# is subset S a subspace of R3?

12 views (last 30 days)
Hannah Blythe on 7 Oct 2017
Edited: Cedric Wannaz on 8 Oct 2017
S - {(2x-y, xy, 7x+2y): x,y is in R} of R3
Three requirements I am using are i. S is nonempty ii. For any u,v in S, u+v is in S iii. For any c in R and u in S, cu is in S So far I have proved the first condition. For the second condition I said a,b,c,d are in R and that u=(2a-b, ab, 7a+2b) and v=(2c-d, cd, 7c+2d). When I added them I got {2(a+c)-(b+d), (ab)+(cd), 7(a+c)+2(b+d)}. What should I do from here?
Cedric Wannaz on 7 Oct 2017
Edited: Cedric Wannaz on 7 Oct 2017
Good start. As (a+c) in R and so is (b+d), you demonstrated that the first component of u+v is compatible with the definition of S.
What about the second component? In other words, can you develop/expand/factor ab+cd into a product of (a+c) and (b+d)?
If you can, check the third component. If you cannot, well unless it's obvious it's not a proof (maybe it's just that you are unable to do it, but it could be possible). Yet you may be able to use what you understand about the mismatch for building a counter-example, and you only need one to show that it doesn't work.

Cedric Wannaz on 8 Oct 2017
Edited: Cedric Wannaz on 8 Oct 2017
Now that you have probably finished your exercise, here is S with the origin displayed as a red dot:
Looking at this, you understand why, when you pick two vectors defined by points that are part of S, their sum is likely not part of S.
And here is the little bit of code that produced this figure (minus the rotations that I did by hand), in case your exercise was related to MATLAB:
[U,V] = meshgrid( -10:10, -10:10 ) ;
X = 2*U-V ; Y = U.*V ; Z = 7*U+2*V ;
for k = 1 : 4
subplot( 2, 2, k ) ;
hold on ; grid on ;
surf( X, Y, Z ) ;
plot3( 0, 0, 0, 'r.', 'MarkerSize', 20 ) ;
end