线性代数提高相似对角化linear-algebra-3:线性代数:3:三-相似理论:填空题:7:82填空题不看本题本题讨论已掌握查看解析收入错题#11081设 AAA 是 3 阶矩阵, b=(3,3,3)Tb = {\left(3,3,3\right) }^{T}b=(3,3,3)T,方程组 Ax=b{Ax} = bAx=b 有通解 k1(−1,2,1)T+k2(0,−1,1)T+(1,1,1)T{k}_{1}{\left(-1,2,1\right) }^{T} + {k}_{2}{\left(0, - 1,1\right) }^{T} + {\left(1,1,1\right) }^{T}k1(−1,2,1)T+k2(0,−1,1)T+(1,1,1)T, 其中 k1,k2{k}_{1},{k}_{2}k1,k2 是任意常数,则 AAA 的特征值为‾\underline{\hspace{7.5em}}.