# Prove every subspace of $V$ is part of a direct sum equal to $V$

In Linear Algebra Done Right, the book proves that every subspace of $$V$$ is part of a direct sum equal to $$V$$. I generally follow the proof, but do not understand some points.

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

Proof Because $$V$$ is finite-dimensional, so is $$U$$ (proved in 2.26. I am OK). Thus there is a basis $$u_1,...,u_m$$ of $$U$$. Of course $$u_1,...u_m$$ is a linearly independent list of vectors in $$V$$ (I am OK). Hence this list can be extended to a basis $$u_1,...,u_m, w_1,...,w_n$$ of $$V$$ (I am not OK. Does it mean $$w_1,...w_n$$ can only be vectors that extend to the basis of $$V$$? If so, why the title of the proof said EVERY subspace of $$V$$ is part of a direct sum equal to $$V$$).

To prove $$V = U\bigoplus W$$, we need only show that $$V = U+W \text{ and } U \cap W = \{0\}$$ (I am OK).

To prove the first equation above, suppose $$v \in V$$. Then, because the list $$u_1,...,u_m, w_1,...w_n$$ spans $$V$$, there exist $$a_1,...a_m, b_1,...b_m \in \mathbb{F}$$ such that $$v=a_1u_1+...+a_mu_m+b_1w_1+...+b_nw_n$$.

In other words, we have $$v=u+w$$, wheere $$u \in U$$ and $$w \in W$$ are defined as above. Thus $$v \in U + W$$, completing the proof that $$V = U + W$$. (I am also OK.)

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

$$v=a_1v_1+...+a_mv_m = b_1w_1+...+b_nw_n$$. Thus $$a_1u_ + ...+a_mu_m - b_1w_1-...-b_nw_n = 0$$

Because $$u_1,...u_m,w_1,...,w_n$$ is linearly independent, this implies that $$a_1=...=a_m=b_1=...=b_n = 0$$. Thus $$v=0$$, completing the proof that $$U \cap W = \{0\}$$.

Is the proof means for every subspace $$U$$ of finite-dimensional $$V$$, we can find a $$W$$ that is the direct sum of $$V$$?

• It means that every subspace has a complementary subspace. Jan 1 '19 at 23:09
• I'm not entirely sure what you are confused about but it seems like you are under the impression that $V$ only has one basis. Consider for example $\mathbb{R} ^2$. We have the standard basis $\{(1,0),(0,1)\}$ but we can replace $(1,0)$ with $(1,1)$ and it is still a basis. In fact replacing $(1,0)$ with anything in the form $(a,b)$ where $a\neq 0$ will result in a basis. $\{(1,2),(1,3)\}$ is another example of a basis. Jan 1 '19 at 23:13

You have a subspace $$U$$ with basis $$u_1,...,u_m$$.
Extend the basis to a basis of $$V$$ by adding vectors $$w_1,...,w_n$$. There is some freedom to choose the $$w_k$$ but they must be linearly indepdendent and the collection must span $$V$$.
Let $$W=\operatorname{sp} \{w_k\}$$.
Since the whole collection spans $$V$$, we must have $$V = U +W$$. If $$u\in U, w\in W$$ and $$u+w = 0$$, we must have $$u=w=0$$ since the whole collection is linearly independent.
Note that $$W$$ is not unique.