# Dual Space .... Loring Tu, Section 3.1, page 19 .... ....

• MHB
• Math Amateur
In summary, Tu discusses the dual space in his book "An Introduction to Manifolds" and uses a completely general basis for the vector space $V$. There is no such thing as a "standard basis" for an arbitrary vector space. However, when working in $\mathbb{R}^{n}$, the unit vectors in the direction of the coordinate axes can be considered a standard basis. In the case of the Proposition mentioned, Tu uses $e$ as the basis for $V$, even though it is not a standard basis. The notation $e_{1}, e_{2}, ...$ does not necessarily mean they are the standard basis vectors, as shown in the example of $\mathbb{R}^{2} Math Amateur Gold Member MHB I am reading Loring W.Tu's book: "An Introduction to Manifolds" (Second Edition) ... I need help in order to fully understand Tu's section on the dual space ... ... In his section on the dual space, Tu writes the following: View attachment 8792In the above text from Tu, just preliminary to Proposition 3.1 Tu writes the following: " ... ... Let $$\displaystyle e_1, \ ... \ ... \ e_n$$ be a basis for $$\displaystyle V$$. ... ... " Now my question is as follows:Is $$\displaystyle e_1, \ ... \ ... \ e_n$$ a completely general basis ... hence making Proposition 3.1 completely general (in terms of basis anyway) ... or given the notation is $$\displaystyle e_1, \ ... \ ... \ e_n$$ the standard basis where $$\displaystyle e_1 = ( 1, 0, 0, \ ... \ ... \ , 0 )^T , e_2 = ( 0,1, 0, \ ... \ ... \ , 0 )^T , \ ... \ ... \ , \ e_n = ( 0, 0, 0, \ ... \ ... \ , 1 )^T$$ ...I must say I can find no instance in the proof of Proposition 3.1 where the basis $$\displaystyle e_1, \ ... \ ... \ e_n$$ is assumed to be anything but completely general ... but would be appreciative if someone would confirm this to be the case ... ... ( ... hmmm ... pity Tu didn't use $$\displaystyle u_1, u_2, \ ... \ ... \ , u_n$$ as the basis for $$\displaystyle V$$ ... ... )My suspicions about the basis being not general ... but indeed the standard basis ... ... came about on reading the following note on page 11 concerning notation ... View attachment 8793Hope someone can clarify the above ... Help will be appreciated ... #### Attachments • Tu - Start of Section 3.1 Dual SPace ... .png 26.5 KB · Views: 85 • Tu - Standard Basis for R^n ... Section 2.2, Page 11 .png 5.6 KB · Views: 86 Hi Peter, Glad to hear you're taking a look at Tu, I think it's a good book on the subject. Peter said: Is $$\displaystyle e_1, \ ... \ ... \ e_n$$ a completely general basis ... hence making Proposition 3.1 completely general (in terms of basis anyway) ... or given the notation is $$\displaystyle e_1, \ ... \ ... \ e_n$$ the standard basis where $$\displaystyle e_1 = ( 1, 0, 0, \ ... \ ... \ , 0 )^T , e_2 = ( 0,1, 0, \ ... \ ... \ , 0 )^T , \ ... \ ... \ , \ e_n = ( 0, 0, 0, \ ... \ ... \ , 1 )^T$$ ...I must say I can find no instance in the proof of Proposition 3.1 where the basis $$\displaystyle e_1, \ ... \ ... \ e_n$$ is assumed to be anything but completely general ... but would be appreciative if someone would confirm this to be the case ... ... ( ... hmmm ... pity Tu didn't use $$\displaystyle u_1, u_2, \ ... \ ... \ , u_n$$ as the basis for $$\displaystyle V$$ ... ... )My suspicions about the basis being not general ... but indeed the standard basis ... ... came about on reading the following note on page 11 concerning notation ... When$V$is an arbitrary vector space there is no such thing as the "standard basis;" i.e., the basis he selects is completely general. When working in$\mathbb{R}^{n}$there is a standard basis; specifically, the unit vectors in the direction of the coordinate axes (see https://en.wikipedia.org/wiki/Standard_basis). In the case of the Proposition, Tu is just reusing the letter$e$for the basis, even though there is no such thing as the "standard basis" in this context. Peter said: or given the notation is $$\displaystyle e_1, \ ... \ ... \ e_n$$ the standard basis where $$\displaystyle e_1 = ( 1, 0, 0, \ ... \ ... \ , 0 )^T , e_2 = ( 0,1, 0, \ ... \ ... \ , 0 )^T , \ ... \ ... \ , \ e_n = ( 0, 0, 0, \ ... \ ... \ , 1 )^T$$ ... It is worth mentioning here that the fact that we can express the basis vectors as$e_{1}=(1, 0, 0, \ldots, 0)^{T}, e_{2}=(0, 1, 0, \ldots, 0)^{T},$etc. has nothing to do with whether$e_{1}, e_{2}, \ldots$are a "standard basis" for the given vector space or not. For example, consider$\mathbb{R}^{2}.