1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Linear transformations question

  1. Aug 28, 2012 #1
    1. The problem statement, all variables and given/known data
    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]

    2. Relevant 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]


    3. 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: Aug 28, 2012
  2. jcsd
  3. Aug 28, 2012 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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.
     
  4. Aug 28, 2012 #3
    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
     
  5. Aug 28, 2012 #4

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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.
     
  6. Aug 28, 2012 #5
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Linear transformations question
Loading...