum_{i,j}x_{i,j}C_{i,j}$, then
$$T(X)=sum_{i,j}x_{i,j}(lambda_i-mu_j)C_{i,j}.$$
In particular, $T(X)=0$ if and only if $x_{i,j}=0$ whenever $lambda_ineq mu_j$. Therefore,
$$ker T=operatorname{span}big{C_{i,j}:lambda_i=mu_jbig}.$$

In your case, the matrices $A$ and $B$ are real symmetric (you didn't specify the field, so I guess the field is the reals), and so they are diagonalizable. Therefore, you can use my approach.

However, if the field is $Bbb{C}$, then a symmetric matrix may not be diagonalizable. I do not know the answer if $A$ or $B$ is not diagonalizable. But the wiki link given by Tianlalu may help.

After I made some calculations, I found this result. Assume that the characteristic polynomials of $A$ and $B$ split into linear factors over $Bbb{K}$. Suppose that $A$ has $k$ Jordan blocks of sizes $r_1,r_2,ldots,r_k$ with eigenvalues $lambda_1,lambda_2,ldots,lambda_k$, and suppose that $B$ has $l$ Jordan blocks of sizes $s_1,s_2,ldots,s_l$ with eigenvalues $mu_1,mu_2,ldots,mu_l$. That is, we have the following proposition.

Proposition: Let $T:M_n(Bbb{K})to M_n(Bbb{K})$ be defined by $T(X)=AX-XB$, where $A$ and $B$ are fixed elements of $M_n(Bbb{K})$. Then, the linear map $T$ is diagonalizable over $Bbb K$ if and only if both $A$ and $B$ are diagonalizable over $Bbb{K}$.

Write $(a_i^1,a_i^2,ldots,a_i^{r_i})$ for a generalized eigenvector sequence of $A$ in the $i$th block, that is,
$$Aa_i^p=lambda_ia_i^p+a_i^{p+1}$$
for $p=1,2,ldots,r_i$, with $a_i^{r_i+1}=0$. Similarly, write $(b_j^1,b_j^2,ldots,b_j^{s_j})$ for a generalized eigenvector sequence of $B$ in the $j$th block, that is,
$$Bb_j^q=mu_jb_j^q+b_j^{q+1}$$
for $q=1,2,ldots,s_j$, with $b_j^{s_j+1}=0$. Then, for fixed $i=1,2,ldots,k$ and $j=1,2,ldots,l$, the span of $C_{i,j}^{p,q}=a_i^p(b_j^q)^t$ with $p$ and $q$ ranging from ${1,2,ldots,r_i}$ and ${1,2,ldots,s_j}$ is a direct sum of generalized eigenspaces of $T$ with eigenvalue $lambda_i-mu_j$.

This is because
$$big(T-(lambda_i-mu_j)big)^hC_{i,j}^{p,q}=sum_{r=0}^h(-1)^rbinom{h}{r}C_{i,j}^{p+h-r,q+r},$$
where $C_{i,j}^{p,q}=0$ if $p>r_i$ or $q>s_j$. Therefore, if $h>r_i-p+s_j-q$, then $$big(T-(lambda_i-mu_j)big)^hC_{i,j}^{p,q}=0.$$ In characteristic $0$, there are $m_{i,j}=min{r_i,s_j}$ generalized eigenspaces and the dimensions of the generalized eigenspaces are
$$r_i+s_j-1,r_i+s_j-3,r_i+s_j-5,ldots,r_i+s_j+1-2m_{i,j}.$$

This shows that
$$operatorname{span}{C^{r_i,s_j}_{i,j}:lambda_i=mu_jbig}subseteqker Tsubseteq operatorname{span}{C_{i,j}^{p,q}:lambda_i=mu_jbig}.$$
For $Xin M_n(Bbb{K})$, we write $X=sum_{i,j,p,q}x_{i,j}^{p,q}C_{i,j}$. If $Xin ker T$, then $x_{i,j}^{p,q}=0$ whenever $lambda_ineq mu_j$, and so
$$T(X)=sum_{i,j,p,q}x_{i,j}^{p,q}(C_{i,j}^{p+1,q}-C_{i,j}^{p,q+1})=sum_{i,j}^{p,q}(x_{i,j}^{p-1,q}-x_{i,j}^{p,q-1})C_{i,j}^{p,q},$$
with $x_{i,j}^{0,q}=x_{i,j}^{p,0}=0$. This is as far as I can go, but it tells me that, unless both $A$ and $B$ are diagonalizable, we have
$$operatorname{span}{C^{r_i,s_j}_{i,j}:lambda_i=mu_jbig}subsetneq ker Tsubsetneq operatorname{span}{C_{i,j}^{p,q}:lambda_i=mu_jbig}.$$

As stated in the comment from Christoph, here is a direct approach solving $AX=XB$ given both $A$ and $B$ are square matrices of order $k$.

Proposition. Let $A=(a_{ij})_{ktimes k}$, $B=(b_{ij})_{ktimes k}$ and the unknown $X=(x_{ij})_{ktimes k}$. We also let $$Y=(x_{11},x_{21},cdots,x_{k1},x_{12},x_{22},cdots,x_{k2},cdots,x_{1k},x_{2k},cdots,x_{kk})^{T}$$
be the column vector with dimension $k^2$. Then, the solution of $AX=XB$ is given by the solution of the homogeneous equation $QY=0$, where $Q$ is the block matrix
$$Q=begin{bmatrix}
Q_{11}&Q_{12}&cdots&Q_{1k}\
Q_{21}&Q_{22}&cdots&Q_{2k}\
cdots&cdots&cdots&cdots\
Q_{k1}&Q_{k2}&cdots&Q_{kk}\
end{bmatrix}$$
with $Q_{ii}=A-b_{ii}I$ ($i=1,2,dots,k$ and $I$ is the identity matrix of order $k$) and $Q_{ij}=-b_{ji}I$ ($i=1,2,dots,k$).

The proof follows by direct multiplication of matrices. We get the system of linear equations about every elements in $X$ and write the equation as the form $QY=0$. Since there always has solutions for the homogeneous equation $QY=0$, we get all the solutions for $AX=XB$.

Example. Solve $AX=XB$, where $A=begin{bmatrix}-4&1\-9&2end{bmatrix}$ and $B=begin{bmatrix}-2&1\-1&0end{bmatrix}$.

From the proposition, we have
$$Q=begin{bmatrix}
-2&1&1&0\
-9&4&0&1\
-1&0&-4&1\
0&-1&-9&2\
end{bmatrix}xrightarrow{text{row operation}}
begin{bmatrix}
1&0&4&-1\
0&1&9&-2\
0&0&0&0\
0&0&0&0\
end{bmatrix}=P.$$
Since $PY=0$ has the same solution to $QY=0$, we get
$$Y=(-4u+v,-9u+2v,u,v)^T,qquad forall u,vinBbb C.$$
Therefore,
$$X=begin{bmatrix}
-4u+v&u\
-9u+2v&v\
end{bmatrix}=
ubegin{bmatrix}-4&1\-9&0end{bmatrix}+vbegin{bmatrix}1&0\2&1end{bmatrix},qquad forall u,vinBbb C.$$

From the example we can see the steps are

• express $Q$ in terms of $A$ and $B$, apply row operations to obtain $P$,




• solve $Y$ for $PY=0$,




• write down the elements in $X$ (corresponding to the elements in $Y$).