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 $A=0$?
If one takes for $u$ the vector $\begin{pmatrix}x_1&amp;\ldots&amp; x_n \end{pmatrix}$ 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.
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 $\sum_{i &gt;j} A_{ij} + \sum_{i &lt;j} A_{ij} = 0$. Finally, set $u = e_{i} + e_{j}$ where ${e_i}$ are canonical basis vectors to show that $\sum_{i &gt; j}A_{ij}$ (or $\sum_{i &lt; j}A_{ij}$) $ = 0$.