Showing that a bilinear form must be a multiple of another

Question

Let $f:\mathbb{R}^{n}\times\mathbb{R}^{n}\to\mathbb{R}$ be bilinear and symmetric. Here, of course, $n\geq 1$. Moreover, suppose that $f(x,x)=0$ for every $x\in\mathbb{R}^n$ which satisfies $$x_1^2=\sum_{i=2}^nx_i^2.$$ Prove that there is a real number $a$ such that for any $x,y\in\mathbb{R}^n$, we have $$f(x,y)=a\left(-x_1y_1+\sum_{i=2}^n x_iy_i\right).$$ Not sure at all how to start with this one. Any ideas?

Answer 1

The real bilinear form $f$ has an associated real symmetric matrix. Given a basis $\beta = \{v_1, v_2, \ldots v_n\}$, the matrix can be found from the values of $f(v_i, v_j)$. After choosing a basis $\beta$, label the matrix as $[f]_\beta^\beta$. By the spectral theorem, this matrix is diagonalizable, thus under a change of basis to some $\Gamma$, the matrix $[f]_\Gamma^\Gamma$ is diagonal. Suppose its diagonal entries are $\lambda_1, \lambda_2, \ldots \lambda_n$.

By your condition, you also know that for any $x$ with co-ordinates $x_i$, if $-x_1^2 + x_2^2 + \ldots x_n^2 = 0$ then $f(x,x) = \lambda_1x_1^2 + \lambda_2x_2^2 + \ldots \lambda_nx_n^2 = 0$.

Now, the condition is true when $x_1 = x_2 = 1$ and the rest are zero. So, we have $\lambda_1 + \lambda_2 = 0$. Then, to find $\lambda_3$, we set $x_1 = x_3 = 1$ and so on. Thus, we see the matrix diagonal is: $-a, a, a, a, \ldots, a$ for $a = -\lambda_2$.