Imum Coeli
Junior Member
- Joined
- Dec 3, 2012
- Messages
- 86
I am not seeing what should be simple...
Question:
Which of the following sets are subspaces of R^3?
U1 = {t(1,2,3)|t belongs to R}
U2 = {(0,0,0),(1,0,0),(0,1,0),(0,0,1)}
U3 = {x,y,2x+4y)|x,y belong to R}
U4 = {(x,y,z) belongs to R^3|x+y-z=0, 2x+y+z=0}
U5 = {(x,y,z) belongs to R^3|x+y-z=1}
U6 = {(x,y,xy)|x,y belong to R}
Notes:
First, a subspace must contain the zero vector and be closed under addition and scalar multiplication.
Now my (incorrect) attempt at an answer.
U1 is a subspace because it is a line through the origin.
U2 is a subspace because it is a plane through through the origin
U3 is a subspace because it is all of R^3
U4 is a subspace because it is the solution set for two homogeneous linear equations
U5 is not a subspace because it does not contain the zero vector.
U6 is a subspace because it is all of R^3
I believe I am missing something obvious...
Thanks for any advice.
Question:
Which of the following sets are subspaces of R^3?
U1 = {t(1,2,3)|t belongs to R}
U2 = {(0,0,0),(1,0,0),(0,1,0),(0,0,1)}
U3 = {x,y,2x+4y)|x,y belong to R}
U4 = {(x,y,z) belongs to R^3|x+y-z=0, 2x+y+z=0}
U5 = {(x,y,z) belongs to R^3|x+y-z=1}
U6 = {(x,y,xy)|x,y belong to R}
Notes:
First, a subspace must contain the zero vector and be closed under addition and scalar multiplication.
Now my (incorrect) attempt at an answer.
U1 is a subspace because it is a line through the origin.
U2 is a subspace because it is a plane through through the origin
U3 is a subspace because it is all of R^3
U4 is a subspace because it is the solution set for two homogeneous linear equations
U5 is not a subspace because it does not contain the zero vector.
U6 is a subspace because it is all of R^3
I believe I am missing something obvious...
Thanks for any advice.