# Let $a,b \in R^5$ be two independent vectors and let $W$ be the subspace spanned by $a,b$. Prove that $U\cap W$ contains a non-trivial vector

Let $a,b \in \Bbb R^5$ be two independent vectors and let $W$ be the subspace spanned by $a,b$. Let $U=\{(x_1, x_2, x_3, x_4, x_5) \in \Bbb R^5\colon x_1 + 2x_2 + x_3 - 3x_4 + x_5 = 0 \}$.

Prove that $U\cap W$ contains a non-trivial vector.

With the information that $a,b$ are two independent vectors in $\Bbb R^5$ that span a subspace $W$ and the given equation $U$ I need to show that the intersection $U\cap W$ is not trivial: it must have a dimension greater than zero.

• What is $U{}{}{}$? – Git Gud Dec 1 '13 at 17:49
• I have added $U$, I had forgotten to add it to the question. – Ruben Dec 1 '13 at 18:06
• My native language is Dutch. – Ruben Dec 1 '13 at 18:17
• No help. I'll just state the theorem: $\dim(F+G)+\dim (F\cap G)=\dim (F)+\dim(G)$. Is this familiar? – Git Gud Dec 1 '13 at 18:19
• Yes that does seem familiar. I think the dim($U$) is 1 and dim($W$) is 2. However, because I don't know what W is exactly I don't see how I can say anything about dim($U+W$). Maybe I can ponder that a bit more. Thank you for the theorem. – Ruben Dec 1 '13 at 18:25

Remark: For all subspaces $F,G$ of $\Bbb R^5$ the following holds: $$\dim(F+G)+\dim (F\cap G)=\dim (F)+\dim(G).$$
Hints: Let $F=W$ and $G=U$.
1. What does $\dim(W)$ equal to? What about $\dim(U)$?
2. So the RHS of the equality is $\boxed{\_\_\_}$?
3. If $U\cap W=\{0_{\Bbb R^5}\}$, then $\dim (W\cap U)=\boxed{\_\_\_}$?
4. So if $U\cap W=\{0_{\Bbb R^5}\}$, then LHS is just $\dim(U+W)$ and it equals the RHS.
5. Finally use the fact that $U+W\subseteq \Bbb R^5$.