3
$\begingroup$

Is the following argument correct?

Suppose $U$ is a subspace of $V$ such that $V/U$ is finite dimensional. Prove that $V$ is isomorphic to $U\times (V/U)$.


Proof. Let $v_1+U,v_2+U,\dots,v_n+U$ be a basis of $V/U$ and define the map $\Delta:U\times (V/U)\to V$ such that $\displaystyle\Delta\left(u,\sum_{k=1}^{n}\lambda_k(v_k+U)\right) = u+\sum_{k=1}^{n}\lambda_kv_k$.

We leave the proof that $\Delta$ is a linear map to the reader and instead proceed by proving that $\Delta:U\times (V/U)\to V$ is both injective and surjective.


Assume $\displaystyle\Delta\left(u_1,\sum_{k=1}^{n}\lambda_k(v_k+U)\right) = \Delta\left(u_2,\sum_{k=1}^{n}\alpha_k(v_k+U)\right)$ therefore $\displaystyle u_1+\sum_{k=1}^{n}\lambda_kv_k = u_2+\sum_{k=1}^{n}\alpha_kv_k$ equivalently $\displaystyle u_1-u_2 = \sum_{k=1}^{n}(\alpha_k-\lambda_k)v_k$ but then $\displaystyle\left(\sum_{k=1}^{n}(\alpha_k-\lambda_k\right)v_k)\in U$ which by theorem $\textbf{3.85}$ is equivalent to $\displaystyle\sum_{k=1}^{n}\alpha_kv_k+U = \sum_{k=1}^{n}\lambda_kv_k+U$, and so $\displaystyle\sum_{k=1}^{n}\alpha_k(v_k+U) = \sum_{k=1}^{n}\lambda_k(v_k+U)$ but since $v_1+U,v_2+U,\dots,v_n+U$ is a basis of $V/U$ it follows that $\lambda_k = \alpha_k,\forall k$ and so $u_1 = u_2$. Thus the map in question is injective.


Now let $v\in V$, thus the affine subset $v+U\in V/U$ and thus for some scalars $\lambda_1,\lambda_2,\dots,\lambda_n\in\mathbf{F}$ we have $\displaystyle v+U = \sum_{k=1}^{n}\lambda_k(v_k+U)$, equivalently by theorem $\textbf{3.85}$ we have $\displaystyle \left(v - \sum_{k=1}^{n}\lambda_kv_k\right)\in U$, then given the above definition of $\Delta$ is it evident that $\displaystyle\Delta\left(v-\sum_{k=1}^{n}\lambda_kv_k,\sum_{k=1}^{n}\lambda_k(v_k+U)\right) = v$.


$\blacksquare$

NOTE: Theorem $\textbf{3.85}$ is as follows Given a subspace $U$ of the vector space $V$ and vectors $v,w\in V$ the following are equivalent.

  • $v-w\in U$

  • $v+U = w+U$

  • $(v+U)\cap(w+U)\neq\varnothing$

$\endgroup$
  • $\begingroup$ what does theorem 3.85 state? $\endgroup$ – Alvin Lepik Sep 5 '18 at 14:05
  • $\begingroup$ @AlvinLepik I have added theorem 3.85 now sorry for not mentioning it earlier. $\endgroup$ – Atif Farooq Sep 5 '18 at 14:10
1
$\begingroup$

Your proof is correct.

Note that you have implicitly chosen (arbitrary) preimages $v_i$ of the basis elements of $V/U$.
Now let $W:=\mathrm{span}(v_1,\dots v_n)\, \le U$.

Then we have $W\cong U/V$ thus $U\times (V/U) \cong U\times W$, and also $U\times W\cong U\oplus W=V$.

$\endgroup$
  • $\begingroup$ WoW so many isomorphisms, funny enough the subsequent exercises in my book go on to establish these very results. Thanks for the verfication. $\endgroup$ – Atif Farooq Sep 5 '18 at 18:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.