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Write a matrix given the null space

  1. Jan 30, 2017 #1
    1. The problem statement, all variables and given/known data

    Build the matrix A associated with a linear transformation ƒ:3→ℝ3 that has the line x-4y=z=0 as its kernel.

    2. Relevant equations

    I don't see any relevant equation to be specified here .

    3. The attempt at a solution

    First of all, I tried to find a basis for the null space by solving the homogeneous linear system:
    \begin{equation}
    \begin{cases} x-4y=0\\z=0\end{cases} \Leftrightarrow \begin{cases} x=4y\\z=0\end{cases}
    \end{equation}
    The solutions are thus of the kind (4y,y,0)=y(4,1,0). A basis is hence given by the single vector (4,1,0).
    Due to the rank-nullity theorem I know that dim(Im) = dim(ℝ3)-dim(ker(f))=3-1=2. Then, I need two more vectors in order to obtain a basis of 3 that is composed by (4,1,0).
    If I chose two vector such as that the matrix of the coordinates has the wanted kernel, I've concluded. My problem is that I don't see how to consciously pick them.
    Can anyone help me? Thank you very much!
     
  2. jcsd
  3. Jan 30, 2017 #2

    Mark44

    Staff: Mentor

    There is another basis vector for the kernel: <0, 0, 1>
     
  4. Jan 31, 2017 #3
    Sorry but I'm not convinced.
    The basis of the kernel is found by solving the system I wrote, corresponding to the matrix
    \begin{equation} \begin{pmatrix} 1&-4&0\\0&0&1 \end{pmatrix} \end{equation}
    This system ha ##\infty^1## solutions of the type ##(4y,y,0)##, i.e. a basis is ##(4,1,0)## and the kernel is one-dimensional.
     
    Last edited: Jan 31, 2017
  5. Jan 31, 2017 #4

    fresh_42

    Staff: Mentor

    Why don't you use ##f(\vec{x})=A\vec{x} = \vec{0}\,##?
     
  6. Jan 31, 2017 #5
    Let's see: knowing the basis vectors I can write
    \begin{equation}
    \begin{pmatrix}4&a&b\\1&c&d\\0&e&f\end{pmatrix} \begin{pmatrix}1\\-4\\0\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}
    \end{equation}
    Thus I get
    \begin{equation}
    \begin{cases}4-4a+0b=0\\1-4c+0d=0\\0-4e+0f=0\end{cases}
    \end{equation}
    from which ##a=1, c=\frac{1}{4}, e=0##. Repeating the procedure with
    \begin{equation}
    \begin{pmatrix}4&a&b\\1&c&d\\0&e&f\end{pmatrix} \begin{pmatrix}0\\0\\1\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}
    \end{equation}
    I get
    \begin{equation}
    \begin{cases}0+0a+b=0\\0+0c+d=0\\0+0e+f=0\end{cases}
    \end{equation}
    from which ##b=0, d=0, f=0##.
    Concluding, I get the matrix
    \begin{equation}
    \begin{pmatrix}4&1&0\\1&\frac{1}{4}&0\\0&0&0\end{pmatrix}
    \end{equation}
    By doing so, though, the kernel becomes 2-dimensional with ##(0,0,1),(-\frac{1}{4},1,0)## as basis.
     
    Last edited: Jan 31, 2017
  7. Jan 31, 2017 #6

    fresh_42

    Staff: Mentor

    Of course. You have only one linear independent vector to span the image, but you need two. The matrix ##A'## of the linear equations in post #3 already had two (row) vectors to span the image. Why don't you expand it with a third row? What does ##A' x = 0## tell you? I mean, reread your own comment.
     
  8. Jan 31, 2017 #7
    First of all, thank you for your patience.
    If I add a vector linearly dependent to the other two I get a matrix which kernel is ##(4,1,0)##. For example
    \begin{equation}
    \begin{pmatrix}1&-4&0\\0&0&1\\0&0&2\end{pmatrix}
    \end{equation}
    Then this is a solution for the excercise? As you might guess I'm not very sure about the conceptual meaning of what I'm doing...
     
  9. Jan 31, 2017 #8

    fresh_42

    Staff: Mentor

    You get a matrix which kernel is ##\{(x,y,z)\,\vert \,(1,-4,0)\cdot \vec{x} = 0 \wedge (0,0,1) \cdot \vec{x} = 0 \}##, i.e. the linear equations you started with. I would have chosen ##0## as the last entry, but ##2## is o.k., too. The first two row vectors are now perpendicular to ##(4,1,0)##, which you were looking for in post #1, and also perpendicular to each other. Therefore together they build a nice little basis of orthogonal vectors in ##\mathbb{R}^3##: two in the image of ##f## and one in its kernel.
     
  10. Jan 31, 2017 #9
    Thank you again Sir!
    Just to be sure: ##(1,-4,0),(0,0,1)## is a orthogonal basis of the ##Im(f)## subspace, ##(4,1,0)## is the basis for the one-dimensional nullspace of ##f## and so, by virtue of the rank-nullity theorem, I have an orthogonal basis of the domain of ##f##, i.e. ##\mathbb{R}^3##.
     
  11. Jan 31, 2017 #10

    fresh_42

    Staff: Mentor

  12. Feb 2, 2017 #11

    pasmith

    User Avatar
    Homework Helper

    You don't need to.

    A linear transformation which does the trick is [tex]\mathbf{x} \mapsto \mathbf{x} \times (4, 1, 0)^{T}.[/tex] This can be written as [itex]\mathbf{x} \mapsto A\mathbf{x}[/itex] for some matrix [itex]A[/itex].
     
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