Eigendecomposition of $(a^\top X a)X - Xaa^\top X$

Question

Let $a\in\mathbb{R}^n$ be a nonzero vector, $X\in\mathbb{R}^{n\times n}$ be positive definite. What are the eigenvalues and eigenvectors of $(a^\top X a) X - Xaa^\top X$?

Answer 1

Wlog use an orthonormal basis where $a$ is a multiple of the first basis element: $a = t e_1$.
Write $X$ in block form as $$ X = \pmatrix{x_1 & X_2\cr X_2^\top & X_3\cr}$$ Then $$ (a^\top X a) X - X a a^\top X = t^2 \pmatrix{0 & 0\cr 0 & x_1 X_3 - X_2^\top X_2}$$ One eigenvalue is $0$, the others are $t^2$ times the eigenvalues of $x_1 X_3 - X_2^\top X_2$.

Note that the latter is a rank-$1$ perturbation of $x_1 X_3$, and the Matrix determinant lemma might be useful.

EDIT: By that lemma,

$$ \det(x_1 X_3 - X_2^\top X_2 - \lambda I) = (1 + X_2 (x_1 X_3 - \lambda I)^{-1} X_2^\top) \det(x_1 X_3 - \lambda I)$$ so if $X_2 (x_1 X_3 - \lambda I)^{-1} X_2^\top = -1$ then $\lambda$ must be an eigenvalue of $x_1 X_3 - X_2^\top X_2$. If $v_j$ are an orthonormal basis of eigenvectors of $x_1 X_3$ with corresponding eigenvalues $\mu_j$, that condition is $$ \sum_j (\mu_j - \lambda)^{-1} (X_2 v_j)^2 = -1$$