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anonymous
 one year ago
Please help with very difficult vector proof
anonymous
 one year ago
Please help with very difficult vector proof

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let u and v be nonparallel vectors in R2, and let w be any vector in R2. Show for unique constants s and t, w = su + tv

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It might help to write what I've been trying: I separated the vectors into horizontal and vertical components to getdw:1442812756247:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got dw:1442812854324:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2I think I might have something, but I used matrices to get there. Would that work?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sure I've covered matrices in my class

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0take me back to linear algebra and linear transforms please.. ..

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2ok so I started off letting w = <x,y> be some vector in the plane of R2 I also let u = <a,b> v = <c,d> so that means w = s*u + t*v <x,y> = s*<a,b> + t*<c,d> <x,y> = <s*a,s*b> + <t*c,t*d> <x,y> = <s*a+t*c, s*b+t*d> x = s*a+t*c x = as + ct y = s*b+t*d y = bs + dt In this case, the scalars a,b,c,d are constant while s,t are variables. The system below as + ct = x bs + dt = y turns into this matrix equation \[\Large \begin{bmatrix}a & c\\b & d\end{bmatrix} \begin{bmatrix}s\\t\end{bmatrix} = \begin{bmatrix}x\\y\end{bmatrix}\] a solution exists if and only if the matrix with a,b,c,d in it is invertible. Which only happens if the determinant of that matrix is nonzero Let's say u and v were parallel. That would mean a/c = b/d ad = bc ad  bc = 0 So if u and v were nonparallel, then ad  bc (which is the determinant of the a,b,c,d matrix) is nonzero

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This makes sense but how can you say that s and t and unique constants instead of variables?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2I made s,t variables to allow <x,y> to vary The idea is that you have a fixed set of 2 vectors. Then you can have those vectors generate the entire plane based on a unique s,t solution

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh I get it now it will be unique based on the vector.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2The uniqueness comes from the fact that the solution of the matrix equation is unique

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well done, thanks a lot!

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2you're welcome
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