I'm struggling with the following demonstration:
Given A,B matrix in $R^{nxn}$. If Rank(A) = n-1 then rank(AB) $\geqslant$ rank(B) - 1
I've tried using induction but I'm not sure where should I do it, if with the Matix size or with the rank of B.
Thanks!
In this case induction is not advised.
Consider instead the composition of maps, i.e first you have $$B:\mathbb{R}^n\rightarrow B(\mathbb{R}^n) \subset \mathbb{R}^n$$ and then $$A:B(\mathbb{R}^n)\rightarrow AB(\mathbb{R}^n)\subset \mathbb{R}^n.$$ What is important here, is to notice that we restrict the domain of $A$ to $B\mathbb{R}^n$, whose dimension is by definition equal to the rank of $B$, and to the codomain to nothing but $AB(\mathbb{R}^n)$ whose dimension is equal to the rank of $AB$.
Now apply the dimension formula to the second map, and use the fact that the kernel of a linear map restricted to a smaller domain is most definitely contained in the kernel of the map on the bigger domain, hence the dimension of the first must be less or equal than the dimension of the latter. In symbols: $$f:V\rightarrow W \text{ and } U\subset V \Rightarrow \ker(f|_U)\leq\ker(f)$$
In you case $$\ker(A|_{B(\mathbb{R}^n)})\leq\ker(A)=1.$$
From this you should be able to conclude quite easily