Proof verification for the set of an intersection of two finite-dimensional subspaces.

Question

I need some help checking if my alternate proof to the problem below holds.

Problem Statement

Suppose that $U$ and $W$ are both five-dimensional subspaces of $R^9$. Prove that $U \cap W \neq \{0\}$.

Proof by contradiction.

Suppose $U \cap W = \{0\}$. Let $u_1,...,u_5$ denote vectors belonging to a basis of $U$ and $w_1,...,w_5$ denote vectors belonging to a basis of $W$.

1. $u_1,...,u_5$ is a basis $\implies$ $a_1=...=a_5=0$ for $a_1u_1+...+a_5u_5=0$

This follows from the fact that vectors in a basis are linearly independent by definition.

2. $w_1,...,w_5$ is a basis $\implies$ $b_1=...=b_5=0$ for $b_1w_1+...+b_5w_5=0$

3. $U \cap W = \{0\} \implies 0 = a_1u_1+...+a_5u_5=b_1w_1+...+b_5w_5$

4. $0 = a_1u_1+...+a_5u_5=b_1w_1+...+b_5w_5 \implies 0 = a_1u_1+...+a_5u_5+b_1w_1+...+b_5w_5$

5. $R^9 \implies$ the list of vectors $u_1,...,u_5,w_1,...,w_5$ must be linearly dependent.

No linearly independent list can be greater in length than a spanning list.

6. $u_1,...,u_5,w_1,...,w_5$ must be linearly dependent $\implies 0 = a_1u_1+...+a_5u_5+b_1w_1+...+b_5w_5$ must have nonzero coefficients.

7. $0 = a_1u_1+...+a_5u_5+b_1w_1+...+b_5w_5$ must have nonzero coefficients $\implies a_1u_1+...+a_5u_5=-b_1w_1-...-b_5w_5$ must have nonzero coefficients.

Note, a nonzero $a$ implies a nonzero $b$ because each respective list of vectors for $U$ and $W$ is linearly independent (no two nonzero $a$'s can cancel to 0 if all the other coefficients are 0).

Which means that there is some nonzero element belonging to $U \cap W = \{0\}$ because we can express some $u$ in terms of $w$, giving us a contradiction and proving that $U \cap W \neq \{0\}$.

Thanks.

Answer 1

The idea is good, but you could have noticed it as follows :

Suppose by contradiction that $\textsf U \cap \textsf W =\{0\}$. Let $\beta=\{v_1,v_2,v_3,v_4,v_5\}$ and $\gamma=\{w_1,w_2,w_3,w_4,w_5\}$ be basis for $\textsf U$ and $\textsf W$ respectively.

From here, we will prove that the set $\beta \cup \gamma$ is also linearly independent, for this, suppose that $$\sum_{j=1}^5 a_j v_j +\sum_{j=1}^5 a_{j+5}w_j =0$$ for some scalars $a_1,a_2,\dots,a_{10}$. Then, substracting the second sum to both sides we obtain $$\sum_{j=1}^5 a_j v_j=-\sum_{j=1}^5 a_{j+5}w_j \in \textsf U \cap \textsf W$$ since the left hand side is in $\textsf U$ and the right hand side in $\textsf W$. But $\textsf U \cap \textsf W =\{0\}$, which means $$\sum_{j=1}^5 a_j v_j = 0 = \sum_{j=1}^5 (-a_{j+5})w_j$$ It follows that $$a_1=a_2=\cdots=a_{10}=0$$ since $\beta$ and $\gamma$ are linearly independent sets.

In conclusion, we prove that $\beta \cup \gamma$ (with cardinality 10) is a linearly independent set in $\mathbb R ^9$, but the set $\{e_1,e_2,\dots,e_9\}$ (with cardinality 9) is also linearly independent (it is the standard basis). And this is enough to find the contradiction.

So, $\textsf U \cap \textsf W \neq \{0\}$.