Try to learn more about linear operator,subspace and dimension Prove that for every subspace $F$ and every linear transformation $L$ of linear space $V$ and $\dim V=n$ is $\dim L(F)+\dim\ker L=\dim(F+\ker L)$
I know that $F$ and $\ker L$ are subspaces of the same vector space, so we have $\dim(F+\ker L)=\dim F +\dim\ker L -\dim(F\cap\ker L)$. And if we have some basis for subspace $F, B=\{f_1,f_2,\ldots,f_k\}\ k\le n,$ and they are linearly independent, then $L(f_1), L(f_2),\ldots,L(f_n)$ are linearly independent if $N(L)=0$, because this is  an isomorphic linear transformation (iff they have same dimension, but here we have) then $\dim L(F)=\dim F$,then we have $\dim(F\cap\ker L)=0$, but I do not know is this good how I explain?  

This is a consequence of the isomorphism theorems for modules (in particular, vector spaces). 
We have $$ L(F) \simeq \frac{F}{F\cap kerL} \simeq \frac{F+kerL}{kerL}.$$
Computing the dimensions we have $$dimL(F) = dim(F+kerL) - dimKerL.  $$
