- #1

smithg86

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**[SOLVED] Linear Algebra - Dual Spaces**

## Homework Statement

(

**V**and

**W**are vector spaces.

**F**is a field)

"The space L(

**V**,

**W**) of linear maps from

**V**to

**W**is always a vector space. Take

**W**=

**F**. We then get the space V* := L(

**V**,

**F**) of

**F**-linear maps

**V**-->

**F**. This is called the

**dual space**of

**V**."

1. Let

**V**=

**F**[tex]^{2}[/tex] with basis e[tex]_{1}[/tex], e[tex]_{2}[/tex]. Define elements e*[tex]_{1}[/tex], e*[tex]_{2}[/tex] [tex]\in[/tex]

**V*** by:

e*[tex]_{1}[/tex] (e[tex]_{1}[/tex]) = 1

e*[tex]_{1}[/tex] (e[tex]_{2}[/tex]) = 0

and

e*[tex]_{2}[/tex] (e[tex]_{1}[/tex]) = 0

e*[tex]_{2}[/tex] (e[tex]_{2}[/tex]) = 1

Show that e*[tex]_{1}[/tex], e*[tex]_{2}[/tex] form a basis for

**V***. Deduce that dim (

**F**[tex]^{2}[/tex])* = 2.

## Homework Equations

1. In

**F**[tex]^{2}[/tex], e[tex]_{1}[/tex] = (1,0) and e[tex]_{2}[/tex] = (0,1).

2. if vectors v[tex]_{1}[/tex], ...v[tex]_{n}[/tex] [tex]\in[/tex]

**V**and the list (v[tex]_{1}[/tex], ...v[tex]_{n}[/tex]) is linearly independent, then:

a[tex]_{1}[/tex]v[tex]_{1}[/tex] + ... + a[tex]_{n}[/tex]v[tex]_{n}[/tex] =0 [tex]\Rightarrow[/tex] all a's = 0, such that a [tex]\in[/tex]

**F**

3. The definition of a linear map.

## The Attempt at a Solution

If e[tex]_{1}[/tex], e[tex]_{2}[/tex] form a basis for

**V**, then the list (e[tex]_{1}[/tex], e[tex]_{2}[/tex]) is linearly independent and spans V. If I am to show that e*[tex]_{1}[/tex], e*[tex]_{2}[/tex] form a basis for

**V***, then the list (e*[tex]_{1}[/tex], e*[tex]_{2}[/tex]) must be linearly independent and span

**V***. (?)

Here is where I get confused. From what I understand, e*[tex]_{1}[/tex] 'keeps' the first component of the vector and discards all others; e*[tex]_{2}[/tex] 'keeps' the second component of the vector and discards all others. Since the vector space

**V***has elements that are linear maps, e*[tex]_{1}[/tex] and e*[tex]_{2}[/tex] must be linear maps. I'm not sure how to 'work with' linear maps as vectors.

Since e[tex]_{1}[/tex], e[tex]_{2}[/tex] span

**V**, any v[tex]\in[/tex]V can be written as a linear combination of them:

v = a[tex]_{1}[/tex] e[tex]_{1}[/tex] + a[tex]_{2}[/tex] e[tex]_{2}[/tex]

To show that e*[tex]_{1}[/tex],e*[tex]_{2}[/tex] span, I have to show that any T[tex]\in[/tex]

**V*** can be written as:

T = b[tex]_{1}[/tex] e*[tex]_{1}[/tex] + b[tex]_{2}[/tex] e*[tex]_{2}[/tex]

But what are the a's? Are they in

**F**(i.e., real numbers) or are they in

**V**(are they vectors)? (I'm not sure how to proceed.)

To show that the list is linearly independent, I need to show:

c[tex]_{1}[/tex] e*[tex]_{1}[/tex] + c[tex]_{2}[/tex] e*[tex]_{2}[/tex] = 0 [tex]\Rightarrow[/tex] all c's = 0 (again, what are the c's?)

Can anyone help me? I know the second part of the problem is easy: from the definition of dimension, if

**V*** has a 2-vector-long basis, then dim

**V*** = 2.

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