14.设A是3阶矩阵，a_(1),a_(2),a_(3)是线性无关的3维列向量，且Aa_(1)=a_(1)+a_(2)+a_(3), Aa_(2)=2a_(2)+a_(3), Aa_(3)=2a_(2)+3a_(3),求A的全部特征值.

设 $ P = [a_1, a_2, a_3] $，则 $ P $ 可逆。由题意得 \[ AP = [Aa_1, Aa_2, Aa_3] = [a_1 + a_2 + a_3, 2a_2 + a_3, 2a_2 + 3a_3] = P \begin{bmatrix} 1 & 0 & 0 \\ 1 & 2 & 2 \\ 1 & 1 & 3 \end{bmatrix} = PB. \] 故 $ A $ 与 $ B $ 相似，特征值相同。求 $ B $ 的特征值： \[ \det(B - \lambda I) = (1 - \lambda) \begin{vmatrix} 2 - \lambda & 2 \\ 1 & 3 - \lambda \end{vmatrix} = (1 - \lambda)(\lambda^2 - 5\lambda + 4) = -(1 - \lambda)^2(\lambda - 4). \] 解得特征值为 $ \lambda_1 = \lambda_2 = 1 $，$ \lambda_3 = 4 $。 **答案：** $\boxed{1, 1, 4}$