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Exercise 3.3.4

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$K = U Σ {\bar{V}}^{T}$ , and ${\bar{K}}^{T} = V Σ Ū^{T}$ where $Ū^{T} U = {\bar{V}}^{T} V = I$ , so we have $H = {\bar{K}}^{T} K = V Σ Ū^{T} U Σ {\bar{V}}^{T} = V Σ^{2} {\bar{V}}^{T}$ , the eigenvalues of $H$ is thus $Σ^{2}$ , so $σ_{i} (H) = | σ_{i} (K) |^{2}$ and the singular values of $H$ decay quickly because $K$ ’s eigenvalues decay quickly from the “Sylvester test”

niuers

2020-03-20 00:00

Exercise 3.3.4

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