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eehsun

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Okay, I know that if I can't get n linearly independent eigenvectors out of a matrix A (∈ℝ

(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

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

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 wherever I am mistaken..

Thanks again.. :)

^{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 wherever I am mistaken..

Thanks again.. :)

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