Assume that $A$ and $B$ are subspaces of the vector space $\Bbb{R}^n$. Show that $H=A \cap B$ is also a subspace of the vector space $\Bbb{R}^n$. Show by construction of the example that $R=A \cup B$ is not always a subspace of $\Bbb{R}^n$.
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1Duplicate of various posts, here and here for example. – Zev Chonoles Apr 09 '13 at 03:43
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1http://math.stackexchange.com/questions/298955/intersection-and-union-of-subspaces?rq=1 http://math.stackexchange.com/questions/71872/union-of-two-vector-subspaces-not-a-subspace?lq=1 related (...what @ZevChonoles said) – Stahl Apr 09 '13 at 03:45
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2When I was a math student problems like show that this is a that frustrated my fellow students. I liked them. To solve them, look up the definition of a that. Verify all the elements of the definition. It tests well that you understand the definition of a that. In this case, the challenge is closure: if you add two elements you get another element. – Ross Millikan Apr 09 '13 at 03:53
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First thing is almost trivial: you only have to understand the very definitions. For the second, as counter example, take
$$A:=Span\left\{\;\binom{1}{0}\;\right\}\;,\;\;B:=Span\left\{\;\binom{1}{1}\;\right\}\;,\;\;A,B\subset \Bbb R^2$$
Take now, for example
$$u:=\binom{1}{0}\;,\;\;v:=\binom{1}{1}$$
and assume $\, A\cup B\,$ is a subspace, but then were does $\,u+v\,$ belong...?
Added: Suppose
$$u+v=a\in A\implies v=a-u\in A\ldots\text{contradiction, since it's easy to check that}\;\;v\notin A$$
then it must be that
$$u+v=b\in B\implies u=b-v\in B\ldots\text{again contradiction because}\;\; u\notin B\;!$$
DonAntonio
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I understand that u+v should be contained in both A and B if their union was a subspace. But i struggle in formulating this in a correct mathematical manner. – user71671 Apr 09 '13 at 08:40
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@user71671, check what I added to my answer which is the heart of the proof that $,A\cup B,$ isn't a subspace unless $,A\subset B;;or;;B\subset A,$ – DonAntonio Apr 09 '13 at 10:07