Let $V$ be a (not necessarily finite-dimensional) real vector space, and suppose $B:V\times V\to \Bbb R$ is a symmetric bilinear form on $V$. The index of $B$ is defined to be the maximum dimension of a subspace of $V$ on which $B$ is negative definite.
Suppose $V$ is expressed as a direct sum $V=V_1\oplus V_2$ with $V_1\perp V_2$ with respect to $B$, and $B$ positive definite on $V_2$. I am trying to show that in this case the index of $B$ equals the index of $B|_{V_1}$. I tried to show that if $W$ is a subspace of $V$ on which $B$ is negative definite, then $W$ must be contained in $V_1$, but it does not work well, I think. Any hints?
Best Answer
Let $p:V\rightarrow V_1$ the projection whose kernel is $V_2$.
Let $E$ be a subspace such that the restriction of $B$ to $E$ is definite negative. Let $x\in E$, suppose that we can write $x=x_1+x_2, x_i\in V_i$, suppose that $p(x)=0$, this implies that $x\in V_2$, $B(x,x)\leq 0$ implies that $x=0$ since the restriction of $B$ to $V_2$ is definite positive.
Let $y=p(x),x\in E,\neq 0$, write $x=x_1+x_2, x_i\in V_i, x_1=y$, $B(x,x)=B(x_1,x_1)+B(x_2,x_2)< 0$ implies that $B(y,y)=B(x_1,x_1)< 0$ since $B(x_2,x_2)\geq 0$. This implies that the restriction of $B$ to $p(E)$ is definite negative and the dimension of $E=dim(p(E))$ is inferior to the index of the restriction of $B$ to $V_1$. This implies that the index of $B$ is equal to the index of the restriction of $B$ to $V_1$.