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Let V={(a,b)|a,b is in R, b>0}, define addition by (a,b)+(c,d)=(ad+bc,bd) and define...

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user6978788 | Student, Undergraduate | Honors

Posted February 3, 2013 at 1:26 AM via

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Let V={(a,b)|a,b is in R, b>0}, define addition by (a,b)+(c,d)=(ad+bc,bd) and define a scalar multiplication by t·(a,b)=(tab^(t-1),b^(t))

Let V={(a,b)|a,b is in R, b>0}, define addition by (a,b)+(c,d)=(ad+bc,bd) and define a scalar multiplication by t·(a,b)=(tab^(t-1),b^(t)).

 

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degeneratecircle | High School Teacher | (Level 2) Associate Educator

Posted January 19, 2013 at 8:30 AM (Answer #1)

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Well, this is an interesting and very tedious question. You didn't post the whole problem statement, but from your tags I'm assuming you're supposed to determine whether `V` is a vector space, or possibly more specifically a subspace of a larger space. Since this is a long answer, I'll just say that yes, this is a vector space.

I like this question because it highlights the fact that usually if we're trying to figure out if a set with elements of the form `(a,b)` is a subspace, we just check whether it is closed under addition and scalar multiplication. That's what we do if we already know we're in a larger vector space, and we usually do because we work with the usual rules of addition and scalar multiplication given by `(a,b)+(c,d)=(a+c,b+d)`` ` and `t(a,b)=(ta,tb).` But here the definitions of addition and scalar multiplication are so bizarre that the only way to check if this is a vector space is to verify each and every axiom (there might be a clever, shorter way but I don't see it).

The post would be too long if I wrote out every step, but I'll do some key ones. Check the link or your textbook to fill in any missing steps.

First, we should check that our addition and multiplication are closed as follows:

`(a,b)+(c,d)=(ad+bc,bd).` Is `(ad+bc,bd)in V` ? Since `b,d>0,` it follows that `bd>0,` and all that's required to be in `V` is for the second element in an ordered pair to be greater than 0, so `V` is closed under addition.

For multiplication, just note that the second element of `t(a,b)` is `b^t` , and since ```b` is positive, so is `b^t` .

To check that there is a zero vector, just note that

`(a,b)+(0,1)=(1a+0b,1b)=(a,b),` so `(0,1)` plays the role of the zero vector (if this is the first time you've seen a zero vector that isn't `(0,0),` this can be hard to wrap your head around). Also, `1>0,` so `(0,1)` is indeed an element of `V.`

To check for inverses, just note that

`(a,b)+(-a/(b^2),1/b)=(a/b-(ab)/(b^2),b/b)=(0,1),` which is our zero vector. Also note that `1/b>0,` so this vector is in `V.` I'll do one more, which is to verify the distributive law that `t{(a,b)+(c,d)}=t(a,b)+t(c,d).` Those curly braces should be brackets, but it comes out wrong when I try that...anyway,

`t{(a,b)+(c,d)}=t(ad+bc,bd)=(t(ad+bc)(bd)^(t-1),(bd)^t)`

`=(tad^tb^(t-1)+tcb^td^(t-1),(bd)^t)`` `

`=(tab^(t-1),b^t)+(tcd^(t-1),d^t)=t(a,b)+t(d,c).`

Whew, I told you it's tedious and you'll have to verify that the other vector space axioms hold (and double check my work!). Hope this helps.

 

 

Sources:

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degeneratecircle | High School Teacher | (Level 2) Associate Educator

Posted January 19, 2013 at 8:34 AM (Answer #2)

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The very last math expression should be `t(c,d),` not `t(d,c)` as I wrote. I hope that's the only typo.

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tiburtius | High School Teacher | (Level 3) Associate Educator

Posted January 19, 2013 at 9:28 AM (Answer #3)

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There is an easier way to show that this is a vector space. You only need to show that this is a linear space for which you need to show that for every two vectors `(a,b),(c,d) in V` and every scalar `t in K` (wher `K` is a field), the following holds:

`(a,b)+(c,d) in V` and `t(a,b) in V`

The only constriction you have is that second part of the vector has to be positive (`b>0`).

