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I know that this result is true when the ground field is $\mathbb{C}$. (Though I don't remember why.) Does it also hold for algebraically closed fields? Can someone give me a hint as to why this is true?

Elle Najt
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  • The reuse of $T$ is confusing. Do you mean to say that an arbitrary matrix $A$ can be written as $A = B^TB$? – Ben Grossmann Oct 09 '14 at 01:53
  • Also, do you specifically mean the transpose as opposed to the conjugate transpose (AKA adjoint)? – Ben Grossmann Oct 09 '14 at 01:54
  • @Omnomnomnom I'm pretty sure that it's just the transpose, since one only specifies that the original matrix $A$ is symmetric. – Elle Najt Oct 09 '14 at 01:58
  • @Omnomnomnom This is used in the complex case on page 58 of Miranda's Riemann Surfaces and Algebraic Curves in order to prove that the smooth conics in $P^2$ are isomorphic to $P^1$. I want to prove this for conics in $P^2$ for some arbitrary (algebraically closed) field, so it was natural to ask about this extension. So for my purposes it would really need to be the transpose. – Elle Najt Oct 09 '14 at 02:00
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    This is apparently a consequence of Takagi's factorization, and this is the first I'm hearing of it. Neat stuff. – Ben Grossmann Oct 09 '14 at 02:07

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Your factorization is an immediate consequence of Takagi's factorization, which itself is a corollary of the following theorem:

From Horn and Johnson: (p. 203, first edition)

Let $A \in M_n$ be given. There exists a unitary $U \in M_n$ and an upper triangular $\Delta$ such that $A = U\Delta U^T$ if and only if all the eigenvalues of $A \overline A$ are real and non-negative. Under this condition, the main diagonal entries of $\Delta$ may be chosen to be non-negative.

It seems that the answer lies behind this theorem. The question is fundamentally whether it can be extended to handle $A \in M_n(\Bbb F)$ for arbitrary (algebraically closed fields) $\Bbb F$.

Ben Grossmann
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