# prove that a set V is a vector space

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• Dec 20th 2011, 08:38 AM
alexmahone
Re: prove that a set V is a vector space
Quote:

Originally Posted by agentmulder
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?

Yes. (See post #13.)
• Dec 20th 2011, 08:53 AM
agentmulder
Re: prove that a set V is a vector space
Quote:

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

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

WOW! this algebraic manipulation between operators is beautiful...(Clapping)
• Dec 20th 2011, 09:05 AM
FernandoRevilla
Re: prove that a set V is a vector space
Quote:

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

Write $\displaystyle V=\{\vec{x},\vec{y},\ldots\}$ and $\displaystyle \mathbb{R}=\{a,b,\ldots\}$ . Then, $\displaystyle \vec{0}=1$ , so

$\displaystyle \boxed{a^{**}\vec{0}}=1^a=1=\boxed{\vec{0}}$
• Dec 20th 2011, 09:29 AM
agentmulder
Re: prove that a set V is a vector space
Quote:

Originally Posted by FernandoRevilla
Write $\displaystyle V=\{\vec{x},\vec{y},\ldots\}$ and $\displaystyle \mathbb{R}=\{a,b,\ldots\}$ . Then, $\displaystyle \vec{0}=1$ , so

$\displaystyle \boxed{a^{**}\vec{0}}=1^a=1=\boxed{\vec{0}}$

yes, thank you, i see. Apparantly the positive real numbers are a vector space according to the operations defined in the OP

Another question... ++ is regular multiplication, ** is regular exponentiation

we don't have to worry about the 'object' 0 being in V since no positive base to a real number exponent can give 0.

I don't know what to call 0... a vector?... a scalar? both?

Are not the elements of V bases? The elements of R exponents?

That would mean the bases are vectors the exponents scalars?

is this correct?

(Speechless)
• Dec 20th 2011, 11:59 PM
FernandoRevilla
Re: prove that a set V is a vector space
Quote:

Originally Posted by agentmulder
I don't know what to call 0... a vector?... a scalar? both?

That depends. For example, if you consider the "object" $\displaystyle 1$ as belonging to $\displaystyle V$ , it is a vector, and satisfies $\displaystyle 1++x=x++1=x$ for all $\displaystyle x\in V$ , so $\displaystyle 1$ is the zero vector. If you consider the "object" $\displaystyle 1$ as belonging $\displaystyle \mathbb{R}$ , it is a scalar and satisfies $\displaystyle a\cdot 1=1\cdot a=a$ , so $\displaystyle 1$ is the identity element for the product of scalars.

On the other hand, the "object" $\displaystyle 0$ does not belong to $\displaystyle V$ but belongs to $\displaystyle \mathbb{R}$ so, necessarily is an scalar, satisfying $\displaystyle a+0=0+a=0$ for all $\displaystyle a\in\mathbb{R}$ so, $\displaystyle 0$ is the identity element for the sum of scalars.
• Dec 22nd 2011, 02:51 PM
agentmulder
Re: prove that a set V is a vector space
Quote:

Originally Posted by FernandoRevilla
That depends. For example, if you consider the "object" $\displaystyle 1$ as belonging to $\displaystyle V$ , it is a vector, and satisfies $\displaystyle 1++x=x++1=x$ for all $\displaystyle x\in V$ , so $\displaystyle 1$ is the zero vector. If you consider the "object" $\displaystyle 1$ as belonging $\displaystyle \mathbb{R}$ , it is a scalar and satisfies $\displaystyle a\cdot 1=1\cdot a=a$ , so $\displaystyle 1$ is the identity element for the product of scalars.

On the other hand, the "object" $\displaystyle 0$ does not belong to $\displaystyle V$ but belongs to $\displaystyle \mathbb{R}$ so, necessarily is an scalar, satisfying $\displaystyle a+0=0+a=0$ for all $\displaystyle a\in\mathbb{R}$ so, $\displaystyle 0$ is the identity element for the sum of scalars.

Woderful! A clear and precise explanation, well done sir.

(Clapping)
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