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!

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- Dec 4th 2011, 07:02 AMgotmejerryHow to show this isomorphism?
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!

- Dec 4th 2011, 07:42 AMAmerRe: How to show this isomorphism?
what is the Group $\displaystyle L(V,W)$ ??

- Dec 4th 2011, 07:56 AMgotmejerryRe: How to show this isomorphism?
L(V, W) represents all linear transformations from V to W

- Dec 4th 2011, 08:56 AMHallsofIvyRe: How to show this isomorphism?
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). - Dec 4th 2011, 09:07 AMgotmejerryRe: How to show this isomorphism?
Yes, sorry I meant that

- Dec 4th 2011, 09:44 AMDrexel28Re: How to show this isomorphism?
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.

- Dec 4th 2011, 10:02 AMDevenoRe: How to show this isomorphism?
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).