SVD vs Eigenvalue Decompositon (Diagonalizability)

[eehsun]

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 wherever I am mistaken..
Thanks again.. :)

 

[td21]

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.

 

[HallsofIvy]

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.)

Those are sufficient conditions, not necessary conditions.

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.. :)

 

[td21]

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.

sorry... should be positive semidefinite.