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Linear transformations question

  • Thread starter Rackhir
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  • #1
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Homework Statement


Today in my final i was given this exercise:
Given [itex]β_1=\{v_1,v_2,v_3\}[/itex] and [itex]β_2=\{u_1,u_2,u_3,u_4\}[/itex], basis of the vector spaces [itex]V[/itex] and [itex]U[/itex] respectively.
a) Find the linear transformation [itex]T:U\rightarrow V[/itex] so that [itex]T(v_i)≠T(v_j)[/itex] if [itex]i≠j[/itex], [itex]T(v_1)=u_1+u_2[/itex] and [itex]T[/itex] is injective

b) Find the transformation matrix from [itex]β_1[/itex] to [itex]β_2[/itex], [itex][T]_{β_1 \rightarrow β_2}[/itex]

Homework Equations


If [itex]T[/itex] is injective if and only if [itex]Kernel(T)=\{0\}[/itex], that means that the nullspace of the transformation matrix is [itex]\{0\}[/itex]


The Attempt at a Solution


I thought that finding [itex][T]_{β_1 \rightarrow β_2}[/itex] first would be easier, or at least it made more sense for me. I found this matrix
[tex] \begin{pmatrix}
1&0&0\\
1&0&0\\
0&1&0\\
0&0&1\\
\end{pmatrix} [/tex]

Where the first column, [itex]\begin{pmatrix}
1\\
1\\
0\\
0\\
\end{pmatrix}[/itex]
comes from [itex]T(v_1)=u_1+u_2[/itex], and the other two were chosen so the nullspace of [itex][T]_{β_1 \rightarrow β_2}[/itex] is in fact, [itex]\{0\}[/itex]
My big question, is this right? or it's horribly wrong and i should feel ashamed when i look myself in the mirror?

Then, for a) find the transformation per se, should i solve the following system?
[tex] \begin{pmatrix}
1&0&0\\
1&0&0\\
0&1&0\\
0&0&1\\
\end{pmatrix} \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}a\\b\\c\\d\end{pmatrix}[/tex], where [itex](x,y,z) \in V[/itex] and [itex](a,b,c,d) \in U[/itex]. So i end with [itex] a=x, b=x, c=y, d=z [/itex], but this is highly dependant on the basis chosen, right? shouldn't be independant?

Any help will be highly appreciated, i'd love to know how this is actually solved.
 
Last edited:

Answers and Replies

  • #2
vela
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Your matrix is fine. Feel free to look in the mirror.

For part (a), all you probably needed to do was say what T does to the basis vectors in β1, and then prove that T is injective.
 
  • #3
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Your matrix is fine. Feel free to look in the mirror.

For part (a), all you probably needed to do was say what T does to the basis vectors in β1, and then prove that T is injective.
YEY, i feel relieved :) and what do you mean with "all you probably needed to do was say what T does to the basis vectors in β1"? is [itex]a=x, b=x, c=y, d=z \ \forall \ (x,y,z)∈V \ and \ (a,b,c,d)∈U[/itex] wrong? as you can see, i have more problems finding out what i need to do, rather than doing it
I just hope my teachers dion't mark me wrong because i found the matrix first :P
Santiago
 
  • #4
vela
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Like you said, the calculation you did depends on the basis, but saying that T(v1) = u1+u2 is independent of the basis. Whatever representation you use, T will always map v1 to u1+u2.

Because T is linear, if you specify what it does to a set of basis vectors, you've determined what T will do to any vector in the space, so simply saying what T(v1), T(v2), and T(v3) are is enough to fully describe T.
 
  • #5
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Like you said, the calculation you did depends on the basis, but saying that T(v1) = u1+u2 is independent of the basis. Whatever representation you use, T will always map v1 to u1+u2.

Because T is linear, if you specify what it does to a set of basis vectors, you've determined what T will do to any vector in the space, so simply saying what T(v1), T(v2), and T(v3) are is enough to fully describe T.
Oh i see, so, with the matrix i chose, [itex]T(v_1)=u_1+u_2[/itex], [itex]T(v_2)=u_3[/itex] and [itex]T(v_3)=u_4[/itex]. Well that does make sense now. And since they will be clearly linearly independant, the proof that T is injective is rather obvious.
 

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