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anonymous

  • one year ago

Please help with very difficult vector proof

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  1. anonymous
    • one year ago
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    Let u and v be non-parallel vectors in R2, and let w be any vector in R2. Show for unique constants s and t, w = su + tv

  2. anonymous
    • one year ago
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    It might help to write what I've been trying: I separated the vectors into horizontal and vertical components to get|dw:1442812756247:dw|

  3. anonymous
    • one year ago
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    I got |dw:1442812854324:dw|

  4. jim_thompson5910
    • one year ago
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    I think I might have something, but I used matrices to get there. Would that work?

  5. anonymous
    • one year ago
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    Sure I've covered matrices in my class

  6. DanJS
    • one year ago
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    take me back to linear algebra and linear transforms please.. ..

  7. DanJS
    • one year ago
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    comb

  8. jim_thompson5910
    • one year ago
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    ok 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 non-parallel, then ad - bc (which is the determinant of the a,b,c,d matrix) is nonzero

  9. anonymous
    • one year ago
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    This makes sense but how can you say that s and t and unique constants instead of variables?

  10. jim_thompson5910
    • one year ago
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    I 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

  11. anonymous
    • one year ago
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    Oh I get it now it will be unique based on the vector.

  12. jim_thompson5910
    • one year ago
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    The uniqueness comes from the fact that the solution of the matrix equation is unique

  13. anonymous
    • one year ago
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    Well done, thanks a lot!

  14. jim_thompson5910
    • one year ago
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    you're welcome

  15. DanJS
    • one year ago
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    good one

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