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It is a tough battle between SVD and the concept of eigenvalues/vectors. SVD is only meaningful for linear operators between inner product spaces, whereas eigen
by lqr 4y ago
It is a tough battle between SVD and the concept of eigenvalues/vectors. SVD is only meaningful for linear operators between inner product spaces, whereas eigenvalues/vectors do not even require a norm. On the other hand, eigenvectors are only meaningful for linear operators from a space to itself.
- snovv_crash 4y agoSVD is just an extension of the Eigenvector Decomposition to allow the two orthogonal matrices to not be equal. Think of SVD as Eigenvectors of your data both in a rowwise and colwise perspective and intuitively it works out pretty well.