congruence hermitian matrix $I_n$. Conclusions.

Hermitian matrix $A\in\mathbb{C}^{n,n}$ is congruent to matrix $I_n$. Then:
a. $\forall_{\overrightarrow{x}\in\mathbb{C}^n\setminus\{0\}} \overrightarrow{x}^HA\overrightarrow{x} > 0$
b. all eigenvalues of matrix A are positive real numbers
c. $A=I_n$

c. It is not true.
From congruency we conclude: $A=C^HI_nC=C^HC$ where $C$ is some nonsingular matrix. $\left[ \begin{array}{ccc} 5 & 0 \\ 0 & 5 \\ \end{array} \right]$ Then $A= \left[ \begin{array}{ccc} 25 & 0 \\ 0 & 25 \\ \end{array} \right]\neq I_n$

I have no idea how to solve b. and a. Can you help me ?

If $A=C^HC$ with $C$ non-singular, then for all non-zero $x\in\mathbb{C}^n$ we have $$x^HAx=x^H(C^HC)x=(Cx)^HCx>0$$ since $C$ is non-singular.
If $\lambda$ is an eigenvalue of $A$, choose an eigenvector $x$ for this eigenvalue, and normalize $x$ so that $x^Hx=1$. Then $$\lambda =\lambda(x^Hx)=x^H(\lambda x)=x^HAx>0$$ so $\lambda$ is a positive real number.
• I don't know why $(Cx)^HCx > 0$ – user343207 Aug 5 '16 at 22:01
• $y^Hy=|y_1|^2+\dots+|y_n|^2>0$ for every non-zero vector $y$. Set $y=Cx$. – carmichael561 Aug 5 '16 at 22:06
• $|y_1|$ is the absolute value of the complex number $y_1$. Yes, $Cx$ is non-zero because $C$ is non-singular. – carmichael561 Aug 5 '16 at 22:18
• $|y_1|^2$ is not the same as $y_1^2$. What if $y_1=i$? Given any non-zero vector $x$, it can always be normalized so that $x^Hx=1$, by dividing by $\sqrt{x^Hx}$. And the first part of my answer showed that $x^HAx>0$. – carmichael561 Aug 5 '16 at 22:23
• The first equality is just linearity of the inner product. The second is because $Ax=\lambda x$. – carmichael561 Aug 5 '16 at 22:27