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Scalar multiplication to show that a set with 2 operations is a vect space, easy q

  • Thread starter ryan8642
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  • #1
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u and v are contained in V

Lets say the scalar multiplication is defined as:

ex.

ku=k^2 u or ku = (0,ku2) u=(u1,u2)

does this mean that this is also the same for different scalar m?

mu=m^2 u or mu = (0,mu2) u=(u1,u2)

and does this mean the same for any vector v

kv=k^2 v or kv = (0,kv2) v=(v1,v2)

Is this correct?

Axioms 7,8,9 contain the 2 different scalars as well as vectors. it really confuses me.

Can someone please put me on the right track :s


_____________

so u guys know and im not confusing you guys i showed 2 examples there to help show my problem.

Ex 1.

Lets say the scalar multiplication is defined as:

ku = (0,ku2) u=(u1,u2)

does this then mean that this is also the same for different scalar m?

mu = (0,mu2) u=(u1,u2)

and also this for any vector v

kv = (0,kv2) v=(v1,v2)

_____________________
addition u+v=(u1+v1, u2+v2)

ex.. axiom 8 (to help explain my problem)

using what is described above.

(k+m)u = ku + mu
(k+m)(u1,u2)=k(u1,u2) + m(u1,u2)
(0,(k+m)u2)=(0,ku2) + (0,mu2)
(0,(k+m)u2)=(0,ku2+mu2)
(0,(k+m)u2)=(0,(k+m)u2)

LS=RS therefore axiom 8 holds for the set.

now using just ku=(0,ku2)

(k+m)u = ku + mu
(k+m)(u1,u2) = k(u1,u2) + m(u1,u2)
((k+m)u1, (k+m)u2) = (0,ku2) + (mu1,mu2)
((k+m)u1, (k+m)u2) = (0+mu1, ku2+mu2)
(ku1+mu1,ku2+mu2) = (mu1, ku2 +mu2)

LS ≠ RS so axiom 8 doesnt hold for the set.

hopefully that helps explain my problem...

which way is correct?? please help!!!
 

Answers and Replies

  • #2
Simon Bridge
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Axioms 7,8,9 contain the 2 different scalars as well as vectors. it really confuses me.
Not clear why this should confuse you ... the definitions are general and have to work for any vector in the vector space... so they also work for any pair of vectors each a member of the vector space.
Can someone please put me on the right track :s
Helps if you state the axiom in question - not everyone numbers them the same. eg. http://en.wikipedia.org/wiki/Vector_space#Definition

Lets start by asserting that ##\vec{u}=(u_1,u_2)## and ##\vec{v}=(v_1,v_2)## are both members of the same vector space - so I don't have to keep writing it out. Let's look at your examples:

Lets say the scalar multiplication is defined as:

##k\vec{u} = (0,ku_2)##
OK - and this should work for any other scalar ##m \neq k## and for any other vector ##\vec{v} \neq \vec{u}## ... the definition has to work for any vectors in the space.

using what is described above.

##(k+m)\vec{u} = k\vec{u} + m\vec{u}##
##(k+m)(u_1,u_2)=k(u_1,u_2) + m(u_1,u_2)##
##(0,(k+m)u_2)=(0,ku_2) + (0,mu_2)##
##(0,(k+m)u_2)=(0,ku2+mu_2)##
##(0,(k+m)u_2)=(0,(k+m)u_2)##

LS=RS therefore axiom 8 holds for the set.
You are using the distributive property for scalar multiplication.
The axiom says that ##(a+b)\vec{v}=a\vec{v}+b\vec{v}##

We need to be careful...
##LHS = (a+b)\vec{v}= (0, (a+b)v_2## from the definition.
##RHS = a\vec{v}+b\vec{v} = (0,av_2)+(0,bv_2) = (0, (a+b)v_2) = (a+b)\vec{v}##
Then we can observe that RHS = LHS and so the distributive law holds according to the rule for adding vectors. Explicitly separating the LHS and RHS like this helps to keep confusion at bay.

Notice that it does not matter what we call the scalars - you used k and m, I used a and b: it makes no difference to the math.
After all - the number "2" is a scalar - how does the vector know if that scalar is a k or an m or an a or a b or a whatever?

now using just ##k\vec{u}=(0,ku_2)##

##(k+m)\vec{u} = k\vec{u} + m\vec{u}##
##(k+m)(u_1,u_2) = k(u_1,u_2) + m(u_1,u_2)##
##((k+m)u_1, (k+m)u_2) = (0,ku_2) + (mu_1,mu_2)##
Nonsense! You did not follow the rule for scalar multiplication ... of course you got a different result!

Just as the rule works for k and m so it also works for any other scalar, and (k+m) is a scalar and so is m.
The letters are just labels and don't have any intrinsic meaning to themselves. You could as easily use a and b and (a+b) or fruit and vegetable names or geometric shapes.
If we have a scalar called "bob" then I'd write ##(bob)\vec{u}=(0, (bob)u_2)##
 
Last edited:
  • #3
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okk,

Thanks for the detailed explanation!
That clarifies what i thought
 
  • #4
Simon Bridge
Science Advisor
Homework Helper
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1,642


No worries: it is easy to get tied up in knots about this stuff.
 

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