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: It's not stated explicitly at that point, but presumably <math>A</math> is already taken to be Hermitian (as it needs to be for the Lanczos algorithm to work), which means it ''has'' an eigendecomposition of the form stated. Instead using the SVD decomposition in this argument won't work, because the entire point is that <math>U' U = I</math> so that the product telescopes! Possibly it would be clearer to just use <math>A = U \operatorname{diag}(\sigma_i) U^{-1}</math>, i.e., hold off on requiring orthogonality — the reason being that the paragraph in question is about the plain power method, which applies in a greater generality. [[Special:Contributions/130.243.68.202|130.243.68.202]] ([[User talk:130.243.68.202|talk]]) 13:01, 2 May 2017 (UTC)
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