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Linear Algebra help

  1. Apr 12, 2012 #1
    Let f : P3 → M2x2 be given by f(a + bx + cx^2 + dx^3) =
    (a + d 0)
    (0 b − c)

    1. Determine the nullspace and nullity of f and specify a basis for the nullspace.
    -I came up with N(h)= {a+bx+cx^2+dx^3/ a+d=0 & b-c=0}=
    ={( 0 b), a,d are elements of R}
    ( c 0)
    2. Determine the rangespace and rank of f and specify a basis for the rangespace.
    -R(h)= ( a+d 0) are elements of matrix 2x2/ a,b are elements of R}
    ( 0 b-c)
    3. Is f one-to-one? Is f onto? Why or why not?
    -f is not one to one because the vectors share partners in the m2x2. f is onto because it spans from the same dimensions.

    If someone could show some guidance it would be appreciated. I don't think I have the range space and null space correct.
     
  2. jcsd
  3. Apr 12, 2012 #2

    Dick

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    Homework Helper

    N(h)= {a+bx+cx^2+dx^3/ a+d=0 & b-c=0} is correct for the nullspace. You could also write that as a*(1-x^3)+b*(x+x^2) for any a and b. It's not [[0,b],[c,0]]. That part is wrong. The nullspace is a set of polynomials, not a set of matrices. Let's start with that.
     
  4. Apr 13, 2012 #3

    HallsofIvy

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    This is correct.
    This is not. A set of polynomials is not a set of matrices!
    But it would be better to write the answer as N(h)= {a+ bx+ bx^2- ax^3}= {a(1- x^3)+ b(x+ x^2)}.
    Also, you haven't said what the nullity is.

    All you have done is copy the definition of the transformation. You haven't specified a basis or the dimension. Start by noting that
    [tex]\begin{bmatrix}a+ d & 0\\ 0 & b- c\end{bmatrix}= a\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}+ b\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}+ c\begin{bmatrix}0 & 0 \\ 0 & -1\end{bmatrix}+ d\begin{bmatrix}1 & 0\\ 0 & 0\end{bmatrix}[/tex]
    How many independent matrices are used there?

    What vectors are you talking about? And what do you mean by "share partners". Actually, it is sufficient to observe that the nullity is not 0. Finally, the set of cubic polynomials and the set of 2 by 2 matrices both have dimension 4. If the transformation is not one to one, it can't be "onto".

     
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