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2阶矩阵在ℂ不可对角化 求$a$

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hbghlyj posted 2023-5-27 04:47 |Read mode
Exercise. Find all $a \in \mathbb{C}$ that make the matrix $A=\begin{pmatrix}a & 2-a \\ a-2 & 1\end{pmatrix} \in M_2(\mathbb{C})$ non-diagonalizable.
解:
$(a-1)^2+4(2-a)(a-2)=0\Leftrightarrow a=\frac53$ or $a=3$.
$A=\lambda I\Leftrightarrow a=1$. 所以当$a=\frac53$或3时$A$不可对角化.
$\therefore a=\frac53$ or $a=3$.
Defective matrix

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