Homogeneous quadratic in $n$ variables has nonzero singular point iff associated symmetric matrix has zero determinant.

Question

I am looking for help on the following problem:

Prove that a homogeneous quadratic polynomial $f \in \mathbb{C}[x_1, \ldots, x_n]$ has a nonzero singular point $p$ if and only if the determinant of the associated symmetric coefficient matrix is $0.$

Here is what I know and have tried to explore:

It is well known that associated to every homogeneous quadratic polynomial $f$ there is an associated symmetric matrix $A$ so that $f(x) = x^TAx$ for every $x \in \mathbb{C}.$ Furthermore, $A$ can be put into diagonal form, so without loss of generality, we may suppose (?) that $A$ is diagonal. Therefore, our problem is reduced to showing that $x$ is singular iff $A$ has a zero on its diagonal.

What I am having trouble with is associating the partial derivatives of the polynomial $f$ with the symmetric matrix. That $A$ has a nonzero determinant would ensure that $f$ has a root, but I am having trouble showing that all the partials would vanish at this root.

Any help would be appreciated.

EDIT: I’ve found this related question/answer here: Definition of non-singular quadratic form. Is this definition of singularity used in this answer equivalent to the one I am using above (that $f$ and all of its first order partials vanish at some nonzero point $p \in \mathbb{C}^n.$)

Answer 1

If $A=(a_{ij})$, then $f(x_1,\ldots, x_n)=\sum a_{ij}x_ix_j$. It may be assumed that $A$ is diagonal (there is a matrix $C$ such that $C^TAC$ is diagonal, and if we will choose coordinates $y=C^{-1}x$, then we would have $f(y)=y^T(C^TAC)y$), in this case $f(x_1,\ldots, x_n)=\sum a_{ii}x_i^2$. If $A$ is singular we may assume that also $a_{11}=0$, then it is easy to see that $f$ vanishes at the point $(1,0,\ldots, 0)$ together with all partial derivatives. It is also easy to see that, if all $a_{ii}$ are nonzero, there is no point where all partial derivatives vanish, except for zero.