Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If I'm not mistaken, if a matrix M has its conjugate M*=M, then M is Hermitian.

In this case then, am I asked to show that (A*A)*=A*A ? What does it have to do with A being invertible though?

And positive definite? How do I show that it's positive definite?

Thanks!

share|improve this question
    
For every $n\times n$ complex matrix $A$, $A^* A$ is Hermitian and positive. Invertibility of $A$ is necessary to prove that $A^\ast $ is definite. –  Vahid Shirbisheh Feb 6 '13 at 19:34
add comment

1 Answer

up vote 6 down vote accepted

In general, we have $(AB)^{*} = B^*A^*$. Hence, we get that $$(A^*A)^* = A^* (A^*)^* = A^*A$$ We need that $A$ is invertible to prove that $(A^*A)$ is positive definite. Consider $x \in \mathbb{C}^n$, we then have $$x^*(A^*A)x = (Ax)^*(Ax) = \Vert Ax \Vert_2^2 \color{red}{\geq} 0$$ If $A$ is invertible, then the nullspace of $A$ is trivial i.e. if $x \neq 0$, then $Ax \neq 0$. Hence, we have that if $x \neq 0$, then $$\Vert Ax \Vert_2^2 \color{blue}{>} 0$$ This shows that if $A$ is invertible, then $A^*A$ is positive definite.

share|improve this answer
    
ohh right I see... and excuse me but what does the notation ∥Ax∥$^2$_2 mean? thank you very much! –  dessskris Feb 6 '13 at 19:40
    
@dessskris $\Vert y \Vert_2$ denotes the two norm of the vector $y$, $\Vert y \Vert_2^2$ denotes the square of the two norm of the vector $y$. –  user17762 Feb 6 '13 at 19:43
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.