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Linear Algebra (Transform a Basis Set)

  1. Jul 15, 2009 #1
    1. The problem statement, all variables and given/known data

    If [tex]T:U\rightarrow V[/tex] is any linear transformation from U to V and [tex]B=\left\{u_1,u_2,\text{...},u_n\right\}[/tex] is a basis for U, then the set [tex]T(B)=\left\{T\left(u_1\right),T\left(u_2\right),\text{...} T\left(u_n\right)\right\}[/tex]

    a. spans V.
    b. spans U.
    c. is a basis for V.
    d. is linearly independent.
    e. spans the range of T.

    2. Relevant equations
    None


    3. The attempt at a solution

    It seems to me like all of these things are true (which is wrong of course). But in the example below, the result of the transformation does in fact meet all of the above criteria. What am I missing here?

    [tex]
    u_1=\left(
    \begin{array}{c}
    1 \\
    0 \\
    0
    \end{array}
    \right)
    [/tex]
    [tex]
    u_2=\left(
    \begin{array}{c}
    0 \\
    1 \\
    0
    \end{array}
    \right)
    [/tex]
    [tex]
    u_3=\left(
    \begin{array}{c}
    0 \\
    0 \\
    1
    \end{array}
    \right)
    [/tex]
    [tex]
    T(x,y,z)=(x,2y,3z)
    [/tex]
    [tex]
    T\left(u_1\right)=\left(
    \begin{array}{c}
    1 \\
    0 \\
    0
    \end{array}
    \right)
    [/tex]
    [tex]
    T\left(u_2\right)=\left(
    \begin{array}{c}
    0 \\
    2 \\
    0
    \end{array}
    \right)
    [/tex]
    [tex]
    T\left(u_3\right)=\left(
    \begin{array}{c}
    0 \\
    0 \\
    3
    \end{array}
    \right)
    [/tex]
     
  2. jcsd
  3. Jul 15, 2009 #2

    Mark44

    Staff: Mentor

    T(B) couldn't very well span U if U and V are different sized vector spaces, so that lets out b. And T(B) might or might not span V, depending on the dimension of the kernel of T. For example, if T(u) = 0, T(B) couldn't possibly span V, and the vectors in T(B) couldn't possibly be linearly independent, and therefore couldn't be a basis for V.
     
  4. Jul 15, 2009 #3
    Great point. Thinking of T(u) = 0 is very helpful. That basically eliminates everything except e doesn't it.

    Okay, I think I got it. The reason why e is the correct answer is because we know that the transformation is linear, correct?
     
  5. Jul 15, 2009 #4

    Mark44

    Staff: Mentor

    I think there's more to it than just that the transformation is linear. Seems like I remember a theorem that says that if T:U --> V is a linear transformation, and B = {u1, u2, ..., un} is a basis for U, then B = {T(u1), T(u2), ..., T(un)} spans the range of T.
     
  6. Jul 15, 2009 #5
    Okay. It makes sense, too.

    Thanks! I am posting one more question as well (a True/False this time). Maybe you could take a look?
     
  7. Jul 16, 2009 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Since [itex]\{u_1, u_2, \cdot\cdot\cdot, u_n\}[/itex] form a basis for U, any vector in U can be written as a linear combination [itex]u= a_1u_1+ a_2u_2+ \cdot\cdot\cdot a_nu_n[/itex]. And that means that any vector in the range of T is of the form [itex]T(u)= a_1T(u_1)+ a_2T(u_2)+ \cdot\cdot\cdot+ a_nT(u_n)[/itex]. That says that [itex]\{T(u_1), T(u_2), \cdot\cdot\cdot, T(u_n)\}[/itex] spans the range of T.
     
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