tylerc1991 said:Homework Statement
Suppose [itex]U[/itex] and [itex]W[/itex] are subspaces of [itex]V[/itex], and [itex]U \, \bigcup \, W[/itex] is a subspace of V. Show that [itex]U \subseteq W[/itex].
The Attempt at a Solution
I have been working on this one for a bit and have not made any headway. I wish I could post anything, even a start to this one but I can't seem to see which direction to head in. If anyone could give me some direction I would be grateful!
Shackleford said:The professor did this in class today. He said it's a bit of a sophisticated proof. He used the proof by contradiction method.
Since this is an iff statement, one direction was fairly trivial, but the other required the contradiction method. He assumed the conclusion failed which violated our assumptions.
tylerc1991 said:Exactly, the original question was an iff statement. I was posting the part that I was having trouble with. Oddly enough, proof by contradiction was the road I was going down but couldn't get anywhere with. Was there any special trick in particular that made this proof so sophisticated?
Shackleford said:It's in the notes. This is actually a homework problem for me too. I kind of stopped paying attention about mid-way through the proof in class. I kind of followed along, but I need to go through it myself. I'm still pretty rusty on linear algebra. I took the sophomore-level class three years ago.
Dick said:The conclusion should really be if U union W is a subspace, then either U is contained in W OR W is contained in U. If U is not contained in W and W is not contained in U, then pick a nonzero element u of U and a nonzero element w of W such that u is not in W and w is not in U. Can u+w be in the union of U and W?
A subspace in linear algebra is a subset of a vector space that satisfies three properties: closure under vector addition, closure under scalar multiplication, and contains the zero vector.
To prove that a subset is a subspace, we need to show that it satisfies the three properties of a subspace: closure under vector addition, closure under scalar multiplication, and contains the zero vector. This can be done by showing that for any two vectors in the subset, their sum and scalar multiple are also in the subset, and that the zero vector is also in the subset.
Yes, a subspace can contain only one vector. This is because a subspace can be defined as a subset of a vector space that satisfies the three properties, and a single vector can satisfy these properties.
A span is the set of all possible linear combinations of a given set of vectors, while a subspace is a subset of a vector space that satisfies three specific properties. In other words, a span is a set of vectors, while a subspace is a type of set that follows certain rules.
To prove that a set of vectors spans a subspace, we need to show that every vector in the subspace can be expressed as a linear combination of the given set of vectors. This can be done by setting up a system of equations and solving for the coefficients of the linear combination.