Vector

**igodspeed** · April 29th 2009, 09:45 AM

- Probably the hardest question in the text, could someone please solve this. I have absolutely no idea how to show $T$ as a linear transformation.
Here is the Question.
Let ${({{ R}^n})^*} = \{ T:{{ R}^n} \to {{ R}^n}|T{\text{ is linear\} }}$ . For $T$ and $S$ in ${({{ R}^n})^*}$ and $C$ in ${{{ R}}}$ define $(T+S)(u)=T(u)+S(u)$ for all $u$ in ${{{ R}^n}}$ . Show that ${({{R}^n})^*}$ is a vector space.
Where ${({{R}^n})^*}$ is a set of linear transformation.

**Gamma** · April 29th 2009, 01:08 PM

Originally Posted by igodspeed

Let ${({{ R}^n})^*} = \{ T:{{ R}^n} \to {{ R}^n}|T{\text{ is linear\} }}$ . For $T$ and $S$ in ${({{ R}^n})^*}$ and $C$ in ${{{ R}}}$ define $(T+S)(u)=T(u)+S(u)$ for all $u$ in ${{{ R}^n}}$ . Show that ${({{R}^n})^*}$ is a vector space.
Where ${({{R}^n})^*}$ is a set of linear transformation.

I assume you meant to add $CT(u)=T(Cu)$
This is just the dual space to $\mathbb{R}^n$

$(T+CS)(u)=T(u)+CS(u)=T(u)+S(Cu)$

Now all you gotta do is show it is actually a linear transformation itself, ie $(\mathbb{R}^n)^*$ showing its closed under scalar multiplication and addition.

$(T+CS) (au+bv)=T(au+bv)+S(C(au+bv))=aT(u)+bT(v)+CaS(u)+Cb S(v)$
$=aT(u)+aCS(u)+bT(v)+bCS(v) = a(T+CS)(u)+b(T+CS)(v)$

So it indeed a linear transformation of $\mathbb{R}^n$ , QED.

**igodspeed** · April 29th 2009, 05:16 PM

- Yes you are right. I guess i need to show that $T+S$ is a linear transformation.
Someone told me that i need to check there's a zero, that it's closed under addition/subtraction, and that "there's an $R$ -action on it. Now, you don't have any $R$ -action hypothesis, so this is not a vector space." So i am a little confused. Thanks for your help.

**Gamma** · April 29th 2009, 05:25 PM

Yeah, that's what I checked. C is an element in the reals. That is why I checked T + CS is a linear transformation, that checks additivity and the R action all in one step. It shows it is closed under addition and scalar multiplication.

Sure the trivial transformation 0 is linear so it is in there. I think maybe you need to review your definitions and/or reread my proof until you understand what you are supposed to do.

**igodspeed** · April 29th 2009, 05:58 PM

- Would it be possible to get the basis for ${({{ R}^n})^*}$ ? It is so abstract.
Thanks,
Kelly

**Gamma** · April 29th 2009, 06:59 PM

If you are interested in learning more about this topic, I suggest checking out Dual space - Wikipedia, the free encyclopedia. They have a fairly extensive overview of the Dual Space and it includes the basis for the dual space. What is possibly even more interesting is that when you consider the space dual to the dual space (the double dual if you will) Linear functionals on the linear functionals of the ground vector space, you get a space that is canonically isomorphic to the ground vector space. This is particularly important because the isomorphism requires absolutely no reference to a basis whatesoever.

Note all those above facts are only necessarily true for finite dimensional vector spaces. I believe the maps are still injective, but may not necessarily by surjective if it is not finite dimensional.

**igodspeed** · April 29th 2009, 09:16 PM

It is way beyond me. I don't even understand half of it. I am in an introductory course, so if you could, could you please explain it to me in layman's terms?
thanks in anticipation,
KC

**Gamma** · April 29th 2009, 09:34 PM

Yeah man, its some abstract stuff.

Basically your basis for the dual space to an n dimensional vector space is just the set of n linear functionals such that the ith one spits out the coefficient on the ith basis element for the vector space.

**igodspeed** · April 29th 2009, 11:10 PM

so how can i find the basis for ${({{ R}^n})^*}<br />$

**Gamma** · April 30th 2009, 09:33 AM

This is the point, it depends on the basis, there are an infinite number of possible choices for basis of $\mathbb{R}^n$ each of these yields a different basis for $(\mathbb{R}^n)^*$ .

To give you the idea though for any given basis for $\mathbb{R}^n$ call it $\{e_1,e_2,...,e_n\}$ , you get n linear functionals in $(\mathbb{R}^n)^*$ that will form the basis, say they are $\{\epsilon^1,\epsilon^2,...,\epsilon^n \}.$

Let the n functionals be defined as follows.

$\epsilon^i(a_1e_1 +a_2e_2 + ... + a_ie_i + ... +a_ne_n)=a_i$

Yeah, there is your basis. If you took the standard basis for $\mathbb{R}^n$ the ith coordinate function would simply pick out the ith coefficient.

**igodspeed** · May 3rd 2009, 09:04 PM

Originally Posted by Gamma

I assume you meant to add $CT(u)=T(Cu)$
This is just the dual space to $\mathbb{R}^n$

$(T+CS)(u)=T(u)+CS(u)=T(u)+S(Cu)$

Now all you gotta do is show it is actually a linear transformation itself, ie $(\mathbb{R}^n)^*$ showing its closed under scalar multiplication and addition.

$(T+CS) (au+bv)=T(au+bv)+S(C(au+bv))=aT(u)+bT(v)+CaS(u)+Cb S(v)$
$=aT(u)+aCS(u)+bT(v)+bCS(v) = a(T+CS)(u)+b(T+CS)(v)$

So it indeed a linear transformation of $\mathbb{R}^n$ , QED.

I was wondering how i can show that there is a zero vector 0 in V such that u+0=0.

**Gamma** · May 3rd 2009, 09:35 PM

Is V the dual space you are talking about? Just in general for future use, V is typically reserved for the vector space. Generally you need to denote the dual space $V^*$ . It is the vector space dual to V.

If this is indeed what you are talking about, the identity under addition of the vector space of linear transformations is just the trivial linear transformation. It takes any vector and sends it to 0, call it Z for zero map. It is easy to check this is indeed a linear transformation, and it clearly satisfies (T + Z)(v)=T(v) + Z(v)=T(v) + 0= T(v) for all v, so it is the identity.

**igodspeed** · May 3rd 2009, 10:03 PM

I mean the Identity not the dual space. Could you also please show me how the sum of u and v, denoted by u+v is in V. Where V is vector space, nonempty set of vectors.
thanks
KC

**igodspeed** · May 3rd 2009, 11:11 PM

As you said depending on the choice of basis used. How would i find the basis based on standard matrix. I do understand what you meant but i don't know how i can actually use this to find the basis?
Thanks
KC

**Gamma** · May 3rd 2009, 11:43 PM

Are you talking about the other question on the other Thread? I am really confused as to what you are talking about? Again I cannot stress enough, there is no "THE basis" there is no preferred basis, it is just a basis.

If you need to find a basis for $\mathbb{R}^n$ ,just pick the standard basis for $\mathbb{R}^n$ .
{(1,0,...,0), (0,1,0,...,0),... (0,0,...,1)}

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