特殊线性群（行列式为1矩阵）中矩阵平方根高效计算方法问询

Absolutely—since we’re working exclusively with matrices in the special linear group ( \text{SL}(n) ) (where ( \det(M) = 1 )), we can lean into this determinant constraint to skip costly steps required by generic matrix square root algorithms (like diagonalization). Below are targeted approaches that deliver faster computation by leveraging SL-specific properties:

1. Direct Algebraic Solution for 2x2 SL Matrices #

For ( 2 \times 2 ) SL matrices, this is hands down the fastest method—no diagonalization needed at all. Let ( M = \begin{pmatrix} a & b \ c & d \end{pmatrix} ) with ( ad - bc = 1 ). We want a square root ( S \in \text{SL}(2) ) (so ( \det(S) = 1 )) such that ( S^2 = M ).

Key SL-specific insight: The trace of ( S ) satisfies ( \text{tr}(S)^2 = \text{tr}(M) + 2\det(S) ). Since ( \det(S) = 1 ), this simplifies to ( \text{tr}(S)^2 = \text{tr}(M) + 2 ). Let ( t = \text{tr}(S) ), so ( t = \pm \sqrt{\text{tr}(M) + 2} ).

You can then solve directly for the elements of ( S ):

• If ( b \neq 0 ): ( S = \begin{pmatrix} \frac{a + t}{2} & \frac{b}{t} \ \frac{c(a + t) - 2c}{2b} & \frac{d + t}{2} \end{pmatrix} ) (simplified using ( ad - bc = 1 ) to eliminate redundant terms)
• If ( c \neq 0 ), use a symmetric approach with ( c ) instead of ( b )
• If ( b = c = 0 ), ( M ) is diagonal with ( a = d^{-1} ), so ( S = \begin{pmatrix} \sqrt{a} & 0 \ 0 & \sqrt{d} \end{pmatrix} ) (choosing roots such that their product is 1)

This runs in ( O(1) ) time, a massive improvement over the ( O(n^3) ) cost of diagonalization for small matrices.

2. Simplified Jordan Form Construction #

For higher-dimensional SL matrices, we can exploit the fact that the product of all eigenvalues (counted with multiplicity) is 1. When computing square roots via Jordan canonical form:

• For diagonal blocks ( J_1(\lambda) ): We only need to choose square roots ( \lambda^{1/2} ) such that the product of all chosen roots is 1 (ensuring ( \det(S) = 1 )). Generic methods would require checking arbitrary roots, but here we can normalize the final set of roots in one step instead of tracking det constraints throughout.
• For Jordan blocks ( J_k(\lambda) ) (where ( k > 1 )): Since ( \det(J_k(\lambda)) = \lambda^k = 1 ), ( \lambda ) must be a ( k )-th root of unity. This lets us precompute valid square roots of ( \lambda ) and solve for the off-diagonal elements of the square root Jordan block using simplified equations (no need to handle arbitrary ( \lambda ) values, which reduces computational overhead).

This cuts down on the number of calculations needed compared to generic Jordan form-based square root methods.

3. SL-Specific Matrix Decompositions #

We can decompose SL matrices into factors that are easier to take square roots of, using the det=1 constraint to avoid extra normalization steps:

• LU Decomposition: For ( M \in \text{SL}(n) ), we can perform LU decomposition where ( \det(L) = \det(U) = 1 ) (since ( \det(M) = \det(L)\det(U) = 1 )). Taking square roots of ( L ) and ( U ) (both unipotent or upper triangular with determinant 1) is faster than taking the root of ( M ) directly, as triangular matrices have straightforward square roots (compute diagonal roots, then solve for off-diagonal elements recursively).
• Orthogonal/Symplectic Factorizations: For real SL matrices, we can decompose ( M ) into a product of Householder reflections and Givens rotations (all with determinant ±1). By pairing reflections to ensure the total determinant is 1, we can take square roots of each factor individually (rotations have trivial square roots, reflections can be combined to simplify) and recombine them—this avoids the cost of full diagonalization.

All these methods rely on the ( \det(M) = 1 ) constraint, so they won’t work for generic matrices, but they outperform diagonalization by eliminating redundant checks and leveraging SL-specific structure.

内容的提问来源于stack exchange，提问作者shai horowitz