已知矩阵A=(aij)n×n的秩为n-1，求A的伴随矩阵A*的特征值和特征向量．
参考答案：由r(A)=n-1可知|A|=0，且A的列向量中有n-1个是线性无关的．
又由A*A=|A|E=0，知A的n-1个线性无关的列向量都是齐次线性方程组A*x=0(即(0E-A*)x=0)的解向量，故A=0至少是A*的n-1重特征值．而上述的n-1个列向量即为对应的线性无关的特征向量．
另外，A*的全部特征值之和等于A*的主对角线元素之和A11+A22+…+Ann(其中Aij为A中元素aij的代数余子式，i，j=1，2，…，n)，所以A*的第n个特征值为[*]由r(A)=n-1，可证得r(A*)=1，不妨设A*的第一列[*]为A*的列向量组的极大无关组，则A*的其他列可由第一列线性表示，即存在常数k2，…，kn使得
[*]
于是有
[*]
[*]
因此，[*]，为A*的属于特征值[*]的特征向量．
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