Mathematica 问题：以符号方式求解矩阵方程 AX=\lambdaBX

Question

我是 Mathematica 的新手，我正在尝试以

的形式求解矩阵方程

AX = \lambda BX

这里，A和B是下面的4*4矩阵，\lambda是一个值，X是特征向量- 4*1 矩阵.

A = {{a1 + b1,  c,  d, f},
     {c,  a2 + b2 , f , e},
     {d , f , a3 + b1 , c},
     { f,  e , c,  a4 + b2}}

B = {{1,  0,  0 , 0},
     {0,  1 , 0 , 0},
     {0 , 0 , -1 , 0},
     {0,  0 , 0,  -1}}

我想求解这个矩阵方程并使用 a1,a2,a3,a4,b1,b2,c,d,e,f 等得到 \lambda 的符号解

谁能告诉我，将不胜感激

此致，

麦克

Answer 1

参见 Wolfram: Matrix Computations - 特别是 'Generalized Eigenvalues'.

部分

For n×n matrices A, B the generalized eigenvalues are the n roots of its characteristic polynomial, p() = det(A - B). For each generalized eigenvalue, λ ∊ λ(A, B), the vectors, , that satisfy

A χ = λ B χ

are described as generalized eigenvectors.

使用符号值的示例：

matA = {{a11, a12}, {a21, a22}};
matB = {{b11, b12}, {b21, b22}};

Eigenvalues[{matA, matB}]

{(1/(2 (-b12 b21+b11 b22)))(a22 b11-a21 b12-a12 b21+a11 b22-Sqrt[(-a22 b11+a21 b12+a12 b21-a11 b22)^2-4 (-a12 a21+a11 a22) (-b12 b21+b11 b22)]),(1/(2 (-b12 b21+b11 b22)))(a22 b11-a21 b12-a12 b21+a11 b22+Sqrt[(-a22 b11+a21 b12+a12 b21-a11 b22)^2-4 (-a12 a21+a11 a22) (-b12 b21+b11 b22)])}

Eigenvectors[{matA, matB}]

...

Mathematica 问题：以符号方式求解矩阵方程 AX=\lambdaBX

Mathematica Issue: solving matrix equation AX=\lambdaBX Symbolically

matrix

wolfram-mathematica

equation-solving