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I found this proof in Serge Lang, Linear Algebra enter image description here

enter image description here

I couldn't understand how $$U^{-1} AU$$ is diagonalizable?

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  • $\begingroup$ Note that $U$ is a change of basis matrix from $\mathcal{B}'$ to $\mathcal{B}$, hence $U^{-1}AU=M_{\mathcal{B}'}^{\mathcal{B}'}(F)$ which is shown by diagonal by applying Theorem 4.3. $\endgroup$ – Melody Dec 14 '18 at 5:21
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    $\begingroup$ Thank you Melody. That's what I originally thought, but $U^{-1}AU = M_{B}^{B}(F)$ where $A=M_{B'}^{B'}(F)$ $\endgroup$ – Alvis Nordkovich Dec 14 '18 at 5:30

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