Exercise 13, Section 2.C - Linear Algebra Done Right. Exercise: Suppose $U$ and $W$ are both $4$-dimensional subspaces of $C^6$. Prove that
there exist two vectors in $U \cap W$ such that neither of these vectors is a
scalar multiple of the other.
My attempt at a proof is as follows.
Proof: Let $u_1,. . .,u_4$ be a basis of $U$ and let $w_1,. . .,w_4$ be a basis of $W$. Then, $u_1,. . .,u_4,w_1,. . .,w_4$ spans $U+W$. Because $U+W$ is a subspace of $C^6$, the $\dim(U+W)\le 6$. Thus, $u_1,. . .,u_4,w_1,. . .,w_4$ can be reduced to a basis of $U+W$. In the process, none of the $u's$ get removed as $u_1,. . .,u_4$ is linearly independent. Thus, some of the $w's$ get removed in the process. Because $\dim(U+W)\le 6$, at least two of the $w's$ get removed. These are the $w's\in U\cap W$. Because $w_1, . .,w_4$ is linearly independent, none of these two vectors are a scalar multiple of each other.
Is the proof correct?
Edit: I implicitly use that theorem that every spanning list in a vector space can be reduced to a basis of that vector space. In the process, we remove those vectors that are in the span of the previous ones. Thus, if we have the list $v_1,. . .,v_k$. We remove $v_j$ only if $v_j$ is in the span of $v_1,. . .,v_{j-1}$.
Edit 2: I have come to know that the proof is wrong. For future readers, I am writing another proof that is also suggested as a hint in the answers.
Proof 2: Using the formula $\dim (U+W)=\dim(U)+\dim(W)-\dim(U\cap W)$, we see that $\dim(U\cap W) \ge 2$. This is because  $\dim(C^6)=6$ and $U+W$ is a subspace of $C^6$. Thus, $\dim(U+W)\le 6$. Let $j\in Z^+$ with $2\le j\le 6$. Let $\dim(U\cap W)=j$. Let $v_1,. . .,v_j$ be a basis of $U\cap W$. Then we have that $v_1,v_2\in U\cap W$ are not scalar multiples of each other as they are linearly independent. Completing the proof.
 A: Define $T:U×W\to \Bbb{C^6}$ by $$T(u, w) =u+w$$
Then

*

*$T$ is a linear map.


*$\ker T=U\cap W$


*$\operatorname{Im} T=U+V$


*$\dim \operatorname{Im} T\le \dim \Bbb{C}^6=6$
Now $\begin{align}\dim (\ker T) &=\dim(U×W) -\dim \operatorname{Im} T\\&=\dim U+\dim W -\dim \operatorname{Im} T\\&\ge 4+4-6\\&\ge 2\end{align}$
Hence $\dim(U\cap V) \ge 2$ . Now you can choose two linearly vectors from $U\cap W$ .
Note:
$U×W$ is a vector space of $\dim U+\dim W$ . Because $\{(u_1,0),(u_2,0),\ldots ,(u_m, 0),(0,w_1),(0,w_2),\ldots,(0,w_n)\}$
is a basis of $U×W$ where $\{u_1, u_2, \ldots, u_m\}$ and $\{(v_1, v_2, \ldots, v_n\}$ are basis of $U$ and $W$ respectively.
A: Hint
It suffices to prove $U\cap W$ has dimension at least $2$.
Use $$\dim (U+W)=\dim U+\dim W-\dim (U\cap W)$$.
A: Because $4\le \dim(U+W)\le 6$, and $\dim(U\cap W)=\dim(U)+\dim(W)-\dim(U+W)$ it is implied that $$2\le \dim(U\cap W)\le 4,$$ which suggests that the subspace is neither zero-space $\{0\}$ nor a line of dimension $1$. So, you can always choose two linearly independent vectors $u$ and $v$ in $U\cap W$ such that $u\ne kv$.
A: If by $w'$ you mean some of the $w_1, \dots, w_4$, then no, your proof is not correct. Those of $w_1, \dots, w_4$ which you remove do not have to lie in $U$. It's easy to construct an example where none of $w_1, \dots, w_4$ lie in $U$. For example, let $e_1, \dots, e_6$ be a basis of $\mathbb{C}^6$;
$u_i=e_i$ for $i=1,\dots, 4$; $w_1 = e_5$; $w_2=e_6$; $w_3=e_1+e_5$; $w_4=e_2+e_6$.
A: 
Thus, u1,...,u4,w1,...,w4 can be reduced to a basis of U+W. In the process, none of the u′s get removed as u1,...,u4 is linearly independent. Thus, some of the w′s get removed in the process.

That seems problematic to me. Reduction of a linearly dependent set of vectors to a linearly independent one is not in general a unique process; there are many different combinations of vectors you can remove to achieve independence. The fact that the u's are independent suggests that there exists a process that doesn't remove them, but your wording implies that no process removes them, which is not correct. I'm also not sure your instructor would consider it trivial that once one w is removed, the set is still linearly dependent.
I think a better approach is to first prove that there are non-zero vectors in U∩W, then pick one, then take the subset of U that is perpendicular to that vector, and the subset of W perpendicular to the vector, and then show that the intersection of those two subsets has a nonzero vector.
A: Define $A: U \times W \to \mathbb{C}^6$ by $A(u,w) = u+w$. Since $\dim {\cal R} A \le 6$, we have $\dim \ker A \ge 2$. In particular, there are linearly independent $(u_1,w_1), (u_2,w_2)$ in $\ker A$.Since they are in the kernel, we see that $w_1=-u_1, w_2 = -u_2 \in U \cap W$ and linear independence implies that $(u_1,-u_1), (u_2,-u_2)$ are not scalar multiples of each other.
