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因为实对称矩阵的特征值都是实数,而秩为2的实对称矩阵必有两个特征值为0。这可以通过矩阵的奇异值分解来证明:设矩阵为$A=U\\Sigma V^T$,其中$\\Sigma$是对角矩阵,对角线上的元素为矩阵$A$的奇异值。由于矩阵$A$是实对称矩阵,$U$和$V$可以取为正交矩阵,即$U^{-1}=U^T$,$V^{-1}=V^T$。因此,$$A^TA = V\\Sigma^T U^TU\\Sigma V^T = V\\Sigma^T \\Sigma V^T$$可以看出,矩阵$A^TA$也是实对称矩阵,并且它的特征值为矩阵$A$的奇异值的平方。由于矩阵的秩为2,$\\Sigma$只有两个非零奇异值$\\sigma_1$和$\\sigma_2$,因此矩阵$A^TA$有两个非零特征值$\\sigma_1^2$和$\\sigma_2^2$,另外还有特征值为0,即$A^TA$的秩为1的特征值。由于$A$和$A^TA$有相同的非零特征值,因此矩阵$A$也有两个非零特征值$\\sigma_1$和$\\sigma_2$,另外还有特征值为0。