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Some linear transformation

  1. Jan 16, 2008 #1
    Question 1
    Let T: P2 -> M22 be a linear transformation such that

    [tex]
    T(1+t)=\left[\begin{array}{cc}1&0\\0&0\end{array}
    \right];[/tex]
    [tex]
    T(t+t^{2})=\left[\begin{array}{cc}0&1\\1&0\end{array}
    \right];[/tex]
    [tex]
    T(1+t^{2})=\left[\begin{array}{cc}0&1\\0&1\end{array}
    \right];[/tex]
    Then find[tex] T(1),T(t),T(t^{2})[/tex]

    My attempt
    All I know is that [tex] 1,t,t^{2}[/tex] are basis of P2, what do I do next?
    How do I find them from given matrices?


    Question 2

    let dim(v)=n and dim(W)=m and P:V->W be a linear transformation, i.e P(v)=0 for all v in V. Show that the matrix of P with respect to any bases for V and W is the mxn zero matrix.
    My attempt
    Let S be a basis of V S={v1,v2,.........vn}
    Let v a vector in v
    [tex]v=c1v1+c2v2+ ....cnvn[/tex]

    [tex]P(v)=c1w1+c2w2+ ............+cnwm=0[/tex]
    Because vectors of S are linearly independant c1,c2 ...... cn are all 0
    So the resultant matrix of P is a zero matrix



    Question 3

    Let L:V->W be a linear transformation. show that L is one to one if and only if dim(range L)=dim(V)
    My attempt:
    We know that dim(V)=dim(range L)+dim(ker L) (1)
    if dim(V)>dim(range L) then dim(ker L) is not 0 and L is not One to one
    if dim(V)=dim(range L) then dim(V)-dim(range L) = dim(ker L)
    and dim(ker L)=0 hence L is one to one.
     
  2. jcsd
  3. Jan 16, 2008 #2

    siddharth

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    Homework Helper
    Gold Member

    Lets take one at a time. For the first one, notice that T is a linear transformation. What does that imply?
     
    Last edited: Jan 16, 2008
  4. Jan 16, 2008 #3

    HallsofIvy

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    Staff Emeritus
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    You just stated [itex]T(1), T(2), T(t^2)[/tex]!? Don't you mean "find T(p) where p is any member of P2"? As Siddharth said, T is linear. Any member of P2 can be written at2+ bt+ c. What is T(at2+ at+ b)?


    [/quote]My attempt
    All I know is that [tex] 1,t,t^{2}[/tex] are basis of P2, what do I do next?
    How do I find them from given matrices?


    Question 2

    let dim(v)=n and dim(W)=m and P:V->W be a linear transformation, i.e P(v)=0 for all v in V. Show that the matrix of P with respect to any bases for V and W is the mxn zero matrix.
    My attempt
    Let S be a basis of V S={v1,v2,.........vn}
    Let v a vector in v
    [tex]v=c1v1+c2v2+ ....cnvn[/tex]

    [tex]P(v)=c1w1+c2w2+ ............+cnwm=0[/tex]
    Because vectors of S are linearly independant c1,c2 ...... cn are all 0
    So the resultant matrix of P is a zero matrix[/quote]
    Your final equaiton, P(v)= c1w1+ c2w2+ ...+ cnwn= 0, is in W- it says NOTHING about "the vectors of S". If it were true that "c1, c2, ..., cn are all 0", then v would be the 0 vector- and that is not, in general true. Remember that you can write a linear transformation, L:V->W, in given bases for V and W by applying L to each basis vector in V in turn, then writing the result in the basis in W. The coefficients then form a column for the matrix. If {v2, v2, ..., vn} is a basis for V, what is P(v1)? What is P(v2)?


     
    Last edited: Jan 16, 2008
  5. Jan 16, 2008 #4
    Question 1 is formulated correctly. They want [tex]T(1);T(t);T(t^{2})[/tex].
    Some one has suggested that :
    [tex]T(1)=1/2( T(1+t) - T(t+t^{2}) + T(1+t^{2}) ))[/tex]
    becaused T being a linear transformation when the RHS expression is developed it yields [tex]T(1)[/tex]. The problem is solved by replacing T(...) in the RHS expression by their given matrices.
    In the same way we can find [tex] T(t) [/tex] and [tex] T(t^{2})[/tex]
     
  6. Jan 16, 2008 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    My mistake. I misread. You are NOT given T(1), T(t), and T(t2) as I thought. You are given T(1+ t), T(1+ t2) and T(t+ t2).

    Yes, that would work, although I would be inclined to wonder HOW you noticed that
    [tex]T(1)=1/2( T(1+t) - T(t+t^{2}) + T(1+t^{2}) ))[/tex]!

    Siddharth's original suggestion was to use linearity to say that
    [tex]1: T(1+t)= T(1)+ T(t)= \left[\begin{array}{cc}1&0\\0&0\end{array}\right];[/tex]
    [tex]2: T(t+t^{2})= T(t)+ T(t^2)= \left[\begin{array}{cc}0&1\\1&0\end{array}\right];[/tex]
    [tex]3:T(1+t^{2})=T(1)+ T(t^2)= \left[\begin{array}{cc}0&1\\0&1\end{array}\right];[/tex]
    Now treat those as three equations in the three unknown matrices, T(1), T(t), T(t2). For example, adding (1) and (3) gives the equation 2T(1)+ T(t)+ T(t2)= a matrix. Subtracting (2) from that gives 2T(1)= a matrix, giving the equation you have. You can similarly solve for T(t) and T(t2).
     
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