Given $\displaystyle A \in L(V, W)$. How can I show that the$\displaystyle V/kerA$ qutioent space is isomorphic to im A? Thank you very much!
Then A is a linear transformation while V/ker(A) is a vector space. What do you mean by an isomorphism between a linear transformation and a vector space?
I suspect you mean an isomorphism between the image of V under A and V/ker(A).
This is just the first isomorphism theorem for vector spaces. Try showing that $\displaystyle V/\ker A\to\text{im }A: x+\ker A\mapsto A(x)$ is a well-defined isomorphism. Now (and I mention this only because [despite the strangeness] some classes do this FIRST) if you know the rank-nullity theorem and you're in finite dimensions you know that $\displaystyle \dim V=\dim \ker A+\dim\text{im }A$ and so $\displaystyle \dim \text{im }A=\dim V-\dim \ker A=\dim V/\ker A$ and so the isomorphism follows.
if you start with a basis $\displaystyle \{u_1,\dots,u_k\}$, of ker(A) you can extend this to a basis $\displaystyle \{u_1,\dots,u_k,v_{k+1},\dots,v_n\}$ for V.
prove that $\displaystyle \{v_{k+1} + \text{ker}(A),\dots,v_n + \text{ker}(A)\}$ is a basis for V/ker(A), and that$\displaystyle \{A(v_{k+1}),\dots,A(v_n)\}$
is a basis for im(A) (this proves the rank-nullity theorem, and gives you drexel28's isomorphism, at the same time).