I am trying to relate eigenvalues with singular values. In particular,

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The discussion focuses on the relationship between eigenvalues and singular values of a matrix A. It establishes that for any eigenvalue of A, the absolute value lies between the smallest and largest singular values of A, specifically stating that smallestSingularValue(A) <= |anyEigenValue(A)| <= largestSingularValue(A). The user mentions using Schur decomposition to order eigenvalues similarly to singular values but struggles to prove their relationship. Additionally, it is noted that singular values are derived from the square roots of the eigenvalues of A^T.A.

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I am trying to relate eigenvalues with singular values. In particular, I'm trying to show that for any eigenvalue of A, it is within range of the singular values of A. In other words,

smallestSingularValue(A) <= |anyEigenValue(A)| <= largestSingularValue(A).

I've tried using Schur decomposition, and then permuting the matrix so that the eigenvalues are ordered like the singular values. But I can't determine their relationship. Any help would be appreciated.
 
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The singular values of A are equal to square roots of the eigenvalues of A^T.A.

Eigenvalues only exist for square matrices. Singular values exist for rectangular matrices as well as square ones.
 


well, this is true:
smallestSingularvalue(T)*|v| <= |Tv| <= largestsingularvalue(T)*|v|
try proving that
 

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