每週問題 May 4, 2015

證明 n\times n 階冪零 (nilpotent) 矩陣 A 滿足 A^n=0

A square matrix A is said to be nilpotent if A^k=0 for some k>0. Show that if A is an n\times n nilpotent matrix, then A^n=0.

 
參考解答:

p 為最大正整數使得 A^p\neq 0,則存在一 n 維向量 \mathbf{x} 使得 A^p\mathbf{x}\neq\mathbf{0}A^{p+1}=0。如果我們能夠證明 \{\mathbf{x}, A\mathbf{x},\ldots,A^p\mathbf{x}\} 為一線性獨立集,則 p<n (因為 n 維向量空間不可能存在 n+1 個線性獨立向量),也就推得 A^n=0。考慮線性組合

\displaystyle  c_0\mathbf{x}+c_1A\mathbf{x}+\cdots+c_pA^p\mathbf{x}=\mathbf{0}

k 為最小整數滿足 c_k\neq 0k < p,則 A^k\mathbf{x}=\sum_{i=k+1}^p\frac{c_i}{c_k}A^i\mathbf{x}。因此,A^{p-k}(A^k\mathbf{x})=A^p\mathbf{x}\neq \mathbf{0},但對於 i>kA^{p-k}\left(\frac{c_i}{c_k}A^i\mathbf{x}\right)=\frac{c_i}{c_k}A^{p+i-k}\mathbf{x}=\mathbf{0}。這造成了矛盾,故證得所求。

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