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?


  • $\begingroup$ 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. $\endgroup$ – user56706 Feb 6 '13 at 19:34

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.

  • $\begingroup$ ohh right I see... and excuse me but what does the notation ∥Ax∥$^2$_2 mean? thank you very much! $\endgroup$ – dessskris Feb 6 '13 at 19:40
  • 1
    $\begingroup$ @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 $y$. $\endgroup$ – user17762 Feb 6 '13 at 19:43

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