# prove that a set V is a vector space

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• Dec 19th 2011, 11:14 PM
WeeG
prove that a set V is a vector space
Hello,

I am trying to prove that a set V is a vector space, I managed to prove most of the process, but stuck with a couple of things...

The set V is the set of positive real numbers, and the field is R.
the two operations are:

x++y = xy
a**x = x^a

(where ++ is the addition, ** is the scalar multiplication, and a is a real scalar)

I find it hard to prove that for 2 scalars r and s, and a member of V, x
(r++s)**x = (r**x) + (s**x)

basically what I don't understand, is what do to when having

r**s, in other words, when I have to multiply 2 scalars, do I do it using the regular operator, or using the new X*scalar operator

any help would be appreciated...
• Dec 19th 2011, 11:29 PM
alexmahone
Re: prove that a set V is a vector space
Quote:

Originally Posted by WeeG
Hello,

I am trying to prove that a set V is a vector space, I managed to prove most of the process, but stuck with a couple of things...

The set V is the set of positive real numbers, and the field is R.
the two operations are:

x++y = xy
a**x = x^a

(where ++ is the addition, ** is the scalar multiplication, and a is a real scalar)

I find it hard to prove that for 2 scalars r and s, and a member of V, x
(r++s)**x = (r**x) + (s**x)

basically what I don't understand, is what do to when having

r**s, in other words, when I have to multiply 2 scalars, do I do it using the regular operator, or using the new X*scalar operator

any help would be appreciated...

$(r++s)**x=x^{rs}$

$(r**x)++(s**x)=x^rx^s=x^{r+s}$

$(r++s)**x\neq(r**x)++(s**x)$

So, $V$ is not a vector space.

If $a**x=a^x$, then $V$ would be a vector space. Are you sure you've copied the question correctly?
• Dec 19th 2011, 11:43 PM
FernandoRevilla
Re: prove that a set V is a vector space
Quote:

Originally Posted by WeeG
I find it hard to prove that for 2 scalars r and s, and a member of V, x (r++s)**x = (r**x) + (s**x)

The equality is not correctly written. Take into account that $++$ is the sum of vectors (not of scalars). You have to prove $(r+s)^{**}x=r^{**}x++s^{**}x$ . Then,

$(r+s)^{**}x=x^{r+s}=x^rx^s=x^r++x^s=r^{**}x++s^{** }x$
• Dec 19th 2011, 11:47 PM
FernandoRevilla
Re: prove that a set V is a vector space
Quote:

Originally Posted by alexmahone
So, $V$ is not a vector space.

Look at my previous post.
• Dec 19th 2011, 11:51 PM
alexmahone
Re: prove that a set V is a vector space
Quote:

Originally Posted by FernandoRevilla
$(r+s)^{**}x=r^{**}x++s^{**}x$

How come you're using 2 different kinds of addition in your equation? The OP's symbol for addition is ++, so why shouldn't the equation be $(r++s)^{**}x=r^{**}x++s^{**}x$?
• Dec 20th 2011, 12:02 AM
FernandoRevilla
Re: prove that a set V is a vector space
Quote:

Originally Posted by alexmahone
How come you're using 2 different kinds of addition in your equation. The OP's symbol for addition is ++, so why shouldn't the equation be $(r++s)^{**}x=r^{**}x++s^{**}x$?

Let us see, the field is $(\mathbb{R},+,\cdot)$ with the usual operations sum and product. The operation defined on $V$ is $++$ . So, for two scalars $r,s$ its sum is $r+s$ , not $r++s$ .
• Dec 20th 2011, 02:30 AM
WeeG
Re: prove that a set V is a vector space
thank you both !

let me get is straight, you are saying, that the operators ++ and ** are between a scalar and a member of V (even if V is number too), and when dealing with 2 scalars, I use the "regular" operators ?

sounds logical I suppose
• Dec 20th 2011, 03:07 AM
alexmahone
Re: prove that a set V is a vector space
Quote:

Originally Posted by WeeG
thank you both !

let me get is straight, you are saying, that the operators ++ and ** are between a scalar and a member of V (even if V is number too), and when dealing with 2 scalars, I use the "regular" operators ?

sounds logical I suppose

++ is between two members of V.

** is between a scalar and a member of V.
• Dec 20th 2011, 04:58 AM
WeeG
Re: prove that a set V is a vector space
yes, and what happen when I have 2 scalars, do I use the "regular" operators ?
• Dec 20th 2011, 05:27 AM
FernandoRevilla
Re: prove that a set V is a vector space
Quote:

Originally Posted by WeeG
yes, and what happen when I have 2 scalars, do I use the "regular" operators ?

Yes, notice that the problem says and the field is $\mathbb{R}$ without mentioning "strange operations", so the interpretation must be the usual operations.
• Dec 20th 2011, 08:52 AM
agentmulder
Re: prove that a set V is a vector space
Quote:

Originally Posted by WeeG
Hello,

I am trying to prove that a set V is a vector space, I managed to prove most of the process, but stuck with a couple of things...

The set V is the set of positive real numbers, and the field is R.
the two operations are:

x++y = xy
a**x = x^a

(where ++ is the addition, ** is the scalar multiplication, and a is a real scalar)

I find it hard to prove that for 2 scalars r and s, and a member of V, x
(r++s)**x = (r**x) + (s**x)

basically what I don't understand, is what do to when having

r**s, in other words, when I have to multiply 2 scalars, do I do it using the regular operator, or using the new X*scalar operator

any help would be appreciated...

How can V as defined above be a vector space? The zero vector is missing from set V...

Have i misunderstood the question?

(Wondering)
• Dec 20th 2011, 08:57 AM
FernandoRevilla
Re: prove that a set V is a vector space
Quote:

Originally Posted by agentmulder
How can V as defined above be a vector space? The zero vector is missing from set V... Have i misunderstood the question?

For all $x\in V$ we have $x++1=1++x=x$ so, $1$ is the zero vector of $(V,++)$ .
• Dec 20th 2011, 08:58 AM
alexmahone
Re: prove that a set V is a vector space
Quote:

Originally Posted by agentmulder
How can V as defined above be a vector space? The zero vector is missing from set V...

Have i misunderstood the question?

(Wondering)

I think the zero vector in this case is $1$ because $x++1=x$ and $a**1=1$.
• Dec 20th 2011, 09:04 AM
agentmulder
Re: prove that a set V is a vector space
Quote:

Originally Posted by alexmahone
I think the zero vector in this case is $1$ because $x++1=x$ and $a**1=1$.

suppose x is 2...then 2++1=2?

oh wait...i see now... x++y=xy so 2++1=(2)(1)=2
• Dec 20th 2011, 09:30 AM
agentmulder
Re: prove that a set V is a vector space
let me ask you if my interpretation of a**x = a^x is correct...

suppose 1 plays the role of the 0 vector then a**1 is the zero vector since scalar multiplication on the 0 vector gives the zero vector.

now, 1^a = 1 for real number scaler a and real number 1

since 1 is the 0 vector...it checks out

is this correct?

but somehow the vectors are being mixed up with the scalers... or am i thinking too much?

(Itwasntme)
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