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Section 41.5 Theory

The fact that a quadratic form can always be represented by a symmetric matrix allows us to state the following version of Theorem 40.6.13.
In the two-variable case, we can also be specific about when a level set of a quadratic form corresponds to an ellipse in the plane, and when it corresponds to a hyperbola.

Aside: Special case.

Proof idea.
  1. We know that there exists a \(2 \times 2\) orthogonal matrix \(P\) so that \(\utrans{P} A P\) is diagonal, and a change of variables \(\uvec{x} = P \uvec{w}\) will diagonalize \(q_A\text{,}\) so that
    \begin{equation*} q_A(\uvec{x}) = q_{\inv{P} A P} (\uvec{w}) \text{.} \end{equation*}
    The first column of \(P\) is a unit vector, so it must point from the origin to a point on the unit circle:
    \begin{equation*} \uvec{p}_1 = \begin{bmatrix} \cos \theta \\ \sin \theta \end{bmatrix} \text{.} \end{equation*}
    As discussed in Remark 41.4.5, the second eigenvector must be orthogonal to the first, so we can choose \(\uvec{p}_2\) to be either of
    \begin{equation*} \begin{abmatrix}{r} - \sin \theta \\ \cos \theta \end{abmatrix} \text{,} \qquad \begin{abmatrix}{r} \sin \theta \\ - \cos \theta \end{abmatrix}\text{.} \end{equation*}
    Both of these vectors are unit vectors and are orthogonal to \(\uvec{p}_1\text{.}\) Also, eigenspaces are subspaces, and subspaces are closed under taking negatives, so it doesn’t matter which one of the two options above we choose. If we choose the first option for \(\uvec{p}_2\text{,}\) we get rotation matrix
    \begin{equation*} P = \begin{abmatrix}{rr} \cos \theta \amp - \sin \theta \\ \sin \theta \amp \cos \theta \end{abmatrix} \text{.} \end{equation*}
  2. This statement should be obvious from the fact that the eigenvalues of \(A\) will be the coefficients in the diagonalized quadratic form.
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