SVD vs Eigenvalue Decompositon (Diagonalizability)

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Discussion Overview

The discussion revolves around the concepts of Singular Value Decomposition (SVD) and Eigenvalue Decomposition, particularly focusing on the conditions for diagonalizability of matrices. Participants explore the differences between these decompositions and the implications of having linearly independent eigenvectors.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants assert that a matrix is not diagonalizable if it does not have n linearly independent eigenvectors, with conditions such as being symmetric or having distinct eigenvalues being mentioned as sufficient for diagonalizability.
  • Others argue that SVD can be performed on any matrix, regardless of the state of its eigenvectors, suggesting that this implies every matrix is diagonalizable in some sense.
  • It is noted that the diagonal entries of the SVD are the square roots of the eigenvalues of A*A' or A'A, not the eigenvalues of A itself, and that they coincide only if A is normal.
  • Some participants clarify that the conditions mentioned for diagonalizability are sufficient but not necessary.
  • There is a question raised about why Σ in the SVD is not considered a diagonalized form of A, indicating some confusion about the definitions involved.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between SVD and diagonalizability, with no consensus reached on whether every matrix can be considered diagonalizable due to the existence of SVD.

Contextual Notes

There are limitations in the discussion regarding the definitions of diagonalizability and the conditions under which SVD and eigenvalue decomposition apply, which remain unresolved.

eehsun
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Okay, I know that if I can't get n linearly independent eigenvectors out of a matrix A (∈ℝnxn), 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=UΣVT) 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.. :)
 
Last edited:
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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.
 
eehsun said:
Okay, I know that if I can't get n linearly independent eigenvectors out of a matrix A (∈ℝnxn), 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=UΣVT) 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 said:
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.
 
Last edited:

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