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Homework Help: Linear transformation ranges

  1. Apr 26, 2010 #1
    L: R^3 -> R^3 is a linear transformation defined by L(v) =A(v)
    A is given as -1 2 0 and w= 1
    1 1 1 2
    2 -1 1 -1

    is w in the range of L?

    My understanding is that if a vector exists such that the product of A and this vector = W then it is in the range. I then have the following equations:

    -X +2Y = 1
    X+Y+Z= 2
    2X -Y+Z= -1

    no solutions exists, so w is not in the range. Can someone confirm this or offer any other insight?

  2. jcsd
  3. Apr 27, 2010 #2


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    bit hard to read is A given as:
    [tex] A = \begin{bmatrix} -1 & 2 & 0\\ 1 & 1 & 1 \\ 2 & -1 & 1 \end{bmatrix} [/tex]
    Last edited: Apr 27, 2010
  4. Apr 27, 2010 #3


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    what you've done sounds reasonable, though i haven't checked the math,

    note so what you are trying to solve is:
    [tex] A \textbf{x} = \textbf{x} [/tex]

    If A was invertible a solution must exist given by:
    [tex] \textbf{x} = A^{-1} \textbf{x} [/tex]
    so you could check that [itex] det(A) = 0[/itex]

    Now, consider the action of A on a unit vector:
    [tex] \textbf{e}_1 = (1,0,0)^T [/tex]
    [tex] A\textbf{e}_1 = \textbf{a}_1 [/tex]
    where [itex] \textbf{a}_1 [/itex] is the first column vector of [itex] A [/itex]

    Now every vector can be wirtten in terms of the basis of unit vectors, so the range of A is in fact the column space of A. So in effect you are checking w is not in the column space of A.
  5. Apr 27, 2010 #4


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    Internet readers do not respect spaces so better to use LaTex.

    [tex]A= \begin{bmatrix}-1 & 2 & 0 \\ 1 & 1 & 1 \\ 2 & - 1 & 1\end{bmatrix}[/tex]
    [tex]w= \begin{bmatrix} 1 \\ 2 \\ -1\end{bmatrix}[/tex]

    is w in the range of A?
    That will be true if and only if there exist x, y, z such that
    [tex]\begin{bmatrix}-1 & 2 & 0 \\ 1 & 1 & 1 \\ 2 & - 1 & 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix} 1 \\ 2 \\ -1\end{bmatrix}[/itex]
    (Click on those to see the code.)

    That can be written as the three equations, -x+ 2y= 1, x+ y+ z= 2, 2x- y+ z= -1.

    IF those three equations have a solution (not necessarily unique) then w is in the range of A.

    It might be simpler to check the determinant of A. If the determinant is not 0, the range of A is all of [itex]R^3[/itex] and so any w is in it.

    If the determinant of a is 0, then you would have to check if w is in the "columns space". In that case, it is probaly simpler to solve the three equations above.
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