`(a,b)+(c,d)=(ad+bc,bd)`

Since `b>0` and `d>0` it follows that `bd>0` thus `(ad+bc,bd) in V`.

`t(a,b)=(tab^(t-1),b^t)`

Since `b>0` it follows that `b^t>0` thus `(tab^(t-1),b^t)in V`

Also notice that in order for `b^t>0` to make sense `t` has to be real.

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degeneratecircle | High School Teacher | (Level 2) Associate Educator

Posted January 19, 2013 at 4:13 PM (Answer #4)

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But when you check only those two conditions, you assume that `V` is a subspace of a larger vector space and inherits its operations from that space. You can't use the fact that `RR^2` is a vector space under the usual operations `(a,b)+(c,d)=(a+c,b+d)` and `t(a,b)=(ta,tb)` to say anything about a subset of `RR^2` under different operations.

You would have to show that `RR^2` is a vector space under these strange operations, which means checking each axiom.

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tiburtius | High School Teacher | (Level 3) Associate Educator

Posted January 19, 2013 at 4:42 PM (Answer #5)

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If you look a bit closer you'll see that I'm not using usual operations of addition and multiplication and there is no assumption about `V` being subspace of `RR^2`. I'm only showing that `V` is closed under those two operations which is enough to show that `V` is linear and thus vector space. 

Thic is sometimes called alternative axiomatization of vector space, but some say that this is vector space only if its elements are vectors (meaning not matrices, functions, operators...) which is also true in this case.

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degeneratecircle | High School Teacher | (Level 2) Associate Educator

Posted January 19, 2013 at 5:22 PM (Answer #6)

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There are about 8 or so (depending on how much the author breaks them up) axioms for a vector space. Since we're not considering a subset of a known vector space, how can you verify that this set is a vector space by checking only two?

How do you know addition is associative? How do you know `1(a,b)=(a,b)?` These are easily checked, but not immediately obvious. I think this discussion (and the problem itself) is instructive, but if it's better suited to a post on a discussion board I'd be fine with that.

user6978788, it's a good idea to talk about this problem with your TA or professor if you're still not sure.

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degeneratecircle | High School Teacher | (Level 2) Associate Educator

Posted January 19, 2013 at 5:50 PM (Answer #7)

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A simple example of a set that's closed under addition and scalar multiplication is the set of all `n x n` matrices, where the operations are

`A+B=AB` and `tA` is defined as the standard scalar-matrix multiplication. This set is closed under both these operations, but addition is not commutative and it isn't a vector space.

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tiburtius | High School Teacher | (Level 3) Associate Educator

Posted January 19, 2013 at 10:38 PM (Answer #8)

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You've made a mistake addition of matrices is commutative `A+B=B+A`. It is multiplication of matrices that is not commutative that is `AB ne BA`.

I agree with idea of moving this to discussion board. 

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degeneratecircle | High School Teacher | (Level 2) Associate Educator

Posted January 19, 2013 at 11:24 PM (Answer #9)

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Ok, I'll message user6978788 and see what she thinks. It would be nice to know if this is helpful to her, since it can be tricky for a beginner, or if we're just confusing her more.

As for my matrix example, the usual definition of matrix addition is commutative, but if we try to make a vector space where the operation of "addition" is the usual matrix multiplication, then this version of addition isn't commutative.

You can translate my example to look more like user's by considering

`V=RR^4={(a,b,c,d): a,b,c,d in RR}.` Define

`(a,b,c,d)+(e,f,g,h)=(ae+bg,af+bh,ce+dg,cf+dh)` and

`t(a,b,c,d)=(ta,tb,tc,td).` This set is obviously closed under addition and scalar multiplication, but it is not a vector space because, for example,

`(0,1,0,1)+(1,1,0,0)=(0,0,0,0),` but

`(1,1,0,0)+(0,1,0,1)=(0,2,0,0),` so this definition of addition is not commutative.

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