How would I prove that a span of certain vectors is equal to the span of their scalar multiples. The proof would include proving both sets are subsets of one another?

Span{x,y,z} = Span{ax,by,cx}

where x,y,z are vectors and a,b,c are non-zero scalars.

  • $\begingroup$ @user Really? This seems true to me . . . $\endgroup$ – Noah Schweber Jan 9 '16 at 17:30
  • $\begingroup$ @NoahSchweber Sorry, I read the question to mean something completely different.... $\endgroup$ – user296602 Jan 9 '16 at 17:30
  • 2
    $\begingroup$ @user Phew, I was really worried for a second there . . . :P $\endgroup$ – Noah Schweber Jan 9 '16 at 17:31
  • $\begingroup$ I would recommend writing down what it means for a vector to be in the span of a set. The result follows directly from the definition $\endgroup$ – Matt Dyer Jan 9 '16 at 17:38

HINT: Start small. Do you see why $Span\{x\}=Span\{ax\}$?


So as you mentioned, a natural thing to try when you want to prove set equality is to prove containment in both directions. In this case, we don't need to show both directions since if we can show that $$span\{x,y,z\}\subseteq span\{ax,by,cz\}$$ for arbitrary vectors $x,y,z$ and arbitrary non-zero scalars $a,b,c$, it follows that $$span\{ax,by,cz\}\subseteq span\{\frac{1}a ax,\frac{1}b by,\frac{1}c cz\}=span\{x,y,z\}$$ Thus in this case we need only show one direction.

If a vector $v\in span\{x,y,z\}$ then we know that $$v=rx+sy+zt$$ for some scalars $r,s,t$ since $v$ is a linear combination of $x,y,z$. Then $$v=(\frac{1}a\cdot r)ax+(\frac{1}b\cdot s)by+(\frac{1}c\cdot t)cz$$ a linear combination of $ax, by$ and $cz$. Thus $v\in span\{ax,by,cz\}.$ Thus $$span\{x,y,z\}=span\{ax,by,cz\}.$$

  • $\begingroup$ No problem. If you think my answer sufficiently satisfies your inquiry feel free to accept it. If not, let me know and I will be glad to expand on anything. $\endgroup$ – Sean English Jan 12 '16 at 14:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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