## Positive Definite Matricies

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)$\Rightarrow$c) he writes:
"There exists a V$\in$Mn(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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 Recognitions: Gold Member Homework Help Science Advisor 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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