(adsbygoogle = window.adsbygoogle || []).push({}); Prove that if U and V are subspaces of R^{n}, then so is [tex]U \cap V[/tex]. Show by example that [tex]U \cup V[/tex] is not necessarily a subspace of R^{n}.

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.

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# Homework Help: Linear Algebra Proof

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