 Solution

By the spectral theorem for real symmetric matrices, choose an orthonormal eigenbasis $u_1,u_2,u_3$ with $A{u}_i={\lambda }_i{u}_i$. If $x={\sum }_i{y}_i{u}_i$, thenThus every point of the ellipsoid has distance at least ${\lambda }_1^{-1/2}$ from the origin. Equality holds precisely whenBoth coefficients are strictly positive, so $y_2=y_3=0$. The surface equation then gives $y_1=\pm {\lambda }_1^{-1/2}$. ThereforeExactly two points attain the minimum. The strict ordering makes the largest-eigenvalue eigenspace one-dimensional; this is the equality case in nearest points to the origin on a positive-definite ellipsoid.