Dear list members,
I would like to share some doubts about eigenvalues theory and singular value decomposition with you.
Let me suppose to have a matrix called M, with M (n by n), real and not singular (full rank).
Eigenvalue theory says that I can decompose M as
M = Q D Q^(-1)
where
Q is the (n by n) matrix whose columns are the eigenvectors of M. Q is not necessarily orthogonal.
D is the diagonal matrix whose diagonal elements are the corresponding eigenvalues. Not necessarily all the eigenvalues are real.
This decomposition can be obtained only if the n eigenvectors are linearly independent. My conditions on M are not sufficient. In fact, for example the matrix M=
1 1
0 1
is square, real and not singular but it cannot be decomposed according to eigenvalues theory.
In fact the characteristic polynomial is (1-x)^(2) and therefore the algebraic multiplicity is 2, while the
geometrical multiplicity is rk(M-I) = 1. (I is the identity matrix because both eigennvalues are 1)
Singular value decomposition says that M can be decomposed as
M = U S V^(T)
where
U and V are orthogonal matrices
S is a diagonal matrix with singular values of M
Can singular value decomposition always be applied, according to my conditions on M? Why?
If yes, how can the small example above reported be developed?
If yes, can I always deal with S instead of M? S would be much much simpler.
thank you for your attention and your help
Stefano
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