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I would not call the SVD an "extension" of the eigen decomposition. The SVD is also defined for square matrices and is not equal to the eigen decomposition exce
by blt 6y ago
I would not call the SVD an "extension" of the eigen decomposition. The SVD is also defined for square matrices and is not equal to the eigen decomposition except for the case of symmetric matrices.
This leads to the question: for a non-symmetric square diagonalizable matrix, what is the difference between the eigenvalues and the singular values? A good heuristic for applications is that the eigenvalues capture the asymptotic behavior of repeating the map, i.e. the behavior of A^k as k goes to infinity, while the singular values capture the transient behavior of applying the map once.
For example, consider the matrix
0.9 x
0 0.9.
As x goes to infinity, its largest singular value becomes infinity, while its eigenvalues remain smaller than one. This means that there exists a vector (0 1) whose image under A is huge, but whose image under A^k still approaches the zero vector for large k.
A good book on these topics is "Spectra and Pseudospectra" by Trefethen and Embree.
- solinent 6y agoGood point, while I don't need this clarification others might. I shouldn't have implied that they are the same for square matrices implicitly, hence my failed attempt at using "similar". However, they're still an extension in the sense that singular values do correspond to (square of the) eigenvalues of the square matrix A A^T and A^T A.
- blt 6y agoYeah I guess "extension" is not a precisely defined word. I wanted to make sure other readers didn't interpret like "generalization", meaning that the original narrow concept is precisely a special case of the extension.
- solinent 6y agoYeah, I wouldn't have used extension for a math audience, probably just "related concept", or "application of". In the software context it usually means the opposite--something specific that makes use of some existing functionality.