Let $A$ be ab $d \times d$ invertible matrix. We define the norm and conorm as follows:
$$m(A) = \inf \{ \frac{|Av|}{|v|}: v \neq 0 \}, \hspace{0.5cm}\|A\|=\sup \{ \frac{|Av|}{|v|}: v \neq 0 \}.$$
We denote by $\sigma_1 \geq \sigma_2 \ldots \geq \sigma_{d}$ the singular values of $A$. It is known that $$\|A^{\wedge \ell}\|=\sigma_{1}\ldots \sigma_{\ell},$$ where $\ell \leq d.$ Is it true that $$m(A^{\wedge \ell})=\sigma_{d} \ldots \sigma_{d -\ell}?$$
Yes. More generally show that the singular values of $\wedge^{\ell} A$ are the $\ell$-tuples of singular values of $A$ (with distinct indices) (this follows straightforwardly from the existence of SVD plus the observation that exterior powers of an orthogonal / unitary remain orthogonal / unitary). The norm and conorm are the largest and smallest singular values so we want the largest and smallest $\ell$-tuple of singular values of $A$.