1. ## conjugates

is it true that A multiplied by its conjugate A* is always greater than zero?

and if so how do i prove/show this?

2. Originally Posted by mathisthebestpuzzle
is it true that A multiplied by its conjugate A* is always greater than zero?

and if so how do i prove/show this?
depends on what $A$ is, but in general, no.

counter-example: take $A = \sqrt{2} - 2$. Then $A^* = \sqrt{2} + 2$. So that $A \cdot A^* = 2 - 4 = -2 < 0$

3. Originally Posted by Jhevon
depends on what $A$ is, but in general, no.

counter-example: take $A = \sqrt{2} - 2$. Then $A^* = \sqrt{2} + 2$. So that $A \cdot A^* = 2 - 4 = -2 < 0$
That's a strange sort of conjugation. I would have taken the conjugate of $\sqrt{2} - 2$ to be $-\sqrt{2} - 2$, in which case AA* is positive.

Originally Posted by mathisthebestpuzzle
is it true that A multiplied by its conjugate A* is always greater than zero?

and if so how do i prove/show this?
If A is meant to be a matrix and A* is its hermitian adjoint then AA* is always positive. Reason: $\langle AA^*x,x\rangle = \langle A^*x,A^*x\rangle = \|A^*x\|^2\geqslant0$.

4. another example is through the usage of complex numbers:

A=(a+j), A* = (a-j)
A.A*=(a^2 - aj + aj- j^2)
= a^2-(-1^(1/2))^2
=a^2 - (-1)
= a^2 + 1 which is always larger than 0

Complex Numbers

5. Originally Posted by Opalg
That's a strange sort of conjugation. I would have taken the conjugate of $\sqrt{2} - 2$ to be $-\sqrt{2} - 2$, in which case AA* is positive.
indeed. what i did is how we conjugate stuff on my planet. apparently on earth they do it differently. good stuff! (excuse: i was drunk -- that's always my excuse)

could we not use something like $A = 2 + \sqrt{145}$ then?

6. The word conjugate is one of those words in mathematics which has many different meanings. There is a standard algebraic/number theoretic definition. Let $F$ be an extension field over a smaller field $K$. Given two algebraic elements $\alpha,\beta\in F$ we say $\alpha,\beta$ are conjugates iff they has the same minimal polynomial. So $a+bi$ and $a-bi$ ( $b\not = 0$) are conjugates because they have the same minimal polynomials.

Though that definition above is applied to fields we can extend it a little and apply it to $\mathbb{Z}$ (which is not a field) as well. Given two algebraic number $\alpha,\beta$ we say $\alpha$ and $\beta$ are conjugates if they have the same minimal polynomial over $\mathbb{Z}$. In your case $\alpha = \sqrt{2} - 2$ and $\beta = -\sqrt{2} - 2$ have $x^2+4x+2$ as their minimal polynomial.

7. these conjugates are over a complex field.