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Possible Duplicate:
Simultaneous diagonalization

Assume $V$ is a n-dimension vector space over a number field $\Bbb{K}$, $\mathscr{A}$ and $\mathscr{B}$ are two linear transforms in $V$, and satisfy $$\mathscr{A}\mathscr{B}=\mathscr{B}\mathscr{A}$$, If both of them are diagonalizable . show that there exist a basis$\{\gamma_1,\gamma_2,\cdots,\gamma_n\}$, such that $$\mathscr{A}(\gamma_1,\gamma_2,\cdots,\gamma_n )=(\gamma_1,\gamma_2,\cdots,\gamma_n )\begin{bmatrix}a_1 & &\\ & \cdots& \\&&a_n\end{bmatrix}$$ and $$\mathscr{B}(\gamma_1,\gamma_2,\cdots,\gamma_n )=(\gamma_1,\gamma_2,\cdots,\gamma_n )\begin{bmatrix}b_1 & &\\ & \cdots& \\&&b_n\end{bmatrix}$$


my approach:

Assume $\lambda_1,\lambda_2,\cdots,\lambda_k$ are all distinct eigenvalue of $\mathscr{A}$ then $$V=V_{\lambda_1}\oplus V_{\lambda_2} \oplus \cdots \oplus V_{\lambda_k}$$

choose maximum linear independent vectors from each $V_{\lambda_i}$, denote by $\{\alpha_1,\alpha_2,\cdots,\alpha_n\}$ ,under this basis,the corresponding matrix of $\mathscr{A}$ is diagonal,but the matrix of $\mathscr{B}$ is only quasi diagonal.

it seems i can't go any further.

thanks very much.

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