Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
3  
Only if both $A$ and $B$ are (separately) linearly independent. –  Robert Israel May 30 '12 at 7:46
4  
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

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.