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Let $V$ a vector space over a field $F$. Let $U_1$ and $U_2$ subvector spaces of $V$. Prove or disprove that $U_1\cup U_2$ is a subvector space aswell.

Consider $V=\mathbb R^2$, $U_1=$span$((0,1)^T)$, $U_2=$span$(1,0)^T$. Then $U_1\cup U_2$ is not closed under addition because $(0,1)^T,(1,0)^t\in U_1\cup U_2$, but $(0,1)^T+(1,0)^T=(1,1)^T \notin U_1\cup U_2$.

I do not understand this solution. Here $U_1\cup U_2=\mathbb R^2$ and clearly $(1,1)^T\in \mathbb R^2$. Could someone explain this to me?

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  • $\begingroup$ Union does not mean span. A vector is in the union of $U_1,U_2$ iff it is in one or the other (or both). $\endgroup$ – lulu Apr 3 '18 at 14:34
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you claim that $$(1,1)^T\in U_1 \cup U_2.$$

What would that mean? It would mean that either

  1. $(1,1)^T\in U_1$ or
  2. $(1,1)^T\in U_2$,

by definition of the union. But both cases are not true. Hence $(1,1)^T\notin U_1 \cup U_2$.

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It looks like you're confusing vector addition and the union of the two sets. The union of the two sets is $\{(x,0) \,|\, x \in \mathbb{R} \} \cup \{(0,y) \,|\, y \in \mathbb{R} \}$. Clearly $(1,1)$ is not in this set since it does not have a zero in either the x or y co-ordinate.

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