# 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.

• Why are the removed $w$'s in $U$? Commented Jul 22, 2022 at 5:37
• @Cpc because these $w’s$ can’t be written as a linear combination of $w_1,. . .,w_4$. They can only be written as a linear combination of the basis of $U$. Commented Jul 22, 2022 at 5:40
• @Cpc I should add more. I was using a theorem proved previously. That every spanning list can be reduced to a basis. In that process we remove those vectors that are in the span of the previous ones. This is the reason the $w’s \in U$ Commented Jul 22, 2022 at 5:41
• Oh yeah, that works. You might want to mention it though. Good job. Commented Jul 22, 2022 at 5:44
• Technically, reducing $u_1,...,w_4$ does not come with an ordering, albeit the 'natural' order would imply what you have written above. I might invoke Gram Schmidt if I was feeling sufficiently motivated. Commented Jul 22, 2022 at 6:08

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 $$i=1,\dots, 4$$; $$w_1 = e_5$$; $$w_2=e_6$$; $$w_3=e_1+e_5$$; $$w_4=e_2+e_6$$.

• I was using a theorem proved earlier that a spanning list can be reduced to a basis. Because the spanning list of $U+W$ was of length 8 and $\dim(U+W) \le 6$, we know that list is not a basis of $U+W$. And because there is a list of length 6 or less that does span $U+W$, we know that we can’t have a linearly independent list longer than length 6. Thus, some of the $w’s$ are indeed in $U$ Because $w_1,. . .,w_4$ is linearly independent as it’s a basis of $W$. We remove those vectors that are in the span of the previous ones. Commented Jul 22, 2022 at 17:13
• @Seeker What exactly do you mean by $w'$s? What are they in my example? Commented Jul 22, 2022 at 17:17
• Sone of the vectors from $w_1,. . .w_4$. Commented Jul 22, 2022 at 17:18
• @Seeker So in my example that would be $w_3$ and $w_4$, right? They do not lie in $U$. For example, $w_3$ does lie in the span of the previous vectors — i. e., $u_1, u_2, u_3, u_4, w_1, w_2$; but it does not lie in the span of $u_1, u_2, u_3, u_4$. Commented Jul 22, 2022 at 17:23
• Yes, that is a counterexample. Thanks a lot for answering! Commented Jul 22, 2022 at 17:28

Hint It suffices to prove $$U\cap W$$ has dimension at least $$2$$.

Use $$\dim (U+W)=\dim U+\dim W-\dim (U\cap W)$$.

Define $$T:U×W\to \Bbb{C^6}$$ by $$T(u, w) =u+w$$

Then

1. $$T$$ is a linear map.

2. $$\ker T=U\cap W$$

3. $$\operatorname{Im} T=U+V$$

4. $$\dim \operatorname{Im} T\le \dim \Bbb{C}^6=6$$

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.

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 $$u\ne kv$$.

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.

• There was a theorem proved earlier that every spanning list can be reduced down to a basis of the vector space. In that we remove those vectors that are in the span of the previous ones. I should have added that to the proof to make it clear. My bad on that one. Commented Jul 22, 2022 at 5:47

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.