I am stumped on this one problem mentioned in Bhatia's Matrix Analysis book: For an invertible operator, obtain a relationship between $A^{-1}$, ${\bigwedge}^n A$, and ${\bigwedge}^{n-1} A$. The convention here is that $n$ is the dimension of the underlying space. I can obtain some simple relationships just because ${\bigwedge}^k A$ is a restriction of $\bigotimes^k A$ and $(\bigotimes^k A)^{-1} = \bigotimes^k (A^{-1})$, but I don't think that's what Bhatia is hinting at here. I suspect something related to looking at ${\bigwedge}^n A$ as a multiplication by $\det A$ and how ${\bigwedge}^{n-1} \mathcal{H}$ is kind of dual-ish or something to $\mathcal{H}$, but I'm stumped.
FWIW this is not for a class, I am just trying to bone up on my linear algebra.
Let us suppose that you have an inner-product $\langle \cdot,\cdot \rangle$ on $\mathcal{H}$, assuming that it is a Hilbert space. Choose an orthonormal basis, $f_1,\dots,f_n$, which is a choice you make . Then choose a basis for the 1-dimensional vector space $\bigwedge^n \mathcal{H}$ to be $f_1\wedge \cdots \wedge f_n$ divided by its norm (in the tensor product power of your original inner-product). Then there is an isomorphism $\phi_n : \bigwedge^n \mathcal{H} \to \mathbb{C}$ obtained by taking the inner-product with the chosen basis element. Given any $\omega^n \in \bigwedge^n \mathcal{H}$ it is a scalar multiple of of the chosen basis element, and you take $\ell_n(\omega^n)$ to be that scalar. Then you can choose another isomorphism $\phi_{n-1} : \bigwedge^{n-1} \mathcal{H} \to \mathcal{H}$ as follows. For each $\omega^{n-1} \in \bigwedge^{n-1} \mathcal{H}$ make the linear transformation $\ell_{n-1}(\cdot,\omega^{n-1}) : \mathcal{H} \to \mathbb{C}$ obtained as $\ell_{n-1}(f,\omega^{n-1}) = \ell_n(\omega^{n-1} \wedge f)$. Note that $\omega^{n-1}\wedge f$ is in $\bigwedge^n \mathcal{H}$, so applying $\ell_n$ to that gives a scalar. Then since $\ell_n(\cdot,\omega^{n-1})$ is a linear functional, there is a unique vector $g\in \mathcal{H}$ such that $\ell_{n-1}(f,\omega^{n-1}) = \langle f,g\rangle$ for every $f \in \mathcal{H}$. So let $\phi_{n-1}(\omega^{n-1})$ be equal to this $g$. Note that $\phi_n \circ (\bigwedge^n A) \circ \phi_n^{-1}$ is a linear transformation from $\mathbb{C}$ to itself. So it is just a scalar. This scalar is just $\det(A)$ no matter what choices were made (for the orthonormal basis, for example). The claim which is based on the comment by The Sand Reckoner is that $\bigwedge^{n-1} A$ is equal to $\det(A)$ times $\phi_{n-1}^{-1} \circ A^{-1} \circ \phi_{n-1}$. Or in other words $$ \bigwedge\nolimits^{n-1} A\, =\, \left(\phi_n \circ (\bigwedge\nolimits^n A) \circ \phi_n^{-1}\right) \left(\phi_{n-1}^{-1} \circ A^{-1} \circ \phi_{n-1}\right)\, . $$