# Positive Definite Matricies

by BrainHurts
Tags: definite, matricies, positive
 P: 84 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?
 Sci Advisor HW Helper PF Gold P: 4,771 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.