Finding a Basis for Subspace a of R^3

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SUMMARY

The discussion focuses on finding bases for the subspace defined by the equation Y + Z = 0 in R^3. The normal vector to this plane is identified as n = (0, 1, 1), with two orthogonal vectors u = (0, -1, 1) and v = (1, 0, 0) proposed as a basis. The book's suggestion of v = (7, 0, 0) is deemed incorrect, as any vector of the form (x, y, -y) is valid, confirming that the subspace is indeed two-dimensional. The confusion arises from the book presenting a single vector as the only solution.

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  • Understanding of vector spaces and subspaces in R^3
  • Knowledge of normal vectors and their significance in geometry
  • Familiarity with orthogonal vectors and their properties
  • Basic linear algebra concepts, including basis and dimension
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Students and educators in mathematics, particularly those studying linear algebra, vector spaces, and geometry. This discussion is beneficial for anyone seeking to deepen their understanding of subspaces in R^3.

kwal0203
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Homework Statement



Find bases for the following subspace a of r^3

Y+z=0


The Attempt at a Solution



First I found a normal to this plane n=(0,1,1)

Then I found two vectors which are orthogonal to the normal u=(0,-1,1), v=(1,0,0)

Is this correct the answer in my book has v=(7,0,0)

Thanks
 
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You are correct. If that is what your book really says it is wrong. Any vector in [tex]R^3[/tex] is of the form (x, y, z). Any vector for which z= -y is of the form (x, y, -y)= (x, 0, 0)+ (0, y, -y)= x(1, 0, 0)+ y(0, 1, -1). Of course, (7, 0, 0) would work as well as (1, 0, 0) but this subspace is definitely two dimensional.
 
Great thanks for that. I can see why x=7 works it just confused me why they put that as the only answer @.@
 

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