Discriminant of a quadratic form on $V_1 \oplus V_2$

Question

Let $V$ be finite-dimensional $F$-vector space with $char(F) \neq 2$. Assume that $V = V_1 \oplus V_2$, where $V_1$ and $V_2$ are vector subspaces of $V$. Let $q_1$ and $q_2$ be quadratic forms on $V_1$ and $V_2$ respectively and set $q(x) = q_1(x_1) + q_2(x_2)$ for any $x = x_1 \oplus x_2 \in V_1 \oplus V_2$. Furthermore, $V_1 \bot V_2$.

Can someone give me a hint how to show that $disc(q) = disc(q_1) \cdot disc(q_2)$ ?

$disc(q)$ denotes the discriminant of $q$ which is the same as the determinant of the Gram matrix of $q$.

Thanks for your help.

Answer 1

Hint If $(v_1, \ldots, v_k)$ is a basis of $V_1$ and $(w_1, \ldots, w_{\ell})$ is a basis of $V_2$, then $$((v_1, 0), \ldots, (v_k, 0), (0, w_1), \ldots, (0, w_{\ell}))$$ is a basis of $V_1 \oplus V_2$. How can you relate the respective Gram matrices $[q_1], [q_2], [q]$?