If $A ∈M_n(C)$ is such that $(Ax,x) = 0$ for all $x ∈C^n$, then $A = 0$

Question

Which of the following statements are true?

$1.$ If $A ∈M_n(C)$ is such that $(Ax,x) = 0$ for all $x ∈C^n$, then $A = 0$.

$2.$ If $A ∈M_n(C)$ is such that $(Ax,x) ≥ 0$ for all $x ∈C^n$, then $A = A^∗$.

I know that $A$ is skew symmetric Matrix iff $x^tAx = 0$ for all $x ∈R^n$ and $A ∈M_n(R)$.

The question posted by Neha Gupta had an WRONG answer. That's why I posted it again.

Answer 1

Both are true. For the first one, the canonical way is to use the Polarization identity to show that $(Ax,y)=0$ for all $x,y$. A less standard way is to notice that $[x,y]=(Ax,y)$ is a sesquilinear form (since $[x,x]=(Ax,x)=0\geq0$), so one has, using Cauchy-Schwarz for $[\cdot,\cdot]$, $$ \|Ax\|^2=(Ax,Ax)=[x,Ax]\leq [x,x]^{1/2}[Ax,Ax]^{1/2}=(Ax,x)^{1/2}(A^2x,Ax)^{1/2}=0. $$ Thus $Ax=0$ for all $x$.

For the second one, $$ (A^*x,x)=(x,Ax)=\overline{(Ax,x)}=(Ax,x). $$ So $((A^*-A)x,x)=0$ for all $x$.

Answer 2

Some insight into the problems via the field (of values) of $A$ (also called the numerical range of $A$).

Given $A \in M_n(\mathbb{C})$, the field (of values) of $A$, denoted by $F(A)$, is defined by $$ F(A) = \{ z^*Az\mid z^*z=1\}. $$ It is well-known that this set is a compact, convex subset of $\mathbb{C}$ that contains the spectrum of $A$. It is also known, and otherwise easy to establish, that $F(A) = \alpha \in \mathbb{C}$ if and only if $A = \alpha I$. This establishes (1).

Let $$H(A):=(A+A^*)/2$$ and $$S(A) :=(A-A^*)/2.$$ Notice that $A = H(A) + S(A)$ and that $H(A)$ and $iS(A)$ are both Hermitian.

It is known that $F(H(A)) = \text{Re}(F(A))$ and $F(S(A)) = \text{Im}(F(A))$. Thus, if $\langle Ax,x\rangle \ge 0$ for any $x\in \mathbb{C}^n$, then $F(S(A)) = \text{Im}(F(A)) = \{0\}$. Following (1), $S(A) = 0$, i.e., $A = A^*$.

Answer 3

Statement $1$ is true as $(A(x+y),x+y))=0$ gives us $(Ax,y)+(Ay,x)=0$ and $(A(x+iy),x+iy)$ gives us $(Ax,y)-(Ay,x)=0$.
Statement $2$ is also true but I am satisfied with Martin's answer.