= Ellipsoid level sets of a positive-definite quadratic form
{title2=$q(x)=k>0,\quad r_j=\sqrt{k/\lambda_j}$}
For a real <positive-definite matrix> $A$ in finite <dimension>, write $A=Q\Lambda Q^T$ by the <spectral theorem for real symmetric matrices>. In the <coordinates> in an <orthonormal basis> $w=Q^Tx$, the positive <level set> $q(x)=k$ becomes
$$
\sum_{j=1}^n\lambda_jw_j^2=k,\qquad \sum_{j=1}^n\frac{w_j^2}{k/\lambda_j}=1.
$$
This is an <ellipsoid>, with <semi-major and semi-minor axes> in two <dimensions> forming an <ellipse>; each semi-axis has length $r_j=\sqrt{k/\lambda_j}$ and direction given by the corresponding <eigenvector>. In the opening example, the shorter semi-axis is $\sqrt{k}/2$ along $e_+$, while the longer is $\sqrt{k/2}$ along $e_-$. A stiffer direction, meaning a larger <eigenvalue>, reaches the same energy at a smaller displacement. The ratio of longest to shortest semi-axis is $\sqrt{\lambda_{\max}/\lambda_{\min}}$, the square root of the <spectral condition number of a positive-definite matrix>.<Convex Optimization by Stephen Boyd and Lieven Vandenberghe>[§2.2.2, p. 30, equation (2.3): ellipsoids and semi-axis lengths; the book uses the inverse quadratic coefficient matrix]{ref}
Strict positivity matters because $\lambda_{\min}>0$ gives
$$
\lambda_{\min}\|x\|^2\leq q(x)\leq\lambda_{\max}\|x\|^2.
$$
Hence a fixed positive <level set> cannot extend arbitrarily far. The <Cholesky decomposition> $A=LL^T$ gives another view: $q(x)=\|L^Tx\|^2$, so the <ellipsoid> is the inverse image of a <sphere> under an <invertible linear map>. A singular <positive semidefinite matrix> has a nonzero <kernel>; adding a <kernel> <vector> preserves its <quadratic form>, so any nonempty positive <level set> is unbounded. These statements concern finite <dimension>; the <positive definite symmetric operator> entry explains why strict positivity alone need not give a uniform lower bound on an infinite-dimensional <Hilbert space>.
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