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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 ?

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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.

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  • $\begingroup$ I don't know why $(Cx)^HCx > 0$ $\endgroup$ – user343207 Aug 5 '16 at 22:01
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    $\begingroup$ $y^Hy=|y_1|^2+\dots+|y_n|^2>0$ for every non-zero vector $y$. Set $y=Cx$. $\endgroup$ – carmichael561 Aug 5 '16 at 22:06
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    $\begingroup$ $|y_1|$ is the absolute value of the complex number $y_1$. Yes, $Cx$ is non-zero because $C$ is non-singular. $\endgroup$ – carmichael561 Aug 5 '16 at 22:18
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    $\begingroup$ $|y_1|^2$ 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 $x^HAx>0$. $\endgroup$ – carmichael561 Aug 5 '16 at 22:23
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    $\begingroup$ The first equality is just linearity of the inner product. The second is because $Ax=\lambda x$. $\endgroup$ – carmichael561 Aug 5 '16 at 22:27

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