Generalized eigenvector唯一性

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设矩阵$A$的Jordan标准型只有1个Jordan块, $X,Y$是两个modal matrix,
则$B=XY^{-1}$是upper triangular Toeplitz matrix.
当$n=5$时$B$形如($a,b,c,d,e$任意数,$a\ne0$.)
\begin{bmatrix}a&b&c&d&e\\0&a&b&c&d\\0&0&a&b&c\\0&0&0&a&b\\0&0&0&0&a\end{bmatrix}
基本上这是因为:
设$N$为幂零矩阵, $𝐱_5\in\ker N^5\setminus\ker N^4$, 则$𝐲_5=a𝐱_5+b𝐱_4+c𝐱_3+d𝐱_2+e𝐱_1\in\ker N^5\setminus\ker N^4$.

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设$N=A-λI$.

设$X=[\mathbf {x} _{1}…\mathbf {x} _{n}]$的Jordan chain为 $ \mathbf {x} _{n-1}=N\mathbf {x} _{n}, $ $ \mathbf {x} _{n-2}=N\mathbf {x} _{n-1}, $         $ ⋮ $ $ \mathbf {x} _{1}=N\mathbf {x} _{2}, \mathbf {x} _{1}∈\ker N$.

设$Y=[\mathbf {y} _{1}…\mathbf {y} _{n}]$的Jordan chain为 $ \mathbf {y} _{n-1}=N\mathbf {y} _{n}, $ $ \mathbf {y} _{n-2}=N\mathbf {y} _{n-1}, $         $ ⋮ $ $ \mathbf {y} _{1}=N\mathbf {y} _{2}, \mathbf {y} _{1}∈\ker N$.

因为$\dim\ker N=1$, 所以$∃λ_{11}:\mathbf {x} _{1}=λ_{11}\mathbf {y} _{1}$. 设$λ_{22}=λ_{11}$.
$\mathbf {x} _{1}-λ_{11}\mathbf {y} _{1}=0⇒\mathbf {x} _{2}-λ_{22}\mathbf {y} _{2}∈\ker N⇒∃λ_{21}:\mathbf {x} _{2}-λ_{22}\mathbf {y} _{2}=λ_{21}\mathbf {y} _{1}$. 设$λ_{32}=λ_{21},λ_{33}=λ_{22}$.
$\mathbf {x} _{2}-λ_{21}\mathbf {y} _{1}-λ_{22}\mathbf {y} _{2}=0⇒\mathbf {x} _{3}-λ_{32}\mathbf {y} _{2}-λ_{33}\mathbf {y} _{3}∈\ker N⇒∃λ_{31}:\mathbf {x} _{3}-λ_{32}\mathbf {y} _{2}-λ_{33}\mathbf {y} _{3}=λ_{31}\mathbf {y} _{1}$.
        $ ⋮ $

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Output: True

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Commuting matrices#Examples

Jordan blocks commute with upper triangular matrices that have the same value along bands.