Every Subspace of V is a part of a direct sum equal to V

As an application of the result above, we now show that every subspace of a finite-dimensional vector space can be paired with another subspace to form a direct sum of the whole space.

Every subspace of $V$ is part of a new direct sum equal to $V$

Suppose $V$ is finite-dimensional and $U$ is a subspace of $V$. Then, there is a subspace $W$ of $V$ such that $V=U\oplus W$.

Proof. Because $V$ is finite-dimensional, so is $U$. Thus there is a basis $u_1,\dots ,u_m$ of $U$. Of course $u_1,\dots ,u_m$ is a linearly independent list of vectors in $V$. Hence this list can be extended to a basis $u_1,\dots ,u_m,w_1,\dots ,w_n$ of $V$. Let $W=\text{span}(w_1,\dots ,w_n)$.

To prove that $V=U\oplus W$, by the condition for direct sum of two subspaces,we only need to show that

$$V=U+W\text{and}U\cap W=\{0\}$$

To prove the first equation, suppose $v\in V$. Then, because the list $u_1,\dots u_m,w_1,\dots ,w_n$ spans $V$, there exist $a_1,\dots ,a_m,b_1,\dots ,b_n\in \mathbb{F}$ such that

$$v=\underset{u}{\underbrace{a_1{u}_1+\cdots +a_m{u}_m}}+\underset{w}{\underbrace{b_1{w}_1+\cdots +b_n{w}_n}}$$

We have $v=u+w$, where $u\in U$ and $w\in W$ are defined as above. Thus $v\in U+W$, completing the proof that $V=U+W$.

To show that $U\cap W=\{0\}$, suppose $v\in U\cap W$. Then there exist scalars $a_1,\dots ,a_m,b_1,\dots ,b_m\in \mathbb{F}$ such that

$$\begin{array}{rl}v& =a_1{u}_1+\cdots +a_m{u}_m\\ v& =b_1{w}_1+\cdots +b_n{w}_n\end{array}$$

Subtracting the two equations from each other:

$$a_1{u}_1+\cdots +a_m{u}_m-b_1{w}_1-\cdots -b_n{w}_n=0$$

Because the list $u_1,\dots ,u_m,w_1,\dots ,w_n$ is linearly independent, this implies that

$$a_1=\cdots =a_m=b_1=\cdots =b_n=0$$

Thus $v=0$, completing the proof that $U\cap W=\{0\}$.