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Math Help - Linear operators and finding bases

  1. #1
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    Linear operators and finding bases

    Hi, I'm having trouble with a couple of questions.

    The quesiton reads

    A linear operator T on R^4 has the matrix
    A= [some 4x4 matrix]
    with respect to bases Alpha=Beta=E4.

    Find the basis for T(R^4).

    -I understand what to do when the question reads find T(1,2,-4,0) or something like that, I'm not sure what to do when it asks me to find a basis for all of R^4.

    If someone could please explain the steps on how to tackle this problem I would be appreciative of that.


    Secondly, the question later reads

    Find the basis of the kernel of T
    and
    Find the standard basis of the kernel of T.

    If anyone could clarify what is the difference between the two and how the calculation procedures are different I would also appreciate that.


    Thank you very much.
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  2. #2
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    Quote Originally Posted by Torcida1911 View Post
    Hi, I'm having trouble with a couple of questions.

    The quesiton reads

    A linear operator T on R^4 has the matrix
    A= [some 4x4 matrix]
    with respect to bases Alpha=Beta=E4.

    Find the basis for T(R^4).

    -I understand what to do when the question reads find T(1,2,-4,0) or something like that, I'm not sure what to do when it asks me to find a basis for all of R^4.
    Consider the following 4 vectors in R^4

    e_1 = (1,0,0,0), e_2 =(0,1,0,0), e_3=(0,0,1,0), e_4=(0,0,0,1)

    These vectors are linearly independent and span R^4. Thus the vectors form a basis of R^4 called the usual basis of R^4.

    If someone could please explain the steps on how to tackle this problem I would be appreciative of that.


    Secondly, the question later reads

    Find the basis of the kernel of T
    and
    Find the standard basis of the kernel of T.

    If anyone could clarify what is the difference between the two and how the calculation procedures are different I would also appreciate that.


    Thank you very much.
    The standard unit vectors e_1,e_2,...,e_n are linearly independent for if we write c_1e_1+...c_ne_n =0
    The usual or standar basis in component form then we obtain c_1...c_n = (0,...0) , which implies the coeficients are zeroes. Furthermore the vectors span R^n because an arbitrary vector x = (x_1,...,x_n) can be expressed as

    x=x_1(1,0,...0)+x_2(0,1,...,0)+...+x_n(0,0,...,1)=  x_1e_1+x_2e_2+...+x_ne_n

    We call e_1, e_2,...,e_n the usual or standard basis for R^n.

    Edit: I think in your problem to find a basis of the kernel of the map T you must set T(v)=0 where v=(x,y,z,t) and then set corresponding components equal to each other to form a homogeneous system whose solution space is the kernel of T.
    Last edited by Roam; February 19th 2010 at 11:34 PM.
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