Question
Asked Oct 14, 2019
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Consider the set ? = ((x, x+ y, 2y)|x,y are real) of vectors in r3 . Is W a subspace? If yes, prove it and if no explain the reason.

     

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Expert Answer

Step 1

 

We have Set S = {(x, x + y, 2y) | x, y are real}. Let u, v be any two element of S then,

u=(x, 2) , v=(x,x, + y,2y,)
where xy, e R_
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u=(x, 2) , v=(x,x, + y,2y,) where xy, e R_

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Step 2

Let a and b be any two element of R. In order to prove that S is a subspace of V, we have to prove that au + bv S, for which we have to show that au + bv is expressible in the form ...

a(xxy2)+ b(x,.x, +y,2y,)
- aι+bν - (α + bx ) (α + bx) +(ωγ +by ) 2 (α + bv)
au+ bν
( α,α + β, 2β)
where α- a, + bx, and βa) +bν,
Φau + bν
au+bν E S
help_outline

Image Transcriptionclose

a(xxy2)+ b(x,.x, +y,2y,) - aι+bν - (α + bx ) (α + bx) +(ωγ +by ) 2 (α + bv) au+ bν ( α,α + β, 2β) where α- a, + bx, and βa) +bν, Φau + bν au+bν E S

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