- #1

snakebite

- 16

- 0

## Homework Statement

Hello,

I have the following problem:

Suppose A is a hermitian matrix and it has eigenvalue [tex]\lambda[/tex] <=0. Show that A is not positive definite i.e there exists vector v such that (v^T)(A)(v bar) <=0

## The Attempt at a Solution

Let w be an eigenvetor we have the following equalities which are equivalent:

Aw=[tex]\lambda[/tex]w

A(w bar) = [tex]\lambda[/tex] (w bar) (i am not sure about this equality)

(w^t)(A)(w bar) = (w^t)[tex]\lambda[/tex] (w bar)

[(w^t)(A)(w bar)]^T = [(w^t)[tex]\lambda[/tex] (w bar)]^T

(w bar)^T*(A)^T*w = (w bar)^T*[tex]\lambda[/tex]*w

[(w bar)^T*(A)^T*w]bar = [(w bar)^T*[tex]\lambda[/tex]*w]bar

(w^T)(A bar)^T(w bar) = (w^T)[tex]\lambda[/tex](w bar)

(w^T)A(w bar) = (w^T)[tex]\lambda[/tex](w bar)

but i cannot prove why (w^T)[tex]\lambda[/tex](w bar) is negative, assuming taht the 2ndd equality is true.

Thank you