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Check of a problem about nullspace

  1. Feb 4, 2017 #1
    1. The problem statement, all variables and given/known data
    Let ##V\subset \mathbb{R}^3## be the subspace generated by ##\{(1,1,0),(0,2,0)\}## and ##W=\{(x,y,z)\in\mathbb{R}^3|x-y=0\}##. Find a matrix ##A## associated to a linear map ##f:\mathbb{R}^3\rightarrow\mathbb{R}^3## through the standard basis such that its nullspace is ##V##.

    2. Relevant equations

    Nullspace definition

    3. The attempt at a solution

    Using the nullspace definition I get that ##f(0,2,0)=f(1,1,0)=(0,0,0)##. Thus,
    \begin{equation} f(0,2,0) =(0,0,0)\end{equation}
    \begin{equation}f(1,1,0) = (0,0,0) \end{equation}
    \begin{equation}f(0,0,1)=(1,1,1) \end{equation}
    Since the matrix has to be written wrt the standard basis of ##\mathbb{R}^3##, which is ##(1,0,0),(0,1,0),(0,0,1)##, I infer that
    \begin{equation} f(1,0,0)=f(1,1,0)-\frac{1}{2}f(0,2,0)=(0,0,0)-\frac{1}{2}(0,0,0)=(0,0,0) \end{equation}
    \begin{equation}f(0,1,0)=\frac{1}{2}f(0,2,0)=\frac{1}{2}(0,0,0)=(0,0,0)\end{equation}
    while I know from the text of the problem that ##f(0,0,1)=(1,1,1)##. In conclusion,
    \begin{equation} A=\begin{pmatrix} 0&0&1\\0&0&1\\0&0&1\end{pmatrix}\end{equation}
    If I look for a basis of the kernel I effectively get the two vectors ##(1,1,0),(0,2,0)##, so it should be ok. Is it correct?
     
  2. jcsd
  3. Feb 6, 2017 #2

    andrewkirk

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    Does this mean
    1. Let ##V\subset \mathbb{R}^3## be the subspace generated by ##\{(1,1,0),(0,2,0)\}## and let ##W=\{(x,y,z)\in\mathbb{R}^3|x-y=0\}##; or does it mean
    2. Let ##V\subset \mathbb{R}^3## be the subspace generated by ##\{(1,1,0),(0,2,0)\}\cup\{(x,y,z)\in\mathbb{R}^3|x-y=0\}##?
    If it means the second one then ##V=\mathbb R^3##, so ##f## must be the zero map. If it means the first one then ##W## is not used in the problem, which raises the question of why it has been defined.
     
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