anonymous
  • anonymous
A = {(a,b,c) where a + b = 2c} Find a basis for A
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
But then that would make c always 0, wouldn't it? That doesn't seem right...
anonymous
  • anonymous
(a,0,1/2a) (0,b,1/2b)
anonymous
  • anonymous
i think this should do

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
yep
anonymous
  • anonymous
you could make use of the standard basis vectors e1=(1,0,0) e2=(0,1,0) e3=(0,0,0). now first we need to prove this will span A... so we need to prove that a linear combination of these vectors will be in A... so suppose we have ANY vector u which is in A where u=(a,b,c), then we know we could rewrite this as u=ae1+be2+ce3 u=ae1+be2+ce3 u=a(1,0,0)+b(0,1,0)+c(0,0,1) u=(a,0,0)+(0,b,0)+(0,0,c) u=(a,b,c) So we could write any vector u from A as a linear combination of A. so e1 and e2 and e3 spans A. now we only need to show that... c1e1+c2e2+c3e3=0 has only trivial solution... so if we write this in a matrix we'll have \[\left[\begin{matrix}1 & 0 & 0 \\ 0 & 1&0\\0&0&1\end{matrix}\right]\left(\begin{matrix}c1\\ c2\\c3\end{matrix}\right)=\left(\begin{matrix}0\\0\\0\end{matrix}\right)\] now since the determinant of the coefficient matrix is not equal to zero then the system will only have exactly one solution which is the trivial solution. so the vectors e1 and e2 and e3 are linearly independent. Therefore, e1, e2, and e3 will span A
anonymous
  • anonymous
@anonymoustwo44, this does,t span the desired space
anonymous
  • anonymous
@myko please prove that as well as you answer
anonymous
  • anonymous
(a,0,1/2a) (0,b,1/2b) or if you like: (1,0,1/2), (0,1,1/2) is the needed base
anonymous
  • anonymous
@anonymoustwo44 u=ae1+be2+ce3 u=a(1,0,0)+b(0,1,0)+c(0,0,1) u=(a,0,0)+(0,b,0)+(0,0,c) u=(a,b,c) it's wrong!!! your e3 as stated befor is (0,0,0) and even more, (a,b,c) the way you constract it with your base do not necesarly belong to A, since A={(a,b,c)} | c =(a+b)/2 !!!!!!!!!!!!!
anonymous
  • anonymous
my bad... my e3 is (0,0,1)
anonymous
  • anonymous
but even this way it doesn't work.... since A={(a,b,c)} | c =(a+b)/2 !!!!!!!!!!!!!
anonymous
  • anonymous
it will work even if c=(a+b)/2 (a,b,c)=(a,b,(a+b)/2)=(a,0,0)+(0,b,0)+(0,0,(a+b)/2)) =a(1,0,0)+b(0,1,0)+((a+b)/2)(0,0,1) =ae1+be2+((a+b)/2)e3 <--- see, even though c=(a+b)/2 we could still write it in as a linear combination of e1, e2, and e3 so e1 e2 and e3 spans A
anonymous
  • anonymous
why mad?

Looking for something else?

Not the answer you are looking for? Search for more explanations.