The difference between exterior maps is bounded by the norm of the difference?

Question

Let $V,W$ be $d$-dimensional real inner product spaces, and let $A,B:V \to W$ be linear maps. Let $\bigwedge^k A,\bigwedge^k B:\Lambda_k(V) \to \Lambda_k(W)$ be the induced maps on exterior powers.

Is it true that $|\bigwedge^k A-\bigwedge^k B| \le C |A-B|^k$ for some constant $C$?

Here the norms on $\text{Hom}(V,W),\text{Hom}(\Lambda_k(V),\Lambda_k(W))$ are the natural ones induced by the products on $V,W$. (One can also use the operator norms, it doesn't really matter).

Note it's easy to get a bound which is a polynomial of degree $k$ in the variables $x_1=|A-B|,x_2=|A|,x_3=|B|$, using the triangle inequality.

Answer 1

No: for $k=d$ we have $|\det(I+\lambda I)-\det(I)|\approx k\lambda,$ which is bigger than $C\lambda^k$ for small enough $\lambda.$