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Q. Say $U$ and $W$ are subspaces of a a finite dimensional vector space $V$ (over the field of real numbers). Let $S$ be the set-theoretical union of $U$ and $W$. Which of the following statements is true:

a) Set $S$ is always a subspace of $V$

b) Set $S$ is never a subspace of $V$

c) Set $S$ is a subspace of $V$ if and only if $U=W$

d) None of the above

I've researched on this question and found a proof that says set $S$ is not always subspace of $V$ (Is it true?).

If $U=W$ then $S$ is a subspace of $V$ since $U$ and $W$ are closed under addition hence $U \cup W = U$ (or $W$). So can I say that $S$ is a subspace of $V$ if and only if $U = W$?

Also, please can you give an example of how to calculate union of two vector spaces? So what would be the union of vector spaces $\begin{pmatrix} a& 0 &0\end{pmatrix} \cup \begin{pmatrix}0& b& 0\end{pmatrix}$ ?

Thanks

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Sabih Sai
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  • It is true if the underline field is $\mathbb{R}$... as a consequence of Baire category theory... other wise it may not be true...for an example if you consider the underline field as $|mathbb{Z_2}$ then you can find some counter example – Anubhav Mukherjee Aug 26 '15 at 14:24
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    The union of $U$ and $W$ is a subspace of $V$ if and only if either $U\subseteq W$ or $W\subseteq U$. For a proof, suppose that neither subset relation holds. Let $x$ be in $U$ but not in $W$ and let $y$ be in $W$ but not in $U$. Show that $x+y$ is not in the union of $U$ and $W$. – André Nicolas Aug 26 '15 at 14:26
  • have a look in this link http://mathoverflow.net/questions/26/can-a-vector-space-over-an-infinite-field-be-a-finite-union-of-proper-subspaces – Anubhav Mukherjee Aug 26 '15 at 14:29
  • Does this answer your question? Union of two vector subspaces not a subspace? – Anne Bauval Jul 16 '23 at 14:13

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If $U \subset W$ (or vice versa) then $U \cup W = W $ so if W is a subspace so is the union.

Otherwise there is $a \in U ,\ a \notin W$ and $b \in W,\ b \notin U$

So, $a, b \in U \cup W$.

But $a + b \notin U$ otherwise with $(-a) \in U$ then $(-a) + a + b \in U \implies b \in U$

And similarly $a + b \notin W$

Therefore $a + b \notin U \cup W$, i.e. there are vectors $a, b$ in $U \cup W$ whose linear combination is not in $U \cup W$, so $U \cup W$ is not closed.

Tom Collinge
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