$2\times 2$ matrix equation

Question

Let $A$, $B$ and $C$ be $2 \times 2$ matrices. Let $X$ be an unknown $2 \times 2$ matrix. Consider the equation $$AX+XB = C$$

The question is: Why can you not solve this equation as $2 \times 2$ systems simultaneously? Write the matrix as four equations with four unknowns.

I tried something like: Let $A = \begin{bmatrix} a_1 & a_2\\ a_3 & a_4\\ \end{bmatrix}$ , $B = \begin{bmatrix} b_1 & b_2\\ b_3 & b_4\\ \end{bmatrix}$, $C = \begin{bmatrix} c_1 & c_2\\ c_3 & c_4\\ \end{bmatrix}$ and $X = \begin{bmatrix} x_1 & x_2\\ x_3 & x_4\\ \end{bmatrix}$

$AX = \begin{bmatrix} a_1x_1+a_2x_3 & a_1x_2+a_2x_4\\ a_3x_1+a_4x_3 & a_3x_2+a_4x_4\\ \end{bmatrix}$

$XB = \begin{bmatrix} x_1b_1+x_2b_3 & x_1b_2+x_2b_4\\ x_3b_1+x_4b_3 & x_3b_2+x_4b_4\\ \end{bmatrix}$

And thus

$$AX + XB = \begin{bmatrix} a_1x_1+a_2x_3+x_1b_1+x_2b_3 & a_1x_2+a_2x_4+x_1b_2+x_2b_4\\ a_3x_1+a_4x_3+x_3b_1+x_4b_3 & a_3x_2+a_4x_4+x_3b_2+x_4b_4\\ \end{bmatrix} = \begin{bmatrix} c_1 & c_2\\ c_3 & c_4\\ \end{bmatrix}$$

I'm not really sure what I ended up with here, I guess if we assume that $A, B, C$ are known we do have 4 equations with 4 unknowns (just set the first entry in AX+XB equal to $c_1$, second entry $c_2$ and so forth... Is this the correct solution? Is there an easy way to explain why we can't solve it as a double system simultaniously?

Answer 1

You can solve the matrix equation if $ A $ and $ B $ satisfy some appropriate hypotheses. For example,

• Supose that $BX=XB$ and exist $S$ such that $S(A+B)=I$. Then \begin{align} AX + XB = & C \\ AX + B X= & C \\ (A+B)X = & C \\ S(A+B)X = &S C \\ X = &SC \\ \end{align}

• Supose that $AX=XA$ and exist $R$ such that $(A+B)R=I$. Then \begin{align} AX + XB = & C \\ XA + XB= & C \\ X(A+B) = & C \\ X(A+B)R = & CR \\ X = &CR \\ \end{align}

• Suppose $AX=XU$ for some $U$ and exist $R$ such that $(U+B)R=I$. Then \begin{align} AX + XB = & C \\ XU + B X= & C \\ X(U+B) = & C \\ X(U+B)R = & CR \\ X = &CR \\ \end{align}

• Supose that $VX=XB$ and for some $V$ and exist $S$ such that $S(A+V)=I$. Then \begin{align} AX + XB = & C \\ XA + VX= & C \\ (A+V)X = & C \\ S(A+B)X = & SC \\ X = &SC \\ \end{align}

Answer 2

If you know $A,B,$ and $C$, then a solution to $AX + XB = C$ is a solution to the following equation, and vice versa:

$$ \begin{bmatrix} (a_1 + b_1) & b_3 & a_2 & 0 \\ b_2 & (a_1 + b_4) & 0 & a_2 \\ a_3 & 0 & (a_4 + b_1) & b_3 \\ 0 & a_3 & b_2 & (a_4 + b_4) \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ \end{bmatrix} = \begin{bmatrix} c_1 \\ c_2 \\ c_3 \\ c_4 \\ \end{bmatrix}.$$

This reduces to matrix computation we know well.