# Is $V$ isomorphic to direct sum of subspace $U$ and $V/U$?

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 $$c_1 \mathbf{u_1}+ \cdots+ c_m\mathbf{u_m}= \mathbf{0}$$
$$\Rightarrow c_1=0, c_2=0, \cdots,c_m=0$$
$$\Rightarrow x=d_1 \mathbf{v_1}+ \cdots d_{n-m} \mathbf{v_{n-m}}$$
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 $$V \cong \ker F \oplus (V/ \ker F) \cong \ker F \oplus\mathrm{im}F$$ 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.

• I am wondering if the following statement is false for infinite dimensional spaces. "If $V$ is a vector space and $U$ is a subspace of $V$, then there exists another subspace of $V$ called $U^\perp$ such that every element of $V$ can be uniquely expressed as the sum of an element from $U$ with an element from $U^\perp$." (I am thinking a counterexample would be if $U$ was the set of real number sequences with finite support (i.e. eventually zero) and $V$ is the set of all real number sequences.) – irchans Jan 3 at 19:03
• @irchans If we take the axiom of choice, then every subspace of a vector space has a direct sum complement (what you call the perpendicular space, but this language is typically reserved for a space equipped with some bi-linear form). The proof is pretty simple. Let $V$ be a $k$-vector space, with $U$ a vector subspace. Let $\mathcal{B}_{U}$ be a basis for $U$ and extend it to a basis $\mathcal{B}_{V}$ (using the axiom of choice) for $V$. Then let $W = \operatorname{Span}_{k}\left( \mathcal{B}_{V} \backslash \mathcal{B}_{U} \right)$. Then $V = U \oplus W$. – Adam Higgins Jan 3 at 19:13
• @irchans Perhaps the reason you think that your example is a counter example is because of the $\textit{weirdness}$ of bases of infinite dimensional vector spaces. Notice that a subset $S$ of a vector space $V$ is said to be a basis if and only if every element $v \in V$ can be written as a $\textbf{finite}$ linear combination of the elements of $S$, and that there is no finite non-trivial linear relation amongst the elements of $S$. – Adam Higgins Jan 3 at 19:19
• Thank you very much ! – irchans Jan 3 at 19:27
• This set of notes seems relevant math.lsa.umich.edu/~kesmith/infinite.pdf – irchans Jan 3 at 19:38

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) $$0\to U\to V\to V/U\to 0$$

(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 $$v \in \ker(p)\cap\mathrm{im}(i)$$ then there is some $$u\in U$$ 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 $$V = \ker(p)\oplus\mathrm{im}(i)$$, 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 $$V \cong \mathrm{im}(i)\oplus\ker(p) \cong U\oplus (V/U)$$, 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.