Let $k=\mathbb{R}$ or $\mathbb{C}$. What is meant by a quadratic form $q: k^{n} \rightarrow k$ ? Show that there is a basis $\left\{v_{1}, \ldots, v_{n}\right\}$ for $k^{n}$ such that, writing $x=x_{1} v_{1}+\ldots+x_{n} v_{n}$, we have $q(x)=a_{1} x_{1}^{2}+\ldots+a_{n} x_{n}^{2}$ for some scalars $a_{1}, \ldots, a_{n} \in\{-1,0,1\} .$

Suppose that $k=\mathbb{R}$. Define the rank and signature of $q$ and compute these quantities for the form $q: \mathbb{R}^{3} \rightarrow \mathbb{R}$ given by $q(x)=-3 x_{1}^{2}+x_{2}^{2}+2 x_{1} x_{2}-2 x_{1} x_{3}+2 x_{2} x_{3}$.

Suppose now that $k=\mathbb{C}$ and that $q_{1}, \ldots, q_{d}: \mathbb{C}^{n} \rightarrow \mathbb{C}$ are quadratic forms. If $n \geqslant 2^{d}$, show that there is some nonzero $x \in \mathbb{C}^{n}$ such that $q_{1}(x)=\ldots=q_{d}(x)=0$.