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linear transformation


Sarahisme
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ok what i have gotten so far is...

 

let [math] v\epsilon V[/math]

 

[math]

that \ is \ v = a_1v_1+...+a_nv_n

[/math]

 

so

[math]

T(v) = T(a_1v_1 + ... + a_nv_n) = T(a_1v_1) + ... + T(a_nv_n) [/math]

[math]

= a_1T(v_1) + ... + a_nT(v_n)

[/math]

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Construct one that does it and the uniqueness follows (a liner map is uniquely determined by where it sends a basis and specifying the action on the basis uniquely determines the map obviously). do you understand the basic method of constructing linear maps from where they send the basis vecotrs? eg in R^2 if e and f are the basis and I tell you that e maps to 2e+f and f maps to e+2f can you write out the matrix? (you can if you just think about matrix multiplication. where does the 2x2 matrix ith entries a,b in the top row and c,d in the bottom row send the vector (0,1)^t (ie the column vector with 1 at the top and 0 at the bottom) in terms of a,b,c, adn d?

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"a liner map is uniquely determined by where it sends a basis and specifying the action on the basis uniquely determines the map obviously"

 

is that a thm. of some kind, because i havent seen it in my textbook yet. although i can see that it makes sense.

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oh ok i think i see....

 

a linear transformation maps one vector space into another, right?

 

so if T:E --> W is a linear transformation

 

the range of the transformation is depedent on the vectors in V, and if you have a basis for E, then you effectively have all basis for the range of T, as a linear transformation is defined by

T(u+v) = T(u) + T(v)

and

T(cu) = cT(u)

 

is what i just said correct?....or relevant? :P

 

Thanks

 

Sarah :)

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ok heres my attempt at what i think the proof to problem should look like (tell me what you think! :) )

 

ok here goes:

-----------------------------------------------------------------------------------------

-----------------------------------------------------------------------------------------

Define T to be

[math]

T( \sum_{i=1}^n(a_iv_i)) = \sum_{i=1}^n(a_iw_i)

[/math]

 

Since T is defined as transforming the basis vectors of the vector space V, then it is well defined.

 

Now take

[math]

v, p \in V

[/math]

and

[math]

c,d \in \mathbb{R} [/math] (i.e. c,d scalars)

 

So

[math]

v = \sum_{i=1}^n(a_iv_i)

[/math]

[math]

p = \sum_{i=1}^n(b_iv_i)

[/math]

 

[math]

(cv + cp) = c(\sum_{i=1}^n(a_iv_i)) + d(\sum_{i=1}^n(b_iv_i))

[/math]

[math]

= v_i(\sum_{i=1}^n(ca_i + db_i))

[/math]

[math]

= \sum_{i=1}^n(v_i(ca_i + db_i))

[/math]

 

Then

[math]

T(cv + dp) = \sum_{i=1}^n(ca_i + db_i)w_i

[/math]

[math]

= c\sum_{i=1}^na_iw_i + d\sum_{i=1}^nb_iw_i

[/math]

[math]

= cT(v) + dT(p)

[/math]

 

[math]

\therefore [/math] T is linear.

 

 

[math]

\therefore [/math] there is a unique linaer transformation [math] T : V \rightarrow W [/math] such that [math] T(v_i) = w_i [/math] for i = 1,…,n.

-----------------------------------------------------------------------------------------

-----------------------------------------------------------------------------------------

 

well that’s what I think it is, however I am not sure this shows that it is unique (or is it because of the fact that is transforms basis vectors that make it unique??, in which case should I say up the top of my proof “Since T is defined as transforming the basis vectors of the vector space V, then it is well defined AND unique.”)

 

well anyway, tell me what you think! :)

 

Cheers

 

Sarah :)

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Proof of uniqueness of a linear transformation:

 

We can assume that {e1,e2,...} is the basis of the vector space V.

 

Suppose T and T' two linear transformations such that

T(e1) = u1 and T'(e1) = u1

T(e2) = u2 and T'(e2) = u2

T(e3) = u3 and T'(e3) = u3, etc.

Then, for each v in V

 

T(v) = T(k.e1+l.e2+m.e3+...)

= k.u1 + l.u2 + m.u3 + ... for some real k,l,m, ...

and

T'(v)= T'(k.e1 + l.e2 + m.e3+ ...)

= T'(k.e1) + T'(l.e2) + T'(m.e3) + ...

= k.T'(e1) + l.T'(e2) + m.T'(e3) + ...

= k.u1 + l.u2 + m.u3 + ...

Hence, T = T'

 

Hope this answers your problem.

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