If a matrix $M$ is positive definite, I would like to show that there exists a self adjoint matrix s.t. $M=SS^{T}$. I have a proof, and it comes from operators in Linear Algebra Done Right by Sheldon Axler. However, in his proof he bunches a bunch of properties (6 to be total) under a positive operator, saying that each implies the other. His proof goes from the first property, then proves the second, then all the way to 6, then to prove 1. Hence, I do not really have a self contained proof showing that if a matrix is positive definite, then there exist a self adjoint matrix s.t. $M=SS^{T}$. Does anyone know if there is a self contained proof? thank you!

  • $\begingroup$ $M$ should also be Hermitian(or say, symmetric). Then this can be proved by Cholesky decomposition. $\endgroup$ – enigne Apr 24 '14 at 10:30
  • $\begingroup$ This would depend on how you define positive definite...I have a nice short proof but it makes use of spectral decomposition, which requires $M$ to be diagonalizable...this you can only assume if $M$ is Hermitian. Unfortunately there are different definitions of positive definite, some which does not require $M$ to be Hermitian, and then the proof won't work - see this: math.stackexchange.com/questions/473118/… $\endgroup$ – Christiaan Hattingh Apr 24 '14 at 20:36

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