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.

 

[blue_raver22]

I would like to clarify some points regarding the differences between SVD and eigenvalue decomposition (diagonalizability). First of all, both SVD and eigenvalue decomposition are methods used to decompose a matrix into simpler components. However, they have different purposes and applications.

Eigenvalue decomposition is primarily used for diagonalizing a square matrix. This means that the resulting diagonal matrix contains the eigenvalues of the original matrix on its main diagonal, and the corresponding eigenvectors form the columns of the transformation matrix. This is useful for solving systems of linear equations and performing other operations on the matrix.

On the other hand, SVD is used for decomposing any matrix, not just square matrices. It decomposes a matrix into three parts: a diagonal matrix Σ, and two orthogonal matrices U and V. The diagonal matrix Σ contains the singular values of the original matrix, which are related to the eigenvalues but not the same. The matrices U and V contain the left and right singular vectors, respectively, which are not necessarily the eigenvectors of the original matrix.

So, while every square matrix may be diagonalizable using eigenvalue decomposition, not every matrix can be decomposed into a diagonal matrix using SVD. Additionally, the diagonal matrix Σ in SVD is not the same as the diagonal matrix in eigenvalue decomposition, as they contain different values.

In summary, while both SVD and eigenvalue decomposition involve diagonalization, they have different purposes and applications. Therefore, it is important to understand the differences between them and use the appropriate method depending on the problem at hand.