Rank of two linear functions

I have two linear functions $f$ and $g$ in an finite Vectorspace with dimension of n. Also $ f $ and $g$ both $V \to V $

I should proof that if $ f \circ g = 0 $ then $ rank(f) + rank(g) \le n $

Does this mean $f = 0$ and $ g = 0 $ well then the rank always have to be 0, but this what about $f=-1$ and $g = 1 $ what could I say about the rank of the two functions? It has be $ \le n $ for each but how does the operator $ \circ $ affect it? Normaly I would say $ Rank(f+g) = rank(f) + rank(g) $

Solution 1:

$\DeclareMathOperator{\ker}{ker}$ $\DeclareMathOperator{\rk}{rk}$

Let the image $f$ have a basis $(\vec e_1 ,...,\vec e_k)$ and the image of $g$ have a basis $(\vec b_1,...,\vec b_l)$. Then by saying $f \circ g=0$ we are saying that $f(\vec b_i)=0\, \forall \,i$ and thus the kernel of $f$ contains the vectors $(\vec b_1,...,\vec b_l)$. This means that $\dim(\ker(f))\ge \rk(g)=l$ which from the dimension theorem: $$\dim(V)=\rk(f)+\dim(\ker(f))$$ Means that $$\dim(\ker(f))=\dim(V)-\rk(f)$$ $$=n-\rk(f)\ge \rk(g)$$ And hence $$n \ge \rk(g)+\rk(f)$$

Edit

Just to clarify the notation the function $f \circ g$ is the function such that: $$f \circ g(\vec v)=f(g(\vec v))$$ I.e. we first apply the function $g$ to the vector $\vec v$ and then take the result and apply the function $f$.