The bases of two subspaces form the basis of their direct sum - is my proof correct?

Question

This problem is from Linear Algebra Done Right 3rd Edition, exercises 2.B.7:

Suppose $U$ and $W$ are subspaces of $V$ such that $V=U \oplus W$. Suppose also that $u_1,\ldots,u_m$ is a basis of $U$ and $w_1,\ldots,w_n$ is a basis of $W$. Prove that $$u_1,\ldots,u_m,w_1,\ldots,w_n$$ is a basis of $V$.

Now I have seen the proof here, but I wonder if my own proof is correct:

Since $V=U \oplus W$, $V=U+W$ and $U \cap W=\{0\}$. The former implies that $u_1,\ldots,u_m,w_1,\ldots,w_n$ spans $V$.

For the latter, $$ 0 = a_1u_1 + \cdots + a_mu_m\\ 0 = b_1w_1 + \cdots + b_nw_n $$ Subtract both sides, $$ 0 = a_1u_1 + \cdots + a_mu_m - b_1w_1 - \cdots - b_nw_n $$ Since $u_1,\ldots,u_m$ and $w_1,\ldots,w_n$ are bases, $a_1=\cdots=a_m=b_1=\cdots=b_n=0$. Therefore, $u_1,\ldots,u_m,w_1,\ldots,w_n$ is linearly independent.

Above all, $u_1,\ldots,u_m,w_1,\ldots,w_n$ is a basis of $V$.

It looks good. However, suppose $U \cap W$ contains elements other than $0$, by substituting the left sides of the first set of equations with an arbitrary $v \in U \cap W$, I can still subtract both sides and derive the same results. I think there is something wrong with my proof, but what is it?

Answer 1

You tried to prove that $0 = a_1u_1 + \cdots + a_mu_m$ and $0 = b_1w_1 + \cdots + b_nw_n$ imply $a_1 = \ldots = a_m = b_1 =\ldots= b_n = 0$ but this is not sufficient. In fact, this claim is a direct consequence of $\{u_1, \ldots, u_m\}$ and $\{w_1, \ldots, w_n\}$ being linearly independent so you don't have to subtract anything.

What you should do (it is done in the proof you linked) is show that $0 = a_1u_1 + \cdots + a_mu_m + b_1w_1 + \cdots + b_nw_n$ implies $a_1 = \ldots = a_m = b_1 =\ldots= b_n = 0$.

In the case of $U \cap V = \{0\}$ we have that $0 = a_1u_1 + \cdots + a_mu_m + b_1w_1 + \cdots + b_nw_n$ indeed implies $0 = a_1u_1 + \cdots + a_mu_m$ and $0 = b_1w_1 + \cdots + b_nw_n$, which then implies $a_1 = \ldots = a_m = b_1 =\ldots= b_n = 0$. However, if $U \cap V \ne \{0\}$, we can have $u = a_1u_1 + \cdots + a_mu_m$ and $-u = b_1w_1 + \cdots + b_nw_n$ for some $u \in U \cap V$ which doesn't help you at all.