Sylvester Matrix Equation

Worked example: Diagonal Case

Let's consider a specific example. Suppose A is a diagonal matrix with entries a1 = 1, a2 = 2. Let B be a diagonal matrix with entries b1 = 3, b2 = 4. Finally, let C = [[4, 5], [5, 6]]. We want to find a 2x2 matrix X such that AX + XB = C.

Since A and B are diagonal, the equation for the entry X_ij simplifies to: (a_i + b_j) X_ij = C_ij.

1. For X₁₁ (i=1, j=1): We have (a1 + b1) X₁₁ = C₁₁. Substituting the values gives (1 + 3) X₁₁ = 4, so 4 X₁₁ = 4, which means X₁₁ = 1.00.
2. For X₁₂ (i=1, j=2): We have (a1 + b2) X₁₂ = C₁₂. Substituting gives (1 + 4) X₁₂ = 5, so 5 X₁₂ = 5, which means X₁₂ = 1.00.
3. For X₂₁ (i=2, j=1): We have (a2 + b1) X₂₁ = C₂₁. Substituting gives (2 + 3) X₂₁ = 5, so 5 X₂₁ = 5, which means X₂₁ = 1.00.
4. For X₂₂ (i=2, j=2): We have (a2 + b2) X₂₂ = C₂₂. Substituting gives (2 + 4) X₂₂ = 6, so 6 X₂₂ = 6, which means X₂₂ = 1.00.

Thus, the unique solution matrix X has all entries equal to 1.00. The residual is the sum of absolute differences |(a_i+b_j)X_ij − C_ij| from unrounded quotients; it is zero here and is displayed in scientific notation. Entries are displayed to six decimals.

Experimenting with the Sylvester Equation

Open the Sylvester Matrix Equation module to see the decoupling in action.

1. Leave the default values: a1=1, a2=2, b1=3, b2=4, and C=[[4, 5], [5, 6]].
2. Click the "Step" button. The first click reveals the equation for X₁₁: (1 + 3) X₁₁ = 4, resulting in X₁₁ = 1.00.
3. Click "Step" twice more to reveal X₁₂ and X₂₁; the cursor now has three of four equations.
4. Click "Finish" to reveal the fourth equation and see the final matrix X and the residual. The interface will indicate a "Unique solution" since all (a_i + b_j) coefficients were non-zero.
5. Now, change a1 to 1, a2 to 2, b1 to -1, and b2 to 3. Change C11 to 1, C12 to 0, C21 to 0, and C22 to 0. Click "Finish". For X₁₁, the coefficient is a1 + b1 = 1 + (-1) = 0. The equation is 0 * X₁₁ = 1, which has no solution. The system will classify the overall status as "No solution".
6. Change C11 to 0 and click Finish again; this edit has rewound the trace. Now the equation for X₁₁ is 0 * X₁₁ = 0. This means X₁₁ can be any value, making it a "free variable". The system will indicate "Infinitely many solutions (1 free variable)".

Limitations and Properties

This interactive module explicitly limits the scope to diagonal A and B matrices. When A and B are arbitrary, the equations for the entries of X are coupled, and a more complex method (like Bartels-Stewart algorithm or Kronecker products) is required to find the solution.

In the diagonal case, the existence and uniqueness of the solution depend entirely on the sums (a_i + b_j). If a_i + b_j = 0 for some pair (i, j):

• If the corresponding C_ij is not zero, the equation 0 * X_ij = C_ij has no solution, so the entire Sylvester equation has no solution.
• If C_ij is also zero, the equation 0 * X_ij = 0 holds for any value of X_ij. Thus, X_ij is a free variable. If no other entry results in "no solution", there are infinitely many solutions. This module identifies such variables as "free" rather than arbitrarily setting them to 0 and claiming the solution is unique.

A unique solution exists if and only if a_i + b_j is non-zero for all i and j.

References

The SciPy documentation states AX + XB = Q and describes the general Bartels–Stewart method using Schur decompositions. This module instead exploits the explicitly diagonal case and independently checks each scalar coefficient; it does not implement SciPy’s general solver. See scipy.linalg.solve_sylvester.