What Makes a Hermitian Matrix Positive Definite?

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Discussion Overview

The discussion centers on the conditions under which a Hermitian matrix is considered positive definite, specifically in the context of a matrix \( V^*V \) derived from a matrix \( V \). Participants explore the implications of Hermitian properties and the relationship between diagonalizability and positive definiteness.

Discussion Character

  • Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant references a paper asserting that a matrix \( A \) being similar to a Hermitian matrix implies certain equivalences, questioning how \( V^*V \) can be shown to be positive definite.
  • Another participant discusses the existence of an H-orthonormal diagonalizing basis for \( V^*V \), arguing that the eigenvalues must be non-negative and thus suggesting that \( V^*V \) is positive definite due to its invertibility.
  • A participant seeks clarification on whether the basis vectors \( e_i \) are standard basis vectors in \( \mathbb{C}^n \) and expresses confusion regarding the notation used in the previous post.
  • One participant proposes a simpler argument involving the Hermitian product \( H(v,V^*Vv) \) to demonstrate that if \( v \) is non-zero, then \( Vv \) must also be non-zero, implying positive definiteness.
  • A later reply expresses appreciation for the previous contributions while indicating a need for further contemplation on the arguments presented.

Areas of Agreement / Disagreement

Participants appear to have differing views on the sufficiency of Hermitian properties for positive definiteness and the clarity of the arguments presented. There is no consensus reached on the implications of the properties discussed.

Contextual Notes

Participants note the invertibility of \( V \) and the implications of diagonalizability, but the discussion does not resolve the conditions under which \( V^*V \) is definitively positive definite.

BrainHurts
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on page 261 of this paper by J. Vermeer (http://www.math.technion.ac.il/iic/e..._pp258-283.pdf ) he writes

The following assertions are equivalent.
a) A is similar to a Hermitian matrix
b) A is similar to a Hermitian matrix via a Hermitian, positive definite matrix
c) A is similar to A* via a Hermitian, positive definite matrix

anyway the proof of a)\Rightarrowc) he writes:
"There exists a V\inMn(ℂ) such that VAV-1 is Hermitian, i.e. VAV-1=(VAV-1)*=(V*)-1A*V*. We obtain:

V*VA(V*V)-1=A*

V*V is the required Hermitian and positive definite matrix."

My questions is how do we know V*V is positive definite? I know it's Hermitian, i know that V*V has real eigenvalues and I know V*V is unitarily diagonalizable.

I don't think that V*V is Hermitian is enough right? Does this mean that a matrix B being Hermitian is a sufficient but not necessary condition for B to be positive definite?
 
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Let e_i be a H-orthonormal diagonalizing basis for V*V. Here H is the standard hermitian product on C^n. The existence of such a basis is equivalent to diagonalizability of V*V by a unitary matrix because the unitary condition is just that the columns are H-orthonormal.
The ith eigenvalue of V*V is then H(e_i,V^*Ve_i)=e_i^*V^*Ve_i=e_i^*V^*e_ie_i^*Ve_i=(e_i^*Ve_i)^*(e_i^*Ve_i)=H(e_i^*Ve_i,e_i^*Ve_i)=|e_i^*Ve_i|^2\geq 0

But "=0" is not possible since V*V is invertible. Therefor wrt the basis e_i, the matrix of V*V is diagonal with all nonpositive diagonal entries, so it's positive definite.
 
Hi this is really helpful thank you but I have one more question are the ei are the standard basis vectors in ℂn?

you wrote H(ei, V*Vei)=ei*V*Vei

=ei*V*eiei*Vei

I'm a little confused on where this eiei*, this is the matrix with the iith entry being 1 and zeroes everywhere else correct?
 
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Mh! Maybe my argument is flawed. try this much simpler one instead: H(v,V^*Vv) =H(Vv,Vv)=|Vv|^2 for all v. If v is not zero, neither is Vv since V is invertible.
 
i thought the first one was nice, umm let me think about this one for a bit, either way thanks!
 

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