# Spectral Theorem proof explanation

I found this proof in Serge Lang, Linear Algebra

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

• 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. – Melody Dec 14 '18 at 5:21
• Thank you Melody. That's what I originally thought, but $U^{-1}AU = M_{B}^{B}(F)$ where $A=M_{B'}^{B'}(F)$ – Alvis Nordkovich Dec 14 '18 at 5:30