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

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 $$U \cap W \neq \{0\}$$.

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.

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

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

3. $$0 = a_1u_1+...+a_5u_5=b_1w_1+...+b_5w_5 \implies 0 = a_1u_1+...+a_5u_5+b_1w_1+...+b_5w_5$$

4. $$R^9 \implies$$ 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.

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

2. $$0 = a_1u_1+...+a_5u_5+b_1w_1+...+b_5w_5$$ 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.

• you just checked the following result: $\text{dim}(V + W) = \text{dim}V + \text{dim}W - \text{dim}(V \cap W)$ Jul 26, 2019 at 0:38

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