Explain geometrically how you diagonalize a real quadratic form \(q(x) = x^t A x\) on \(\mathbb{R}^n\) (Principal Axis Theorem and Gram-Schmidt orthogonalization).
Solution
Write the real quadratic form as \[ q(x)=x^TAx \] with \(A=A^T\). By the real spectral theorem, there is an orthogonal matrix \[ Q=[v_1\ \cdots\ v_n] \] whose columns are orthonormal eigenvectors of \(A\), with \[ Q^TAQ=\operatorname{diag}(\lambda_1,\dots,\lambda_n). \] With the orthogonal change of coordinates \(x=Qy\), \[ q(x)=\lambda_1y_1^2+\cdots+\lambda_ny_n^2. \] Thus diagonalization is geometrically a rotation/reflection of coordinates to the mutually orthogonal principal directions of the form; the cross terms disappear in those coordinates.
One way to find the principal directions is variational. On the Euclidean unit sphere, the extrema of the Rayleigh quotient \[ x\mapsto x^TAx \] occur at eigenvectors: the Lagrange-multiplier equation is \[ Ax=\lambda x. \] After choosing one unit eigenvector, its Euclidean orthogonal complement is \(A\)-invariant, so the procedure can be repeated there. This is the geometric content behind the principal-axis theorem.
For a positive-definite form, the level set \(q=1\) is an ellipsoid; if \(\lambda_i>0\), the principal semiaxis in the \(v_i\) direction has length \(1/\sqrt{\lambda_i}\). For an indefinite form, positive and negative eigenvalues give the different families of hyperbolic directions, while zero eigenvalues give flat/cylindrical directions. Thus the signs and nullity, not a universal “longest axis,” encode the geometry.
A nonorthogonal diagonalization can also be obtained by completing squares, equivalently by Gram–Schmidt orthogonalization for the symmetric bilinear form associated to \(q\) (when the required pivots are nonzero). Sylvester’s law of inertia states that after an invertible linear change of coordinates the form has the normal form \[ y_1^2+\cdots+y_p^2-y_{p+1}^2-\cdots-y_{p+q}^2, \] with the numbers of positive, negative, and zero directions invariant.