Question

Linear Algebra:

If S = {(a, b) in R^2 | b>0}
and we define addition to be
(a, b) [+] (c, d) = (ad+bc, bd)

I need to show that (a,b) [+] (c, d) belongs to a vector space.

Is it sufficient to say that b>0 and d>0, so bd>0 ? Or do we have to say something about the ad+bc as well?

Answer 1

therer are at least 3 conditions for a vector space

Answer 2

closed under addition
closed under multiplication
contains a zero vector

Answer 3

the rest of the conditions are consequences of these

Answer 4

Oh yes I know what you mean. I meant I wa strying to show the first axiom. Saying that in genral, for vectors x', y' belonging to a vector space V, then x' + y' is in V

Answer 5

i take it these are your generic vectors? <a,b> and <c,d>

Answer 6

yes

Answer 7

and the only conditions are R^2 meaning? the vectors themselves are in R^2? or that their component elements are in R^2 (which makes no sense if thats the case)

Answer 8

I believe they mean the vectors themselves are in R^2

Answer 9

then by the definition they gave of the additive rule would have to be parsed; your assumption the d>0 is superfluous to me

Answer 10

well since (a, b) are in R^2, b>0, and if (c, d) is in R^2, then d>0 also no?

Answer 11

no, d is any real value; only b is a positive number

Answer 12

but aren't they saying that (a, b) is any vector in R^2?

Answer 13

So I thought they meant the second component of any vector is >0

Answer 14

they are saying that any vector in the space of the for <a,b> is in R^2; its a generic vector and all the forms of it are as well

<a,b>
+ <c,d>
------------
<ad+bc , bd>

Answer 15

since addition and multiplication are closed under the Reals the left and right side will be simply of the form <x,y>

if we assume c.d to be the zero vector for addition <0,0> do we get <a,b> back?

Answer 16

not to butt in, but do you know what this is?

Answer 17

the zero vector is tends to be the easiest one to try to figure out

Answer 18

this is openstudy :)

Answer 19

lol

Answer 20

i believe they re trying to define a "group"

Answer 21

i meant this
<a,b>
+ <c,d>
------------
<ad+bc , bd>

Answer 22

but don't we have to restrict d>0, we can't have <0,0> right according to the restriction?

Answer 23

o.O

Answer 24

the binary operator is defined that way

Answer 25

<0,0> is not the identity

Answer 26

what it means? i dunno

Answer 27

hint,
\[\frac{a}{b}+\frac{c}{d}=...\]

Answer 28

if c,d can be the inverse of a,b then we can define an identity from it perhaps?

Answer 29

...thought I had an idea lol. im lost again

Answer 30

adding fraction is what you are doing

Answer 31

identity is <0,b>

Answer 32

ahhh, ok

Answer 33

well actually i guess you need so say something about equivalence classes, so maybe i should be quiet

Answer 34

Wow this question seems more confusing that I even thought lol

Answer 35

the closest ive come to learning eq classes is reading the word in a group theory book ....

Answer 36

Oh I don't know much about group theory. We haven't done it yet. This is only my first linear algebra class =( lol

Answer 37

spose the "vectors" look like this:\[\binom{a}{b}\]
then the binary operation looks more like the addition of fractions

Answer 38

typo there. the identity is <0,b> if you can do the following
<0,b> + <c,d> = <bc,bd>

now with fractions we can "cancel" the b and get that this is <c,d> making <0,b> the identity

Answer 39

so what is missing from this set up is a definition of what
<a,b> = <c,d> would be

Answer 40

how would we know when two of these are equal
with fractions we know it
we know
\[\frac{a}{b}=\frac{c}{d}\iff ad =bc\]

Answer 41

We have do define was <a,b> = <c, d> is? All I have from the question is what the set S was :s

Answer 42

to*

Answer 43

:) sneaky little question indeed

Answer 44

Are we allowed to say though that we are manipulating fractions, even though nothing is mentioned?

