(linear algebra) union of subspaces

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
5 replies · 4K views
dlevanchuk
Messages
29
Reaction score
0
Eh, kind of stuck on this question. I need some suggestions on how to tackle the problem..

Homework Statement


Let U and V be the subspaces of R_3 defined by:

U = {x: aT * x = 0} and V = {x: bT * x = 0} (T means transpose)

where

a = [1; 1; 0] and b = [0; 1; -1]

Demonstrate that the union of U and V is not a subspace of R_3..

Homework Equations


See Above


The Attempt at a Solution


Should I just combine U and V, into something like UuV = {x: aT * x = b*T * x = 0}, since both equations equal to 0, just kind of combine them together..

any tips? am I on the right track?
 
Physics news on Phys.org
describe / visualize what the subspace U is first. Then describe/visualize what V is. finally, what is U union V? It will be obvious after you do this.
 
Describing U:
U: aT *x = [1 1 0]*[x1;x2;x3] = x1+x2 = 0

Describing V:
V: bT *x = [0 1 -1]*[x1;x2;x3] = x2-x3 = 0

union of U and V {x: x1+x2=x2-x3=0}

after moving around x's i get
union of U and V {x: x1+x3 = 0}

But as i test this subspace (zero vector, sum of two arbitrary vectors and scalar multiplication) , the union of U and V passes all of the tests...

I have a feeling that I am doing something wrong when I am adding two arbitrary vectors:
u = [-1; 1; 1] and v = [-2; 2; 2] (both vectors are in subspace of U and V).
when i add them i get u+v = [-3; 3; 3], which satisfies the x1+x3=0 equation, so it passed that test..

am I wrong? :( tired of banging my head against the table!
 
coupla things wrong here:

-you're confused as to what 'union' means. union means 'in V OR U'. not ' V AND U'.

-sum of 2 arbitrary vectors
its supposed to be the sum of any 2 vectors is also inside. not just the random 2 that you happen to pick.this is what i meant by 'describing' earlier:

U is the set of all vectors perpendicular to a. (dot product=0)
V is the set of all vectors perpendicular to b.U union V is the set of all vectors perpendicular to either a or b. now, can you find 2 vectors in [U union V] whose sum is not in [U union V]?
 
Eh, i still don't get it..

So the sum of two arbitrary vectors will need to satisfy either U or V in order to have the new "U union V" to be considered subspace??
 
Last edited: