
#1
Dec912, 12:54 AM

P: 80

on page 261 of this paper by J. Vermeer (http://www.math.technion.ac.il/iic/e..._pp258283.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)[itex]\Rightarrow[/itex]c) he writes: "There exists a V[itex]\in[/itex]M_{n}(ℂ) such that VAV^{1} is Hermitian, i.e. VAV^{1}=(VAV^{1})*=(V*)^{1}A*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? 



#2
Dec912, 08:43 AM

Sci Advisor
HW Helper
PF Gold
P: 4,768

Let e_i be a Horthonormal 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 Horthonormal.
The ith eigenvalue of V*V is then [itex]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[/itex] 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. 



#3
Dec912, 01:15 PM

P: 80

Hi this is really helpful thank you but I have one more question are the e_{i} are the standard basis vectors in ℂ^{n}?
you wrote H(e_{i}, V*Ve_{i})=e_{i}*V*Ve_{i} =e_{i}*V*e_{i}e_{i}*Ve_{i} I'm a little confused on where this e_{i}e_{i}*, this is the matrix with the iith entry being 1 and zeroes everywhere else correct? 



#4
Dec912, 09:19 PM

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HW Helper
PF Gold
P: 4,768

Positive Definite Matricies
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.




#5
Dec912, 09:38 PM

P: 80

i thought the first one was nice, umm lemme think about this one for a bit, either way thanks!



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