Please help with very difficult vector proof

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Please help with very difficult vector proof

Mathematics
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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
It might help to write what I've been trying: I separated the vectors into horizontal and vertical components to get|dw:1442812756247:dw|
I got |dw:1442812854324:dw|

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I think I might have something, but I used matrices to get there. Would that work?
Sure I've covered matrices in my class
take me back to linear algebra and linear transforms please.. ..
comb
ok so I started off letting w = be some vector in the plane of R2 I also let u = v = so that means w = s*u + t*v = s* + t* = + = 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
This makes sense but how can you say that s and t and unique constants instead of variables?
I made s,t variables to allow 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
Oh I get it now it will be unique based on the vector.
The uniqueness comes from the fact that the solution of the matrix equation is unique
Well done, thanks a lot!
you're welcome
good one

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