# $Sp(A)\bigoplus Sp(B) \Leftrightarrow A\cup B$ is linearly independent

Is the following statement true?

Assuming $V$ is a Vector Space and $A,B\subseteq V$ and that $A$ and $B$ are linearly independent:

$Sp(A)\bigoplus Sp(B) \Leftrightarrow A\cup B$ is linearly independent.

And if we make no assumptions about A and B being linearly independent does $Sp(A)\bigoplus Sp(B)$ imply that $A\cup B$ is linearly independent?

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Only if both $A$ and $B$ are (separately) linearly independent. – Robert Israel May 30 '12 at 7:46
Funny, my names Robert and I live in Israel :-) – Robert S. Barnes May 30 '12 at 7:52
What if A and B are not linearly independent separately? Could $Sp(A)+SP(B)$ still be a direct sum? – Robert S. Barnes May 30 '12 at 8:01
Yes, of course. – Robert Israel May 30 '12 at 17:30

Your "iff" statement is right. If $A\cup B$ is linearly independent, then $Sp(A)\cap Sp(B)=\{0\}$. To see this, just think what it would mean for something to be in both spans. To get you started: if $\sum\alpha_i a_i=\sum \beta_j b_j\in Sp(A)\cap Sp(B)$, then $\sum\alpha_i a_i-\sum \beta_j b_j=0$ ... now draw a conclusion about the alphas and betas based on your assumption.
On the other hand, suppose $A\cup B$ is linearly dependent. Pick a nonzero linear combination $\sum\alpha_i a_i+\sum \beta_j b_j= 0$. Since both $A$ and $B$ are individually linearly independent, it must be there is a nonzero $\alpha_i$ and a nonzero $\beta_j$. But then $\sum\alpha_i a_i=-\sum \beta_j b_j$ shows that a nonzero element of $Sp(A)$ is also a nonzero element in $Sp(B)$, so $Sp(A)\cap Sp(B)\neq\{0\}$, and the sum is not direct.
For your second statement, remember that any nonempty subset of a linearly independent set is necessarily linearly independent. So if $A$ is LD, then $A\cup X$ is LD for any other collection $X$ of vectors in your space.