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Linear Algebra Proof

  • Thread starter danago
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  • #1
danago
Gold Member
1,122
4
Prove that if U and V are subspaces of Rn, then so is [tex]U \cap V[/tex]. Show by example that [tex]U \cup V[/tex] is not necessarily a subspace of Rn.

For the first part i can see that elements of [tex]U \cap V[/tex] are elements of both subspaces U and V. If i take any two element from [tex]U \cap V[/tex] 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, [tex]U \cap V[/tex] 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:

[tex]
\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}
[/tex]

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.

[tex]
\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}
[/tex]

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.
 

Answers and Replies

  • #2
Defennder
Homework Helper
2,591
5
Well you can write it in this form:

Suppose [tex]\vec{u_1},\vec{u_2} \in U \cap V[/tex].

Clearly that implies that [tex]\vec{u_1} \ \mbox{and} \ \vec{u_2} \in U[/tex] and this in turn implies that for any [tex]a,b \in \Re , a\vec{u_1} + b\vec{u_2} \in U[/tex] 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 [tex]U \cap V[/tex]?
 
  • #3
104
0
Prove that if U and V are subspaces of Rn, then so is [tex]U \cap V[/tex]. Show by example that [tex]U \cup V[/tex] is not necessarily a subspace of Rn.

For the first part i can see that elements of [tex]U \cap V[/tex] are elements of both subspaces U and V. If i take any two element from [tex]U \cap V[/tex] 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, [tex]U \cap V[/tex] 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?
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
danago
Gold Member
1,122
4
cheers for the replies guys. All pretty clear now :)
 

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