Mfrhtyj

Given a vector space $V$ and a subspace $U$ of $V$.
$$ V cong U oplus(V/U) $$
Does the above equation always hold? Where $oplus$ is external direct sum.
For finite dimensional vector space $V$, here is my attemp of prove:

Let dimension of $U$ be m, dimension of $V$ be $n$.
Find a basis of $U$ : ${ mathbf{ u_1, u_2, cdots ,u_m}}$ and extend it to a basis for $V$ : ${ mathbf{ u_1, u_2, cdots ,u_m, v_1, cdots,v_{n-m} } }$.

For every vector $mathbf{x} in V$, we can write $mathbf{x}= c_1 mathbf{u_1}+ cdots+ c_m mathbf{u_m} + d_1 mathbf{v_1}+ cdots d_{n-m} mathbf{v_{n-m}}$ uniquely. Define a linear map $T$ as $$T(mathbf{x})=(c_1 mathbf{u_1}+ cdots+ c_m mathbf{u_m}, [d_1 mathbf{v_1}+ cdots d_{n-m} mathbf{v_{n-m}}])$$
,where $$ is used to express the equivalent class. We claim $T$ is an isomorphism.

Surjectivity is obvious. As for injectivity,

if $T(mathbf{x})=(mathbf{0},[mathbf{0}])$, then $ c_1 mathbf{u_1}+ cdots+ c_mmathbf{u_m}= mathbf{0} $
$Rightarrow c_1=0, c_2=0, cdots,c_m=0$
$Rightarrow x=d_1 mathbf{v_1}+ cdots d_{n-m} mathbf{v_{n-m}}$

Since $[d_1 mathbf{v_1}+ cdots d_{n-m} mathbf{v_{n-m}}] = [mathbf{0}]$, we have
$ (d_1 mathbf{v_1}+ cdots d_{n-m} mathbf{v_{n-m}}-mathbf{0})in U$, which means $d_1=0, d_2=0, cdots,d_{n-m}=0$
$Rightarrow mathbf{x}=mathbf{0}$, so $T$ is injective.

Is the above proof correct? Does this mean $V cong ker F oplus (V/ ker F) cong ker F oplusmathrm{im}F$ for any linear map $F$, because $ker F$ is a subspace of $V$ ?

The final question is about how should I prove it when the dimension of $V$ is infinite?

It is true that for every finite dimensional vector space $V$ with $U$ a vector subspace that
$$
V cong U oplus (V / U)
$$
I think your proof is essentially correct. And yes it is true that for $T: V rightarrow W$ any linear that we have
$$
V cong operatorname{ker}(T) oplus operatorname{Im}(T)
$$
The rank-nullity theorem is a direct consequence of this.

In more technical language, we say that every "short exact sequence" of finite dimensional vector spaces over a field $k$ $textit{splits}$. What this means is that if $T : U rightarrow V$, $S : V rightarrow W$ are linear maps such that $T$ is injective, $operatorname{ker}(S) = operatorname{Im}(T)$, and $S$ is surjective, then $V cong U oplus W$.

Note then that this directly gives us our result since if $U$ is a subspace of $V$, then the inclusion map $iota : U rightarrow V$ and projection map $pi : V rightarrow (V / U) $ set up exactly a short exact sequence.

In terms of whether or not this extends to infinite dimensional vector spaces, the result does hold again (assuming the axiom of choice), and the proof is essentially the same. All your proof relies on is the ability to extend a basis of a subspace to a basis of your entire space. We can do this with the axiom of choice.

Your proof works. The answer to your second question is yes, that is true. For an infinite dimensional vector space, take any linear map $F: V -> W$. Then $U = ker F$ is a subspace of $V$. Note that we have a short exact sequence (if you don't know what that means, don't worry, the explanation is coming)
$$0to Uto Vto V/Uto 0$$

(That is, there's an injective map $U to V$ (inclusion, I'll call it $i$) and a surjective map $V to V/U$ (the quotient map, I'll call it $q$) such that the image of the injection is the kernel of the surjection).

But there's also a surjective map $V to U$ (projection onto $U$, I'll call it $p$), and note that for any $u in U$, $pi(u) = u$ (since $p$ fixes $u$).

Now, we're going to show that $V$ is the (internal) direct sum of the kernel of $p$ and the image of $i$. First, note that it's the sum of the two: for any $v in V$, $v = (v - ip(v)) + ip(v)$, $ip(v)$ is obviously in the image of $i$, and $p(v - ip(v)) = p(v) - pip(v) = 0$ (with the last equality being due to our note about $p$, since $p(v)in U$.

And further, the intersection is trivial: if $v in ker(p)capmathrm{im}(i)$ then there is some $uin U$ such that $i(u) = v$, and $pi(u) = p(v) = 0$, but $pi(u) = u$, so $u = 0$, hence $v = 0$. Thus, $V = ker(p)oplus mathrm{im}(i)$.

Now, it's clear that $mathrm{im}(i)cong U$ since it's the image of $U$ under an injective map, so we need only show that $ker(p)cong V/U$.

For that purpose, since $q$ is surjective, for any $w in V/U$, there is some $v in V$ such that $w = q(b)$. But since $V = ker(p)oplusmathrm{im}(i)$, there are unique $u in U$, $x in ker(p)$ such that $v = i(u) + x$, so $w = q(b) = q(i(u)+x) = qi(u) + q(x)=q(x)$ (since the image of $i$ is the kernel of $q$, in particular $qi = 0$). Thus, $q|_{ker(p)}: ker(p)to V/U$ is surjective. But also, if $q(v) = 0$, and $v in ker(p)$. But $ker(q) = mathrm{im}(i)$, and $p$ fixes $mathrm{im}(i)$, so the only element that it sends to $0$ is $0$ itself, so we must have $v = 0$, hence $q|_{ker(p)}$ is also injective, so is an isomorphism.

Thus, we have $V cong mathrm{im}(i)oplusker(p) cong Uoplus (V/U)$, as required.

This is precisely a special case of the Splitting Lemma, and my proof is essentially just one part of the proof of that, translated: there are easier ways to prove it, but I thought that it would be useful to see it in a more broadly applicable form.