Question

Which of the following sets of vectors are vector spaces? Justify
a) R^2 U R^3
b) {|x belongs to R} c) {|k is an integer} d) {|x belongs to R} U {<0,y>|y belongs to R} f) Span({<1,3,2>,<-1,0,3>})

Answer 1

e isn't a vector space.

(9,0) is in the space, and (0,7) is in the space, but (9,0)+(0,7)=(9,7) is not in the space.

Answer 2

c is only a vector space if the scalar field is the integers. If it is the Reals, then c is not a vector space.

Answer 3

ok...how about a? i think it's not a vector space. am i right?

Answer 4

a is the union of R2 and R3?

Answer 5

yup

Answer 6

So some elements will be of form \[\left(\begin{matrix}a \\ b\end{matrix}\right)\]and others of form \[\left(\begin{matrix}a \\ b \\ c\end{matrix}\right)\]?

Answer 7

that's right

Answer 8

Hmmm...

I think it would be fine, just what is the way to add a vector from R2 and a vector from R3?

Answer 9

Not too sure...Is it even possible to add <a,b> with <c,d,e>?

Answer 10

and according to your answer for (c), (b) wont be a vector space. why so?

Answer 11

Not normally just like that.

But there are ways around it. For example, \[\mathbb{R}^{2}\subseteq \mathbb{R}^{3}\], and you could just write all of the elements from R2 as \[\left(\begin{matrix}a \\ b \\ 0\end{matrix}\right)\]

Answer 12

so, what you're saying is R2 U R3 is essentially R3 and hence a vector space. right?

Answer 13

Yes.

Answer 14

But b is a vector space. The reason c isn't is that it requires that all vectors be of the form (k, 2k) where k is an integer. But you could multiply a vector by say 7.59378, and then you couldn't get a vector of form (k,2k) with k being an integer.

The one in b has no such restriction.

Answer 15

Yes. Sorry. I misread your explanation for c. How about (f)?
I believe (d) is not a vector space.

Answer 16

I agree that d is not.

About f... Pretty sure it is a vector space.

Answer 17

Great. Thank you.

Answer 18

You're welcome.