Could a quadric be a $k$ dimensional hyperplane? Assume that $Q$ is a  hypersurface in $\mathbb C^n$ of the second degree, given by equation
$
\sum_{i,j=1}^n a_{ij}x_i x_j +2\sum_{i=1}^n b_i x_i +c=0,
$
with $a_{ij}=a_{ji}$ and complex coefficients.
I'm not familiar with algebraic geometry but I want to know whether $Q$ could be an empty set or an affine hyperplane  in $\mathbb C^n$, but different from each  $n-1$ dimensional affine hiperplane
$
H: \ \sum_{i=1}^n a_i x_i+a_0=0,
$
where not all $a_1,...,a_n$ vanish,
which is a quadric since
$
H: \ (\sum_{i=1}^n a_i x_i+a_0)^2=0.
$
I know that in the case of real quadric in $\mathbb R^n$ can be an affine hiperplane of arbitrary dimension $<n$, for example if $k\in \{1,...,n\}$ the equation
$
x_1^2+...+x_k^2=0
$
represents $n-k$-dimensional subspace of $\mathbb R^n$.
Thanks
 A: First of all, because $\Bbb{C}$ is closed, a nonconstant polynomial $f(x_1,\ldots, x_n)$ always has solutions in $\Bbb{C}^n$ (and in fact infinitely many). For instance, given $f(x,y)=x^2+y^2$, set $y=z_0\in \Bbb{C}$ and then $x^2+z_0=0$ has $2$ solutions (with multiplicity). So, for each choice of $y$ there are $2$ solutions in $x$. You can generalize this argument.
I know you said you are not familiar with much algebraic geometry, but I will sketch an argument so that you can look up a few things and hopefully agree.
Now, the projective closure of a hyperplane in $\Bbb{C}^n$ is a hyperplane in $\Bbb{P}^n$. Taking $Q$ as you defined it, we get a projective variety cut out by a degree $2$ equation when we take its projective closure $Z$. Next, we use Bézout's theorem for hypersurfaces which says (in this case) that given a line $L$ in $\Bbb{P}^n$ and a hypersurface $Y=Z(f)$, the number of intersection points with multiplicity is given by $\deg(L)\cdot \deg(f)=1\cdot\deg(f).$ For us, $f$ is degree $2$ so that the intersection with multiplicity of a line with $Z$.
On the other hand, if $Z$ were a hyperplane in $\Bbb{P}^n$, it would be cut out by a degree $1$ equation by definition and hence intersect the line $L$ in one point. So, $Z$ is not a projective hyperplane and $Q$ does not define an affine hyperplane.