$Take$u_{1}$to be the unit vector in the positive$x$-direction and$u_{2}$to be the unit vector obtained by rotating$u_{1}$counterclockwise$45^{\circ}$about the origin. Then$B=\{u_{1},u_{2}\}$is a basis for$\mathbb{R}^{2}$, which is not the standard basis for$\mathbb{R}^{2}.$However, the coordinate vector representations of$u_{1}$and$u_{2}$with respect to the chosen basis$B$are given by$u_{1}=(1,0)^{T}$and$u_{2}=(0,1)^{T},$because$u_{1}=1\cdot u_{1} +0\cdot u_{2}$and$u_{2} = 0\cdot u_{1} +1\cdot u_{2}.$If you prefer, you can add some notation to remind you that you are working with the coordinate vectors with respect to$B$; e.g.,$[u_{1}]_{B}=(0,1)^{T}$and$[u_{2}]_{B}=(0,1)^{T}.$GJA said: Hi Peter, Glad to hear you're taking a look at Tu, I think it's a good book on the subject. When$V$is an arbitrary vector space there is no such thing as the "standard basis;" i.e., the basis he selects is completely general. When working in$\mathbb{R}^{n}$there is a standard basis; specifically, the unit vectors in the direction of the coordinate axes (see https://en.wikipedia.org/wiki/Standard_basis). In the case of the Proposition, Tu is just reusing the letter$e$for the basis, even though there is no such thing as the "standard basis" in this context. It is worth mentioning here that the fact that we can express the basis vectors as$e_{1}=(1, 0, 0, \ldots, 0)^{T}, e_{2}=(0, 1, 0, \ldots, 0)^{T},$etc. has nothing to do with whether$e_{1}, e_{2}, \ldots$are a "standard basis" for the given vector space or not. For example, consider$\mathbb{R}^{2}.$Take$u_{1}$to be the unit vector in the positive$x$-direction and$u_{2}$to be the unit vector obtained by rotating$u_{1}$counterclockwise$45^{\circ}$about the origin. Then$B=\{u_{1},u_{2}\}$is a basis for$\mathbb{R}^{2}$, which is not the standard basis for$\mathbb{R}^{2}.$However, the coordinate vector representations of$u_{1}$and$u_{2}$with respect to the chosen basis$B$are given by$u_{1}=(1,0)^{T}$and$u_{2}=(0,1)^{T},$because$u_{1}=1\cdot u_{1} +0\cdot u_{2}$and$u_{2} = 0\cdot u_{1} +1\cdot u_{2}.$If you prefer, you can add some notation to remind you that you are working with the coordinate vectors with respect to$B$; e.g.,$[u_{1}]_{B}=(0,1)^{T}$and$[u_{2}]_{B}=(0,1)^{T}.\$[

Thanks for your help, GJA ... including the tip regarding Tu ...

Peter

## 1. What is Dual Space?

Dual Space, also known as the dual vector space, is a mathematical concept that is closely related to the original vector space. It is a space of linear functionals, which are functions that map vectors to scalars.

## 2. How is Dual Space related to the original vector space?

The dual space is the space of all linear functionals on the original vector space. This means that for every vector in the original space, there is a corresponding linear functional in the dual space.

## 3. What is the purpose of Dual Space?

Dual Space is used to define and study the properties of linear functionals, which are important in many areas of mathematics and science. It also helps to simplify calculations and proofs in linear algebra and functional analysis.

## 4. How is the dimension of Dual Space related to the original vector space?

The dimension of the dual space is equal to the dimension of the original vector space. This means that if the original space has n dimensions, the dual space will also have n dimensions.

## 5. How is Dual Space represented mathematically?

Dual Space can be represented as the set of all linear functionals on the original vector space, denoted as V*. It can also be represented as the set of all linear combinations of the basis vectors of the original space, denoted as {e1*, e2*, ..., en*}.

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