(a) The candidate is verified to be a two-sided inverse: multiplying it by \(A + t \, u u^*\) collapses, by self-adjointness of \(A^{-1}\), to \(1\) plus a multiple of a rank-one operator whose scalar factor vanishes by the choice of the coefficient. For matrices one product suffices; for operators both products are computed, and together they exhibit \(A + t \, u u^*\) as a unit. Taking traces and using \(\operatorname {Tr}\bigl( (A^{-1} u)(A^{-1} u)^* \bigr) = u^* A^{-2} u\) gives the trace identity.

(b) The first is a strictly positive operator plus a positive one; it is positive, and invertible because it dominates an invertible positive operator. For the second, the hypothesis makes the Sherman–Morrison denominator positive, so (a) shows that \(A - w \, u u^*\) is invertible and exhibits its inverse as a strictly positive operator plus a positive one; an invertible operator whose inverse is positive is itself positive.

(c) A self-adjoint element \(a\) of a C\(^*\)-algebra satisfies \(a \preceq \| a\| \cdot 1\), and \(u u^*\) is self-adjoint with \(\| u u^*\| = \| u\| ^2\). For matrices the formalisation derives \(u u^* \preceq \| u\| ^2 \cdot 1\) from Lemma 2.1(b), since the trace of \(u u^*\) is \(\| u\| ^2\).

Factor \(\beta A\) out and apply the matrix determinant lemma \(\det (A + v w^*) = \det A \, (1 + w^* A^{-1} v)\), which holds because the added matrix has rank one. The dependence on \(\alpha \) is affine; this identity is what replaces the derivative of the determinant.