Complex vector space identity Let $A$ be an $n \times n$ complex matrix and let $u \in \mathbb{C}$ be a column vector. Suppose $u^*Au=0$ for all $u$.  What's the most straightforward way to see that $A=0$?
 A: If one takes for $u$ the vector $\begin{pmatrix}x_1&\ldots& x_n \end{pmatrix}$ then the proposed equation becomes a quadratic form that is zero for all the values of the indeterminates $x_1, \ldots,x_n$ so its coefficients must  all be zero.
A: We first rewrite $A$ into $H+S$ where $H:=\frac{1}{2}(A+A^*)$ is hermitian and $S:=\frac{1}{2}(A-A^*)$ is skew-hermitian. For each $u$, $u^* H u$ is real and $u^* S u$ is purely imaginary. If $u^* A u=0$ is zero for all $u$, then both $u^* H u$ and $u^* S u$ are zero for all $u$. Hermitian and skew-hermitian matrices are normal and hence diagonalizable (over $\Bbb C$). Therefore, either $H$ is the zero matrix or it has some nonzero eigenvalue $\lambda$; the same is true for $S$. In the latter case, $u_\lambda^* H u_\lambda=\lambda \|u\|^2\neq 0$, contradicting the above conclusion, so $H=0$. Similarly, $S=0$ and therefore $A=0$.
A: First, as suggested above, use $u$ as the basis vectors. This gives the diagonal elements, $A_{ii}$ must be zero. Next, use $u = \mathbb{1}$(vector of all ones). This gives $\sum_{i,j}A_{ij} = 0$. Using this with above fact you get $\sum_{i >j} A_{ij} + \sum_{i <j} A_{ij} = 0$. Finally, set $u = e_{i} + e_{j}$ where ${e_i}$ are canonical basis vectors to show that $\sum_{i > j}A_{ij}$ (or $\sum_{i < j}A_{ij}$) $ = 0$.
