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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?

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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.

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"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"- Why is this? (I know only rudimentary linear algebra, but the gist of the rest of the answer makes sense) –  Alyosha Dec 23 '12 at 20:03
    
Great question! See this question for a nice proof that the characteristic polynomial is a similarity invariant: math.stackexchange.com/questions/87699/… –  user7530 Dec 23 '12 at 20:08
    
If you allow all orthogonal transformations of $(x,y,z)$, all the invariants are functions of the coefficients of the characteristic polynomial, because a real symmetric matrix can be diagonalized by an orthogonal transformation, and the characteristic polynomial determines the eigenvalues (with multiplicities), and thus the diagonalized form. –  Robert Israel Dec 23 '12 at 22:00
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If you only allow orthogonal transformations of $(x,y)$ (a $1$-parameter family of transformations), since your $3 \times 3$ symmetric matrix has $6$ degrees of freedom, there should be (locally) $5$ independent invariants. And indeed you have $5$: three coefficients of the characteristic polynomial, and the trace and determinant of the top left $2 \times 2$ matrix. –  Robert Israel Dec 23 '12 at 22:06
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