Irreducible conic implies that the underlying matrix is invertible I guess that it is true that a conic (2nd degree homogeneous equation in complex variables) is irreducible (i.e can't be factorized over polynomials) if and only if the underlying matrix of coefficients for that is invertible (i.e all its eigenvalues are non-zero). 


*

*If the above statement is correct (or whatever is the nearest correct statement!) then I would like to understand its proof. 

*Further is it/(if yes then how) true that there will always exist an unitary transformation to diagonalize that matrix? 

*I think that this matrix can't be called a quadratic form or a bilinear form.
Right? 

*I guess when people talk of conics in $\mathbb{C}^3$ or $\mathbb{CP}^2$ they mean that this matrix has all entries in $\mathbb{R}$. Right? 
 A: A quadratic form is a homogeneous polynomial of degree two, with coefficients in
some field.  I will write the polynomial as $Q(x)$, where $x$ is vector of length $d$.
By some version of Taylor's theorem
$$
 Q(x+y) = Q(x) +Q(y) + x^T By
$$
where $B$ is a $d\times d$ matrix and $x^T By$ is a bilinear form. Since $x+y=y+x$,
the form is symmetric, so $x^TBy=y^TBx$ and therefore $B$ is symmetric. The set of
points $x$ such that $Q(x)=0$ is a projective quadric, and when $d=3$ it is a projective conic.
Over the reals and complexes, $B$ determines $Q$. (And $B$ is what you call the underlying matrix.)
Any proof you have for the real case that a conic is irreducible if $\det(B)\ne0$
will work in the complex case provided it does not use eigenvectors. 
Now suppose our field is $\mathbb{C}$. Then $B$ is a complex symmetric matrix, but we can say very
little about how to diagonalize a symmetric matrix over $\mathbb{C}$, we can only 
unitarily diagonalize a matrix if it is normal ($BB^*=B^*B$) and a complex symmetric matrix need not 
be normal.
We do not call the matrix $B$ a quadratic form, because it is a matrix and not a polynomial.
If people talk about a conic over $\mathbb{C}$, then we assume that the coefficients of
the polynomial and the entries of $B$ may be complex. This is standard.
Using eigenvalues and eigenvectors makes it easier to work with quadratic forms
over the reals. However this approach does not work over $\mathbb{C}$ or over finite fields.
(Edit. In response to the question below.)
Since $Q(x)$ is quadratic, if it factors then it must be a product of two linear factors,
so we can then write it as $(c^Tx)(d^Tx)$ (where $c$ and $d$ could be parallel). Now we can calculate
$$
 Q(x+y) -Q(x) -Q(y) =  (c^Tx)(d^Ty)+(d^Tx)(c^Ty).
$$
If we set $B = cd^T+dc^T$ (which is the sum of two $d\times d$ matrices, then
$$
 x^TBy = (x^Tc)(d^Ty) +(x^Td)(c^Ty) = (c^Tx)(d^Ty)+(d^Tx)(c^Ty).
$$
So the matrix of the quadratic form is $cd^T+dc^T$. Since we are in three dimensions (over $\mathbb{C}$ or $\mathbb{R}$ it does not matter), there is a non-zero vector $h$ say such that $c^Th=d^Th=0$. Therefore $Bh=0$ and $B$ is not invertible.
For the other direction, if $\det(B)=0$ then $B$ has rank at most two. Now if
$B$ has rank two and is symmetric we can write it as $B=cd^T+dc^T$ for two vectors
$c$ and $d$. (This is proved in my book with Royle, it is reasonably standard linear
algebra. There might be a way to simplify this but my dinner's calling...) Now since
$$
 4Q(x) = Q(2x) =Q(x+x) = Q(x)+Q(x) +x^TBx
$$
we see that $x^TBx=2Q(x)$. So $2Q(x)$ factors, because $x^TBx=2(c^Tx)(d^Tx)$.
This is probably a bit dense, but you can see that we are not using properties of the field,
other than it does not have even characteristic.