Answer 45

then maybe i am reading too much in to it, maybe you should just check the vector space axioms. but if we don't know when two are equal, how would we know what the identity is?

Answer 46

All we have to to is show that as long as b and d are positive integers then vectors (a,b) and (c,d) are in the subspace of S because it satisfies b+d>0

Answer 47

Is that wrong?

Answer 48

you need to know what the zero vector is, because it needs to be in your subspace

Answer 49

and in order to know what the zero vector is, you need to know when two vectors are the same

Answer 50

how do i know that 0 + v = v if i don't know what conditions give me
v = v

Answer 51

so the question is, what is the identity under this operation. with fractions it is easy, it is 0/b which translates to <0,b> in this case, but that is because we know that dc/bd = c/d

Answer 52

hm ok I kinda see where you're getting at. But we can't say that once we add the zero vector on the right side, and obtain the left side, then naturally v = v? without needing to define it beforehand?

Answer 53

@Romero how would showing that b+d>0 satisfy the question:

I need to show that (a,b) [+] (c, d) belongs to: a vector space

Answer 54

as long as we dont try to paint the problem with our biased math notions .. we should be good

Answer 55

maybe i am reading too much into it
the question says

I need to show that (a,b) [+] (c, d) belongs to a vector space.

and i am not really sure what this means "belongs to a vector space"

Answer 56

d can = 0 so long as it doesnt mess up the program

Answer 57

question is king of goofy
does "a" belong to a vector space?

Answer 58

lol, i was assuming the elements were composed of real numbers ...

Answer 59

did the question come out of a book? is that the exact wording?

Answer 60

sure but i mean how can you ask if something "is an element of a vector space" ??

Answer 61

by satisfying the axioms that define a vector space

Answer 62

but the question doesn't say "is this a vector space" which i could understand. it says

I need to show that (a,b) [+] (c, d) BELONGS to a vector space.

Answer 63

"I need to show that (a,b) [+] (c, d) belongs to a vector space."
Do you have to do this. Is this what the problem is asking you or are you trying to see if it belongs to a vector space?

Answer 64

Well they had a similar example in the book: they had S = {x in R | x > 0}. Define addition in S by x [+] y = xy. Prove that S is a vector space.

1) Show x [+] y is in S:
x [+] y = xy. Since x >0 and y>0, then x[+]y is in S

Answer 65

you have elements of R^2
that is ordered pairs of real numbers,
and you have a definition of addition. so you could ask "does this set under this operation form a vector space?"

that question would make sense

Answer 66

right, which I read as; take the elements as defined in R^2 and perform a binary operation on them; is the results of that operation conform to the concept of a vector space; or is it only a subset and not a subspace?

Answer 67

holy crap!

Answer 68

i get what the problem is asking, show that it is closed!!

Answer 69

It looks like you are making assumptions about the values of the variables. So can you answer the question with a specific example?

Answer 70

specifics are never "proof"

Answer 71

but unless i am senile that second question you asked is wrong, that does not form a vector space
in a vector space a(x+ y) = ax + ay for all vectors x,y and scalers a

Answer 72

Not trying to prove anything but rather show a contradiction.

Answer 73

and it is certainly not true with "+" defined as multiplication, because no way is a(xy) = axay

Answer 74

contradiction by example are always welcome ;)

Answer 75

should be
Well they had a similar example in the book: they had S = {x in R | x > 0}. Define addition in S by x [+] y = xy. Prove that S is a NOT vector space.

Answer 76

Ohh wait. I'm so sorry. Later on the page, when they are proving the scalar multiplication stuff. They say "Define scalar multiplication to be c [*] x = x^c

Answer 77

ok well i am sorry i butted in, i thought the question was "show this is a vector space" but what it looks like it is asking is "show that it is closed" so you have to show that you end up in the space, which you do since if b > 0 and d > 0 certainly bd > 0