Invariants in a second order equation For $Ax^2+2Bxy+Cy^2+2Dx+2Ey+F=0$, why are $\begin{vmatrix}
A &B \\ 
B &C 
\end{vmatrix}$ and $\begin{vmatrix} A & B & D\\  B & C & E\\  D & E & F \end{vmatrix}$ invariant under an orthogonal transformation?
I was considering simply convincing myself of its self-evidence by through looking at the mechanics of the possible transformations, but the fact that 2 invariants are expressible in determinant form makes it look as if there's a far more elegant scheme underneath.
What is the 'book proof' of their invariance (if there is an elegant one beyond mechanics), and how is it proved that they (and $A+C$) are the only possible invariants for a second order equation (for orthogonal transformations)?
The answer to the following will probably be implicit in the main answer, but how would this proof be extendable into an $n$-ordered equation?
 A: The homogeneous second-order form
$$Ax^2 + 2Bxy + Cy^2 + 2Dxz + 2Eyz + Fz^2=0$$
Can be written as the matrix equation
$$\left[\begin{array}{c}x & y & z\end{array}\right]\left[\begin{array}{ccc}A & B & D \\ B & C & E\\ D & E & F\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right] = 0.$$
Orthogonal transformations of $(x,y,z)$ (of which two-dimensional orthogonal transformations of only $x$ and $y$ are a special case) must, at a minimum, preserve the coefficients of the characteristic polynomial of this matrix, which are similarity invariants. Therefore the determinant, trace, and sum of principal minors must be invariants under orthogonal transformations. Degree $n$ equations must have at least $n$ invariants, for the same reason.
There might be more invariants; I don't know off-hand how to prove they're the only ones. For instance, your allowable transformations are only those of the form $\left[\begin{array}{cc}R &\\& 1\end{array}\right]$ for $R$ a two-dimensional orthogonal matrix; therefore the determinant and trace of the top-left 2x2 matrix are also invariants.
