To prove a sufficient and necessary condition of a quadratic form to equal to zero

Question

The question is the following:

Let X a vector space and $B: V \times V \rightarrow K$ a positive semi-definite form. Show that $q_B(y)=0 \Longleftrightarrow B(x,y)=0$, $\forall$ $x$ $\in$ $V$, where $\ q_B(y)\ $ is the quadratic form $\ B(y,y)\ $.

Is it because $q_B = \sum_{i,j=1}^n a_{ij}x_i \bar{x_j}$? Similar type proof as to proving linear independence?

Answer 1

• $\ B(x,y)=0\ \forall x\in V\implies q_B(y)=0\ $ is trivial (put $\ x=y\ $ in the antecedent of the implication)
• $\ q_B(y)=0\implies B(x,y)=0\ \forall x\in V\ $ follows from the Cauchy-Schwarz inequality: $$|B(x,y)|\le\sqrt{q_B(x)q_B(y)}\ ,$$ which holds for any positive semi-definite bilinear form. The proof is essentially the same as for any inner product.