Fredholm determinant of a finite-rank operator

By [A1], AC supplies the Countable Choice hypothesis needed in [A3] and [A5]. The finite-rank $F$ in the statement is therefore trace class, so the arbitrary-Hilbert determinant theorem applies.

Fix $u,v\in H$, put $\alpha :=⟨u,v⟩$, and define $F(x):=⟨x,v⟩u$. By [A2], $F$ is linear; [A4] gives $\mathrm{\parallel }F(x)\mathrm{\parallel }=\mathrm{\mid }⟨x,v⟩\mathrm{\mid }\mathrm{\parallel }u\mathrm{\parallel }\le \mathrm{\parallel }v\mathrm{\parallel }\mathrm{\parallel }u\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel }$, so it is bounded. Its range lies in $\mathrm{span}\{u\}$, hence it has rank at most one. Also $F(u)=\alpha u$.

Let $E$ be any finite-dimensional subspace containing $\mathrm{ran}F$. For every $x\in E$, $F(x)\in \mathrm{ran}F\subseteq E$, so $E$ is $F$-invariant. By [A5], $D_H(I+zF)={\det }_E(I_E+z(F{\mathrm{\mid }}_E))$ for every $z\in \mathbb{C}$. If $F=0$, this says both sides are $1$ because $I_E$ has determinant $1$; this includes $E=\{0\}$, where the convention is also explicitly in [A5].

If $u=0$, then $F=0$, $\alpha =0$, and $E=\{0\}$ is finite-dimensional and contains the range. Step 2.1 gives $D_H(I+zF)=1=1+z\alpha$ for all $z$; this also covers $H=\{0\}$.

Suppose $u\mathrm{\ne }0$ and set $E=\mathrm{span}\{u\}$. By [A6], $E$ is the set of scalar multiples of $u$. The one-element list $(u)$ is independent: its only coefficient relation is $\lambda u=0$, which forces $\lambda =0$. More generally, an injective finite list into $\{u\}$ has at most one entry, since any two entries would both equal $u$; the empty list is independent, so $\{u\}$ is independent under [A6]. It spans $E$, hence it is a basis and $(u)$ is an ordered basis; thus $E$ has dimension one. The range of $F$ is contained in $E$, and $E$ is invariant because $F(u)=\alpha u$. Relative to $(u)$, [A7] gives the matrix $[\alpha ]$ for $F{\mathrm{\mid }}_E$ and $[1+z\alpha ]$ for $I_E+z(F{\mathrm{\mid }}_E)$. By [A8]–[A9], its ordinary determinant is $1+z\alpha$; step 2.1 therefore yields $D_H(I+zF)=1+z⟨u,v⟩$. If $v=0$ then this is the zero operator and the value is $1$; if $v\mathrm{\ne }0$ but $\alpha =0$, then $F(v)=⟨v,v⟩u\mathrm{\ne }0$ and, for every $x$, $F^2(x)=⟨F(x),v⟩u=⟨x,v⟩⟨u,v⟩u=0$. Thus the nonzero rank-one operator is nilpotent and its one-dimensional restriction is zero, so the determinant is again $1$.

At $z=0$ both sides of the finite-rank identity and rank-one formula equal $1$. The zero-rank case, the zero vector and zero Hilbert space, and the nilpotent rank-one case are covered in steps 2.1, 3.1, and 3.2; nonzero $\alpha$ gives the scalar factor $1+z\alpha$ in step 3.2. There is no interval endpoint parameter. The only assumption is AC [A1]; no choices are made in the finite-dimensional rank-one calculation, and both conclusions are equalities rather than iff claims. [A1, A5, A6, A8, A9, step 2.1, step 3.1, step 3.2] \qed