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Thread: Linear maps

  1. #1
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    Linear maps

    Consider the linear map defined on $\displaystyle R^4$ with values in $\displaystyle R^3$ given by the matrix:
    A =
    ( 1 3 1 -2 )
    ( 3 9 4 -4 ) yes, this is supposed to be a matrix, i'm new
    (-1 -3 -2 0 )
    a) find all x e $\displaystyle R^4$ satisfying the linear equation Ax=b, with
    b =
    ( 1)
    (-2)
    ( 4)

    b) find a basis for ker A

    c) Using the Rank Theorem, find the dimension of im A

    This is a past paper without solutions and would really help me revise, thank you very much !
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  2. #2
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    Well, I don't know if there is an easier way than this, but I suspect
    there is:

    a) If you multiply Ax=b then you will get three equations for which
    3 of the variables in x can be eliminated leaving you with

    $\displaystyle \begin{aligned}
    x & =6+2t\\
    y & =\frac{2}{3}t\\
    z & =-5-2t\\
    t & =t\end{aligned}$


    so $\displaystyle \begin{aligned}
    \mathbf{x} & =\left(\begin{array}{c}
    6+2t\\
    \frac{2}{3}t\\
    -5-2t\\
    t\end{array}\right)\end{aligned}$


    b) the kernel of A can be found in a similar way Ax=0

    $\displaystyle \begin{aligned}
    \mathbf{ker(A)} & =\left(\begin{array}{r}
    0\\
    \frac{4}{3}t\\
    -2t\\
    t\end{array}\right)=\frac{t}{3}\left(\begin{array} {r}
    0\\
    4\\
    -6\\
    3\end{array}\right)\end{aligned}$


    which is a straight line through the origin,

    so a basis could be $\displaystyle $\left(\begin{array}{r}
    0\\
    4\\
    -6\\
    3\end{array}\right)$$which has dimension 1, ie n(A)=1

    c) I'm not sure what the rank theorem is but r(A)+n(A)=dimX means
    that r(A)+1=4 so n(A)=3. But I thought that was kind of obvious anyway,
    so maybe I misunderstood the question.
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  3. #3
    MHF Contributor

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    The "rank theorem" does, indeed, say that, if A is a linear operator from vector space U to vector space V, then the rank of A plus the nullity (dimension of the kernel) of A is equal to the dimension of U. And you have a typo:
    r(A)+ n(A)= 4 and n(A)= 1 so r(A)= 3, not "n(A)= 3". That says that A maps $\displaystyle R^4$ onto $\displaystyle R^3$ which was not entirely "obvious". The rank of A, the dimension of its image in $\displaystyle R^3$ must be less than or equal to 3 but certainly might have been less.
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  4. #4
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    Ah, I see, thanks for clearing that one up HallsofIvy.
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