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SVD vs Eigenvalue Decompositon (Diagonalizability) 
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#1
Jun1511, 06:14 PM

P: 9

Okay, I know that if I can't get n linearly independent eigenvectors out of a matrix A (∈ℝ^{nxn}), it is not diagonalizable
(and that some necessary conditions for diagonalizability in this regard may be being symmetric and/or having distinct eigenvalues.) This is how things are for the usual eigenvalue decomposition (A=SΛS^{1}), right? However, if I am not mistaken, there is not such a restriction present on a matrix for being able to perform the SVD decomposition (A=UΣV^{T}) on it. I mean, every matrix, irrespective of the state of its eigenvectors, can be decomposed into a three parts, one diagonal, two orthogonal, right? So doesn't this mean that every matrix is diagonizable, regardless of its eigenvectors? Thanks.. Edit: I know that the answer to the question is no but I don't understand why we don't consider Σ to be a diagonalized form of A. Please correct me whereever I am mistaken.. Thanks again.. :) 


#2
Jun1611, 12:47 AM

P: 90

hi!
U and V is different generally, so cannot be consider diagonalizable. Also, The diagonal entries of SVD is the square root of eigenvalue of A*A'(A'*A), so not the eigenvalue of A itself. They are the same iff A is normal. 


#3
Jun1611, 06:19 AM

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