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let there be $A,B \in V$.

$$A+B=\{a+b: a\in A,b \in B\}$$

Does $\operatorname{Span}(A+B)=\operatorname{Span}(A)+\operatorname{Span}(B)$?

I think it is right, even if there are vectors that are both in $A$ and $B$.

Michael Hardy
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gbox
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    Hmm.. depends on linearly independent vectors. There maybe vectors in B that are scalar multiples of vectors in A! – wannadeleteacct Aug 03 '14 at 19:30
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    Related. (Also works as a hint). Edit: Nevermind, I misinterpreted the question. Let $V=\mathbb R^2, A=(1,0)$ and $B=(0,1)$ for a counter example. – Git Gud Aug 03 '14 at 19:34
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    The symbol $\in$ is used for a single element. For example, if $\mathbb{R}$ is the set of real numbers then $1 \in \mathbb{R}$ and $\pi \in \mathbb{R}$. We use the symbol $\subset$ to denote a subset, i.e. a set of objects in a set, for example ${1,\pi} \subset \mathbb{R}$. – Fly by Night Aug 03 '14 at 20:02

2 Answers2

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Your notation is unclear. You probably mean $A$ and $B$ are subsets of $V$. If we denote by $A+B=\{a+b: a\in A, b\in B\}$, then a vector $v$ belongs to $\def\span{\operatorname{span}}\span(A+B)$ if it can be written as $$ v=\alpha_1(a_1+b_1)+\alpha_2(a_2+b_2)+\dots+\alpha_n(a_n+b_n) $$ for scalars $\alpha_1,\alpha_2,\dots,\alpha_n$ and $a_1,a_2,\dots,a_n\in A$ and $b_1,b_2,\dots,b_n\in B$. In this case $$ v=(\alpha_1a_1+\alpha_2a_2+\dots+\alpha_na_n)+ (\alpha_1b_1+\alpha_2b_2+\dots+\alpha_nb_n) $$ and this proves $\span(A+B)\subseteq\span(A)+\span(B)$.

If the vector space $V$ is finite dimensional, in order to have the reverse inclusion the condition is \begin{align} \dim\span(A+B)&=\dim(\span(A)+\span(B))\\ &=\dim\span(A)+\dim\span(B)-\dim(\span(A)\cap\span(B)) \end{align}

This is not true in general; an easy example is $A=\{u\}$, $B=\{-u\}$, where $u\ne0$. Then $A+B=\{0\}$, while $\span(A)=\span(B)=\langle u\rangle$ and, clearly, $$ \{0\}=\span(A+B)\ne\span(A)+\span(B)=\langle u\rangle. $$

egreg
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To show they are equal, you have to show span{$ A + B $} $\subseteq$ span{$ A$} + span{$B $}, and span{$ A$} + span{$B $} $\subseteq$ span{$ A + B $}.

For span{$ A + B $} $\subseteq$ span{$ A$} + span{$B $}, let $x \in$ span{$ A + B $}. Then $x = c(a + b)$ for some scalar $c$ and elements $a \in A, b \in B$. But then that means $x = ca + cb$, and $ca \in$ span{$A$}. Similarly, $cb \in$ span{$B$}. So $ca + cb \in $ span{$A$} + span{$B$}.

So we know as we just proved that span{$ A + B $} $\subseteq$ span{$ A$} + span{$B $}. But the other way does not hold. Here is a counter example:

Suppose $V = \mathbb{R}^{2}$. Let $A = \left \{ \begin{bmatrix}1 \\ 0 \end{bmatrix} \right \}$ and $B = \left \{ \begin{bmatrix}0 \\ 1 \end{bmatrix} \right \}$. Then span{$A$} $ = \left \{ \begin{bmatrix}c \\ 0 \end{bmatrix} \mid c \in \mathbb{R} \right \}$. Similarly, span{$B$} $ = \left \{ \begin{bmatrix}0 \\ d \end{bmatrix} \mid d \in \mathbb{R} \right \}$. Then span{$A$} + span{$B$} $= \left \{ \begin{bmatrix}c \\ d \end{bmatrix} \mid c,d \in \mathbb{R} \right \} = \mathbb{R}^{2}$. But span{$A + B$} $= \left \{ \begin{bmatrix}c \\ c \end{bmatrix} \mid c \in \mathbb{R} \right \}$. This clearly is not all of $\mathbb{R}^{2}$, since it only contains vectors where the $x$ and $y$ coordinate are equal. In particular, $\begin{bmatrix} 2 \\ 3 \end{bmatrix} \in \mathbb{R}^{2} = $ span{$A$} + span{$B$}, but $\begin{bmatrix} 2 \\ 3 \end{bmatrix} \not \in $ span{$A + B$}.

layman
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