Proving the existence of a complement subspace $U' \subset V$.

Question

Given a subspace $U \subset V$ of a vector space $V$, show that there exists a subspace $U' \subset V$, such that $U+U'=V$ and $U \cap U'=\{0_V\}$.

I am aware that $U'$ is the complement of $U$ in $V$.

Would it be correct to prove this over the basis extension theorem?

For every subspace $U$ there exists a basis of $U$, let it be $B_U$. It is true that $B_U \subset V$. According to the basis extension theorem one can extend any set of linearly independant vectors in $V$ to a basis of $V$. Thus one can extend $B_U$ to a basis of $V$. Let this basis be $B_V$. It is true, since the vectors in $B_V$ are linearly independant, that $\langle B_U \rangle \cap \langle B_V \setminus B_U \rangle = \{0_V\}$. It is also obviously true that $\langle B_U \rangle + \langle B_V \setminus B_U \rangle =V $. Let $\langle B_V \setminus B_U \rangle=U'$.

Is this correct or am I missing something?

This is correct. If you're early in your linear algebra knowledge, I'd recommend hashing out in more explicit detail the "obvious" part, for practice, but conceptually, you're set. — Amin Idelhaj, Dec 21 '16 at 23:48
Actually I had the notion of a direct sum in mind when I wrote that, nevermind you're good. — Amin Idelhaj, Dec 22 '16 at 00:01
@user0 The basis extension theorem holds for infinite dimensional vector spaces aswell (by Zorn's lemma). — B.Swan, Sep 05 '23 at 22:16

score 2 · Accepted Answer · answered Dec 22 '16 at 00:13

Yep, your proof is absolutely correct, your "obvious" part, in my opinion, is perfectly valid and should be obvious to anyone with a basic understanding of what a set operation is and linear algebra.

One thing to note is that you can avoid the basis extension theorem (though it is perfectly valid) in favour of a mathematical logical argument. Since $U$ is a proper subspace of $V$, there is some $a \in V$ that is not in $U$ and $a \neq \vec{0}$ since the zero vector is in $U$.

Notice that the subspace spanned by $a$, call it $W$ has the property that $U \cap W = \vec{0}$, since if $\vec{a}$ is not in $U$, then $c\vec{a}, c \in \mathbb{R} \neq 0$ also cannot be in $U$, otherwise contradicting the fact that $U$ is a subspace closed under scalar multiplication...

Anytime you add in some vector $a$ that is not in $U$, no linear combination of it can be in $U$. Since if some linear combination is in $U$, but $a$ is not in $U$, this contradicts its subspace assumption.

You can extend this argument to keep adding some vectors not in $U$ and can conclude that indeed, there is some subspace $U'$ such that $U + U' = V$

Proving the existence of a complement subspace $U' \subset V$.

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