Let $A$ be an $n \times n$ symmetric matrix and let $y_1, \dots, y_{r+s}$ be $r+s$ linearly independent $n \times 1$ vectors such that for all $n \times 1$ vectors $x$, $$x'Ax= (y_1' x)^2 + \cdots +(y_r' x)^2 - (y_{r+1}'x)^2 - \cdots - (y_{r+s}'x)^2$$ Prove that rank of $A$ is $r+s$.
Here ' denotes the transpose. I was thinking on the line of Gram-Schmidth orthogonalisation but this will change the given set of linearly independent vectors.
You can find a nonsingular matrix $B$ such that, $B^Ty_i = e_i$ where $e_i$ is the vector with $1$ at the $i$th place and $0$ elsewhere. To see this, since the vectors $y_1.\dots,y_{r+s}$ are linearly independent, you can extend it to a basis of $\mathbb{R}^n$ say $y_1,\dots,y_n$. And then chose $B^T = \begin{bmatrix} y_1 &\dots & y_n\end{bmatrix}^{-1}$.
For any vector $x$ we have $(Bx)^TA(Bx) = \sum_{i=1}^r (x^TB^Ty_i)^2 - \sum_{i=r+1}^{r+s}(x^TB^Ty_i)^2 = \sum_{i=1}^r (x^Te_i)^2 - \sum_{i=r+1}^{r+s}(x^Te_i)^2.$
Which can be expressed as the matrix identity $x^TB^TABx = x^T D x$. Where $D$ is a diagonal matrix whose first $r$ diagonal entries are $+1$ and the next s diagonal elements are $-1$ and the rest are $0$.
Since both $B^TAB$ and $D$ are both symmetric, we must have $B^TAB = D$. Since $D$ has rank $r+s$ and $B^TAB$ has the same rank as that of $A$ since $B$ is non-singular, the result follows.