Let M and N be linear subspaces of a vector space V, and let

\(M\cup N=\{x\in V:\quad x\in M\text{ or } x\in N\}\)

be the union of M and N.

True or false(explain): \(M \cup N \) is a linear subspace of V.

{If true, give a proof. If false, give a counterexample (a example of V, M, N for which \(M\cup N\) is not a linear subspace of V).}

That's what the question says. I do not understand the definition of union of 2 linear subspaces... i.e. \(M\cup N=\{x\in V:\quad x\in M\text{ or } x\in N\}\) <--- this one. Anyone please help me?

MaxWong
Jun 26, 2017

#1**+1 **

I see why you're confused:

A vector space is a set containing some vectors. That set can contain a finite number of vectors, and it can contain an infinite number of vectors.

Of course that set must follow some special rules: most of them are annoying and boring, but i think one of them is special, and more "important" than the others (in this type of questions, questions like: the set W is defined like that, is W a vector space?). That rule is, whenever you multiply a vector by a scalar, or when adding two vectors, the result MUST be a vector from the set. I think that rule is special because you USUALLY use it to disprove/prove that W is a vector space. Checking if the other rules work is boring and annoying, and they probably will anyways.

You were given a set (the union of N and M)- that set contains vectors that are in N or M (can be in N and M, too). You were NOT told whether that set is a vector space or not. So, you have to check it is, by checking all the rules of a vector space. Remember, when a set is given for that purpose, THEY DONT GIVE YOU THE SPAN OF THE SET. They give you a set and they want you to tell whether that set is a vector space (or equally, whether span(set)=set)

Guest Jun 26, 2017

#4**0 **

The union of two subspaces just means that we include all of the elements of each subspace into another space,obeying the same rules as they did before.Here the elements are vectors.( This just like the union of two sets.no different.)

So you have all the vectors in subspace M and all the vectors in subspace N together. Since you are told that the vector x is either in M or N,then the union of M and N must contain ALL the vectors x ,and they must be linearly independent.(Since they are linearly independent in both M AND in N)

Now ,both linear subspaces obey a(x1 + x2) = ax1 + ax2 where x1,x2 are any two linearly independent vectors in either M or N.

Finally,if x is in either M or N,it cannot be outside M or N,so you cannot find vectors x, outside M U N,hence M U N must be a linear subspace of V. (take heart,no-one finds this stuff easy to begin with.)

Guest Jun 26, 2017

#6**+2 **

\(\text{In contrast to what has been said before the union of two linear subspaces of }V\\ \text{ is in general not a linear subspace itself. Take for example the linear vector space}\\ \mathbb{R}^2\text{ of two dimensional vectors. We can now define the two following subspaces:}\\ M=\{(a,0);a\in\mathbb{R}\},\\ N=\{(0,b);b\in\mathbb{R}\}.\\ \text{It should be clear that each individual subspace is a linear subspace of }\mathbb{R}^2. \text{The}\\ \text{union of the two is not a linear subspace because the vector (1,1) is not an element}\\ \text{of the union (which it should be since (1,0) and (0,1) are elements of the union).}\\ \text{Hence the given statement is disproven by means of counterexample.}\)

Honga
Jun 26, 2017