Linear Algebra Proof

1. Jul 18, 2008

danago

Prove that if U and V are subspaces of Rn, then so is $$U \cap V$$. Show by example that $$U \cup V$$ is not necessarily a subspace of Rn.

For the first part i can see that elements of $$U \cap V$$ are elements of both subspaces U and V. If i take any two element from $$U \cap V$$ and form some linear combination, then it must lie within U (since the elements are in U and U is a subspace) and it must also lie within V (since the elements are in V and V is a subspace), hence, $$U \cap V$$ is also a subspace.

Thats what i would have called informal reasoning. How can i actually prove that this is the case? Is my reasoning enough to stand as a solid proof?

For the second part i didnt have any troubles. I just defined:

$$\begin{array}{l} U = \{ k\left( {\begin{array}{*{20}c} 1 \\ 1 \\ \end{array}} \right)|k \in R\} \\ V = \{ k\left( {\begin{array}{*{20}c} 1 \\ 0 \\ \end{array}} \right)|k \in R\} \\ U \cup V = \{ a\left( {\begin{array}{*{20}c} 1 \\ 1 \\ \end{array}} \right),b\left( {\begin{array}{*{20}c} 1 \\ 0 \\ \end{array}} \right)|a,b \in R\} \\ \end{array}$$

I then chose two elements from the union of U and V and formed a linear combination which was not an element of the union of U and V, hence U and V is not a subspace.

$$\begin{array}{l} \left( {\begin{array}{*{20}c} 0 \\ 1 \\ \end{array}} \right),\left( {\begin{array}{*{20}c} 1 \\ 1 \\ \end{array}} \right) \in U \cup V \\ \left( {\begin{array}{*{20}c} 0 \\ 1 \\ \end{array}} \right) + \left( {\begin{array}{*{20}c} 1 \\ 1 \\ \end{array}} \right) = \left( {\begin{array}{*{20}c} 1 \\ 2 \\ \end{array}} \right) \notin U \cup V \\ \end{array}$$

I think thats the way to go about it, so its really just the first part i needed a bit of help with.

Thanks,
Dan.

2. Jul 18, 2008

Defennder

Well you can write it in this form:

Suppose $$\vec{u_1},\vec{u_2} \in U \cap V$$.

Clearly that implies that $$\vec{u_1} \ \mbox{and} \ \vec{u_2} \in U$$ and this in turn implies that for any $$a,b \in \Re , a\vec{u_1} + b\vec{u_2} \in U$$ since U is a subspace of R^n. What can you say about u1,u2 in relation to V now? And how would you relate this to $$U \cap V$$?

3. Jul 18, 2008

n_bourbaki

Yes. It is a perfectly good formal proof. In fact I would much rather see a simple explanation like that than some string of symbols like u,v,a,b with decoration.

4. Jul 18, 2008

danago

cheers for the replies guys. All pretty clear now :)