Anirbit,
this is to answer your last question and your refined question in the comment to JC:
Forget about eigenvalues, those (mostly) get their explicit value gets lost in the projective world and the only thing that matters is whether they are zero or not. $$[x_0:x_1:x_2]\mapsto [\lambda_0x_0:\lambda_1x_1:\lambda_2x_2]$$ with $\lambda_0\lambda_1\lambda_2\neq 0$ is an automorphism of $\mathbb P^2$, so in order to prove that any irreducible conic is isomorphic to $x_0^2+x_1^2+x_2^2=0$ you only need to notice that the matrix defining your conic is symmetric and hence diagonalizable. The irreducibility condition translates to the matrix being of full rank, i.e., invertible, i.e., having non-zero eigenvalues, so using the above automorphism you can get the form $x_0^2+x_1^2+x_2^2=0$.
To see why the statement about irreducubility is true notice that the diagonalizable part does not use anything about irreducibility, so you get something like $\lambda_0x_0^2+\lambda_1x_1^2+\lambda_2x_2^2=0$, which is the equation of two lines (possibly a double line) if and only if $\lambda_0\lambda_1\lambda_2=0$.
There is also a heuristic proof of this: the matrix corresponding to the quadric which is the union of two lines is (sort of) the tensor product of a column matrix with a row matrix and hence cannot be invertible. This is not quite solid reasoning, but it gives you an idea.
