Okay, I know that if I can't get n linearly independent eigenvectors out of a matrix A (∈ℝ(adsbygoogle = window.adsbygoogle || []).push({}); ^{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.. :)

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# SVD vs Eigenvalue Decompositon (Diagonalizability)

**Physics Forums | Science Articles, Homework Help, Discussion**