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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) $

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    $\begingroup$ Oh yes I forgot to mention $ f : V \to V $ - why is this proof trivial? Well I hope it will be trivial to me some day but right now it leaves me puzzled $\endgroup$ – Maximilian Kindshofer Nov 8 '15 at 8:12
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    $\begingroup$ You misunderstand to notion of function (map). $f$ and $g$ are functions, e.g. $f(x)=2x$, $g(x)=-3x$. $f\circ g$ is a composition of functions. In your example, $f=-1$, $g=1$ are not functions, but elements of $V$. $\endgroup$ – A.Γ. Nov 8 '15 at 8:24
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    $\begingroup$ Sorry I was thinking of something else when I said it was trivial, although it is still quite easy, see my answer. $\endgroup$ – Quantum spaghettification Nov 8 '15 at 8:25
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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 $f$.

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