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Finding a basis for the null space and range of a matrix

  1. Sep 14, 2013 #1
    1. The problem statement, all variables and given/known data
    ##S## is a linear transformation and ##\{u_{1},u_{2}\}## is a basis for the vector space.
    $$
    S(u_{1})=u_{1}+u_{2}\\
    S(u_{2})=-u_{1}-u_{2}
    $$
    I would like to find a basis of the null space and range of ##S##.


    2. Relevant equations
    In my text, it says that the proper matrix representation of ##S## is
    $$
    \left(\begin{array}{cc}
    1 & -1 \\
    1 & -1 \\
    \end{array}\right).
    $$


    3. The attempt at a solution
    I understand that this way of representing S with a matrix will preserve equality when we compose two linear transformations in matrix from but it is confusing me. I found a supposed basis for the null space as follows. I first put the matrix in echelon form:
    $$
    \left(\begin{array}{cc}
    1 & -1 \\
    1 & -1 \\
    \end{array}\right) \rightarrow \left(\begin{array}{cc}
    1 & -1 \\
    0 & 0 \\
    \end{array}\right).
    $$
    Then we are left with the equation ##x_{1}-x_{2}=0## so any vector in the null space has the form ##(\lambda,\lambda)## where ##\lambda## is an arbitrary number. From the definition of the null space we should have that ##S\cdot(\lambda,\lambda)=0## and this holds when I put ##(\lambda,\lambda)## into column form and multiply it by the matrix of ##S## but when I try to plug ##(\lambda,\lambda)## into ##S(u_{1},u_{2})=(u_{1}+u_{2},-u_{1}-u_{2})##, things don't seem to work out anymore. Did I do something wrong when finding the basis or am I thinking about the null space in the wrong way?

    I would also like help on finding a basis for the range. I know that it is the same as a basis for the column space but again, do I use the columns in the matrix of ##S## as I have defined it above?
     
    Last edited: Sep 14, 2013
  2. jcsd
  3. Sep 14, 2013 #2

    Dick

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    Science Advisor
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    The book's statement of the matrix for S is clearly wrong. What do you think it should be?
     
  4. Sep 14, 2013 #3
    I thought that the matrix of ##S## should be presented by
    $$
    \left(\begin{array}{cc}
    1 & 1 \\
    -1 & -1 \\
    \end{array}\right)
    $$
    but the texts main reason behind why this was wrong was because say we had another linear transformation T defined by the equations
    $$
    T(u_{1})=2u_{1}+u_{2}\\
    T(u_{2})=u_{1}-u_{2}
    $$
    then we will have the matrix representation for ##T## as
    $$
    \left(\begin{array}{cc}
    2 & 1 \\
    1 & -1 \\
    \end{array}\right)
    $$.
    Then the linear transformation ST would be defined as
    $$
    ST(u_{1})=S(T(u_{1}))=S(2u_{1}+u_{2})=2(u_{1}+u_{2})-u_{1}-u_{2}=u_{1}+u_{2}\\
    ST(u_{2})=S(T(u_{2}))=S(u_{1}-u_{2})=u_{1}+u_{2}-(-u_{1}-u_{2})=2u_{1}+2u_{2}
    $$
    which would then have a matrix representation of
    $$
    \left(\begin{array}{cc}
    1 & 1 \\
    2 & 2 \\
    \end{array}\right).
    $$
    We should get the same matrix if we multiplied the matrix of ##S## and ##T##:
    $$
    \left(\begin{array}{cc}
    1 & 1 \\
    -1 & -1 \\
    \end{array}\right)\left(\begin{array}{cc}
    2 & 1 \\
    1 & -1 \\
    \end{array}\right)=\left(\begin{array}{cc}
    3 & 0 \\
    -3 & 0 \\
    \end{array}\right).
    $$
    On the other hand, if we represented the matrices the way the book wants them we have the matrix for ##S## as
    $$
    \left(\begin{array}{cc}
    1 & -1 \\
    1 & -1 \\
    \end{array}\right)
    $$
    and the matrix for ##T## as
    $$
    \left(\begin{array}{cc}
    2 & 1 \\
    1 & -1 \\
    \end{array}\right).
    $$
    Multiplying these gives us
    $$
    \left(\begin{array}{cc}
    1 & -1 \\
    1 & -1 \\
    \end{array}\right)\left(\begin{array}{cc}
    2 & 1 \\
    1 & -1 \\
    \end{array}\right)=\left(\begin{array}{cc}
    1 & 2 \\
    1 & 2 \\
    \end{array}\right)
    $$
    which would be our composed transformations represented as a matrix the way the book wants it.
     
  5. Sep 14, 2013 #4
    First, I'm willing to work with [tex]S = \begin{pmatrix}
    1 & -1\\
    1 & -1

    \end{pmatrix}[/tex]

    But I cannot make any sense out of S(u1) = u1 + u2S(u2), because the first term is a vector and the second term is a scalar. There may be a typo here. So let's work with the matrix.

    You already concluded that the null space are all the vectors of the form (a,a). Another way of writing that is a(1,1) and (1,1) is the basis of your nullspace. Observe that this is a 1 dimensional space, which matches up nicely with the fact that your echelon form has just one zero row.

    Re the basis for the range the easiest approach in this case is to notice that the range and nullspace taken together are the entire space, which is two dimensional. The range is also a one dimensional space, so any vector whatsoever that is not in the nullspace is a basis for the range.

    The column space is clear in the echelon form of the matrix. It consists of two vectors, (1,0) and (-1,0) which you can see are not independent since the first is just -1 times the second. You can use either of these vectors as your basis for the column space, and they certainly will do as the basis for the range.

    So let u1 = (1,1). From the matrix definition S(u1) = (1-1, 1-1) = (0,0) which is what you should be getting since (1,1,) is a basis for the nullspace. We took u2 = (1,0) so S(u2) = (1,0).
    As long as S(u2) is in the range and not the null space -- as is evidently the case -- I'm happy.
     
  6. Sep 14, 2013 #5
    I fixed the typo. Sorry about that.

    I understand everything but this part. In particular, how do you have S(u1) = (1-1, 1-1)?
     
  7. Sep 14, 2013 #6

    vela

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    This is not true. If you take an arbitrary vector (x,y) and multiply it by the matrix of S, you get
    $$\begin{pmatrix} 1 & -1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x-y \\ x-y \end{pmatrix}$$ You should be able to see by inspection what a basis for the range is.
     
  8. Sep 15, 2013 #7
    Yes, you are right. A basis for the range is also (1,1). The (1,0) or any vector independent of (1,1) is the basis for the subspace of vectors such that [tex] Sx \ne 0[/tex] i.e. those that are not in the nullspace and will thus get mapped into the range. Sorry I mixed them up.
     
  9. Sep 15, 2013 #8
    I am little confused. Is the textbooks way of representing ##S## correct?
     
  10. Sep 15, 2013 #9

    vela

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    Staff Emeritus
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    Yes, it is. You can find the columns of the matrix of a transformation by applying it to the basis vectors. In this case, you had
    $$S(\vec{u}_1) = \vec{u}_1 + \vec{u}_2.$$ In matrix notation, this would correspond to the equation
    $$S\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}.$$ If you don't already see it, you should convince yourself that multiplying a matrix by the vector (1,0) picks off the first column of the matrix. So this says the first column is (1,1).
     
  11. Sep 15, 2013 #10
    Ahh! Ok. That makes sense.
     
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