Exercise I.A.3

Question

If $U$ and $W$ are subspaces of $V$, show that $U\cup W$ need not be a subspace. However, if $U\cup W$ is a subspace, show that either $U \subseteq W$ or $W\subseteq U$.

Toğrul Topal · Accepted Answer

\begin{enumerate}
\item It is easy to verify that the sets $U:=\{(t,2t) \in \mathbb{R}^2: t \in \mathbb{R}\}$ and $W:=\{(t,3t) \in \mathbb{R}^2: t \in \mathbb{R}\}$ are linear subspaces of the real line $\mathbb{R}$. However, $(t,2t) + (t,3t) = (t,5t)$ which is not contained in $U\cup W$.
\item Suppose for the sake of contradiction that there exists $u \in U: u 
ot\in W$ and $w\in W: w 
ot\in U$. From the first condition we conclude that $u+w
ot\in U$ since otherwise $(u+w) - u = w \in U$. Similarly, $u+w
ot\in W$. But this contradicts the fact that $U\cup W$ is closed under linear combinations.
\end{enumerate}

Exercise I.A.3

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Exercise I.A.3

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