$\displaystyle V$ is a linear space. $\displaystyle A,B\in V$ are sets, and $\displaystyle W$ is subspace:
$\displaystyle Span(A\cup B\cup W)=(Span(A)+Span(B))\cup W$
i just need help to figure out if it's true or false.
thanks in advanced!
$\displaystyle V$ is a linear space. $\displaystyle A,B\in V$ are sets, and $\displaystyle W$ is subspace:
$\displaystyle Span(A\cup B\cup W)=(Span(A)+Span(B))\cup W$
i just need help to figure out if it's true or false.
thanks in advanced!
Hi guys, and thank you.
Denovo, that's the first example that crossed my mind, by this example the LHS and RHS are equal:
$\displaystyle Span\left \{ (0,1,0), (0,0,1), (x,0,0)|x\in \mathbb{R} \right \}=\left \{ (x,y,z)|x,y,z\in \mathbb{R}\right \}=\mathbb{R}^3$
is equal to:
$\displaystyle Span(A)=\left \{ (0,y,0)|y\in \mathbb{R}\right \}+Span(B)=\left \{ (0,0,z)|z\in \mathbb{R}\right \}\cup \left \{ (x,0,0)|x\in \mathbb{R}\right \}=\left \{ (x,y,z)|x,y,z\in \mathbb{R}\right \}=\mathbb{R}^3$
but my intuition misled me so many times, and i wasn't sure that proves it.
no, they are not equal. the point (3,4,5) is in Span(AUBUW), since:
(3,4,5) = 4(0,1,0) + 5(0,0,1) + (3,0,0), and (0,1,0) is in A, so in AUBUW, (0,0,1) is in B, so in AUBUW, and (3,0,0) is in W, so is in AUBUW.
is (3,4,5) in Span(AUB)UW? if so, it would have to be in either Span(AUB), or W, or both. is this true?
my bad.
$\displaystyle (3,4,5)$ is neither in $\displaystyle (Span(A)+Span(B))$ nor in $\displaystyle W$, and therefore is not in $\displaystyle (Span(A)+Span(B))\cup W$.
thanks.
this disproves it.
this is all so confusing...