If the inner product of Ax with x is 0 for all x, then A=0. Given matrix $A\in M_{n}(\mathbb{C})$, if $\left<Ax,x\right>=0$ for all $x\in \mathbb{C^n}$, then $A=0_{n}$. (Here $\left<a,b\right> = b^{\ast}a$ where "*" is the conjugate transpose.)
Can anyone help me prove this? By Schur's theorem, A is similar to a strictly upper-triangular matrix, which is nilpotent. I'm not sure if this will help me. 
ETA: In the comments I linked to an answer in which the author states "for $\lambda\neq 0$ we get $\left<Ax,y\right>=-\frac{\bar{\lambda}}{\lambda}\left<Ay,x\right>$. For fixed $x,y$ we can vary $\frac{\bar{\lambda}}{\lambda}$, so that $\left<Ax,y\right>=0$. Can someone help me understand that last statement, i.e. how we can get $\frac{\bar{\lambda}}{\lambda}$ to be $0$?
 A: You have the right idea. Let $A = (a_{i,j})_{1 \leq i,j \leq n} \in \mathcal{M}_{n}(\mathbb{C})$. From Schur theorem, there exist an orthonormal basis $\mathcal{B} = \big( \varepsilon_{1},\ldots,\varepsilon_{n})$ of $\mathbb{C}^{n}$ such that, if $a$ denotes the endomorphism of $\mathbb{C}^{n}$ associated to $A$, then : $\mathrm{Mat}(a,\mathcal{B}) = T = \mathrm{Diag}\big( \lambda_{1},\ldots,\lambda_{n} \big)$ where $T$ is an upper triangular matrix. $\mathcal{B}$ is such that : $\forall j \in \lbrace 1,\ldots,n \rbrace, \; \mathrm{Vect}(\varepsilon_{1},\ldots,\varepsilon_{j})$ is $a$-stable, meaning that : $\forall j \in \lbrace 1,\ldots,n \rbrace, \; A\varepsilon_{j} \in \mathrm{Vect}(\varepsilon_{1},\ldots,\varepsilon_{j})$. We can say that :
$$ \left\langle A\varepsilon_{1},\varepsilon_{1} \right\rangle = \left\langle \lambda_{1}\varepsilon_{1},\varepsilon_{1} \right\rangle = \lambda_{1} \Vert \varepsilon_{1} \Vert^{2} = \lambda_{1} = 0. $$
and
$$ \forall j \in \lbrace 2,\ldots,n \rbrace, \; \left\langle A\varepsilon_{j},\varepsilon_{j} \right\rangle = \left\langle \lambda_{j}\varepsilon_{j} + \underbrace{e}_{\in \mathrm{Vect}(\varepsilon_{1},\ldots,\varepsilon_{j-1})},\varepsilon_{j} \right\rangle = \lambda_{j} \Vert \varepsilon_{j} \Vert^{2} = \lambda_{j} = 0. $$
because $\mathcal{B}$ is an orthonormal basis.
As a consequence : $\forall j \in \lbrace 1,\ldots,n \rbrace, \; \lambda_{j} = 0$. Therefore, $A$ is a matrix which is similar to the null matrix. It proves that $A=0$.
