Hi,
problem:
Sometimes a subset of a vector space is itself a vector space (with respect to the linear operations already given). Consider, for example, the vector space and the subsets consisting of those vectors for which
(a) is real,
(b) ,
(c) either ,
(d) ,
(e)
In which of these cases is a vector space?
attempt:
(a)
Not a vector space since multiplication by a scalar yields .
is not in .
(b)
Is a vector space if . If I think of this in terms of , I would have the plane if I'm not sure if I should be thinking like this..
(c)
Again, if I think of it as , I would get the plane and the plane.
This is not closed under addition, ,i.e., not a vector space.
I do not see which axiom for a vector space is violated though..
(d)
Is a vector space.
(e)
Hm, I would say no since it looks like it's not closed under addition.
Sorry for my pathetic attempts, can't do much better..
Thanks.
I assume that is the vector space of ordered triples of complex numbers with the usual addition and scalar multiplication.
[quote] and the subsets consisting of those vectors for which
(a) is real,
(b) ,
(c) either ,
(d) ,
(e)
In which of these cases is a vector space?
attempt:
(a)
Not a vector space since multiplication by a scalar yields .
is not in .[quote]
Are you assuming that ? If so, you should say so. And you mean that is not real, not that it is not in V. V is a set of vectors in and is a single number. A simpler example is to take .
[quote]
(b)
Is a vector space if . If I think of this in terms of , I would have the plane if I'm not sure if I should be thinking like this..[/tex]
If you are left with vectors of the form . What do you get if you add two such things? Multiply by some complex number?
Both (0, 1, 1) and (1, 0, 1) are in this subset. Is their sum?(c)
Again, if I think of it as , I would get the plane and the plane.
This is not closed under addition, ,i.e., not a vector space.
I do not see which axiom for a vector space is violated though..
Yes, that is correct. I assume you have shown that.(d)
Is a vector space.
Again, both (1, 0, 1) and (0, 1, 0) are in this subset. Is their sum?(e)
Hm, I would say no since it looks like it's not closed under addition.
Sorry for my pathetic attempts, can't do much better..
Thanks.
The question was taken from Finite-Dimensional Vector Spaces by Halmos. I guess he assumes the same thing as HallsofIvy..
Thanks for the heads up!
I assume that .
.
The set is not closed under scalar multiplication and is therfore not a vector space.
Is this better?
If , the set consisting of vectors of the form is closed under both scalar multiplication and addition. It is a vector space.
No it's not. The book lists a couple of axioms valid for a vector space. When I first learned about vector spaces I was told that they are closed under scalar multiplication and vector addition.
Could you tell me which axiom fails here?
Now that you have shown me (e), I should be able to show this in a nices way.
Both are in the subset. Their sum is also in the subset.
Their sum is not in the subset.
Thanks mate!