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Math Help - Linear maps

  1. #1
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    Linear maps

    Consider the linear map defined on R^4 with values in 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 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

    \begin{aligned}<br />
x & =6+2t\\<br />
y & =\frac{2}{3}t\\<br />
z & =-5-2t\\<br />
t & =t\end{aligned}


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


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

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


    which is a straight line through the origin,

    so a basis could be $\left(\begin{array}{r}<br />
0\\<br />
4\\<br />
-6\\<br />
3\end{array}\right)$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 R^4 onto R^3 which was not entirely "obvious". The rank of A, the dimension of its image in 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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