线性代数提高向量组与秩线性代数做题本（二）解答版选择题查看解析已掌握收入错题本题讨论不看本题#10438设 A=[1254],P=[1101],B=P−1A100PA = \left\lbrack \begin{matrix} 1 & 2 \\ 5 & 4 \end{matrix}\right\rbrack,P = \left\lbrack \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}\right\rbrack,B = {P}^{-1}{A}^{100}PA=[1524],P=[1011],B=P−1A100P,则 B+EB + EB+E 的线性无关的特征向量可以为‾\underline{\qquad}.A. [01],[75]\left\lbrack \begin{matrix} 0 \\ 1 \end{matrix}\right\rbrack,\left\lbrack \begin{matrix} 7 \\ 5 \end{matrix}\right\rbrack[01],[75]B. [−21],[−35]\left\lbrack \begin{matrix} - 2 \\ 1 \end{matrix}\right\rbrack,\left\lbrack \begin{matrix} - 3 \\ 5 \end{matrix}\right\rbrack[−21],[−35]C. [1−1],[25]\left\lbrack \begin{matrix} 1 \\ - 1 \end{matrix}\right\rbrack,\left\lbrack \begin{matrix} 2 \\ 5 \end{matrix}\right\rbrack[1−1],[25]D. [01],[32]\left\lbrack \begin{matrix} 0 \\ 1 \end{matrix}\right\rbrack,\left\lbrack \begin{matrix} 3 \\ 2 \end{matrix}\right\rbrack[01],[32]
