# Rank of symmetric matrix

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.

• Where's the bilinear form? Commented Apr 22, 2020 at 6:25

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.