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Finding Kernel and Image of Matrix transformation

  1. Oct 16, 2011 #1
    1. The problem statement, all variables and given/known data

    Matrix A =
    0 1 0
    0 0 1
    12 8 -1

    Let E1 = a(A)(A+2I)2
    Let E2 = b(A)(A-3I)

    For each of these, calculate the image and the kernel

    2. Relevant equations

    I found a(A) to be 1/25
    and b(A) to be 1/25*(A-7I)
    Also, if I am not mistaken, I think KernelE1 = ImageE2 and vice versa

    Matrix E1 =
    4 4 1
    12 12 3
    36 36 9

    Matrix E2 =
    21 -10 1
    12 29 -11
    -132 -76 40


    3. The attempt at a solution

    Um... well if v1, v2, v3 are the column vectors of E1 respectively, and w1, w2, w3 are those of E2, isn't {w1, w2, w3}α the Kernel of E1 and Image of E2 (and the other way around)???

    Part two says to find a new basis such that the linear transformation corresponding to A is represented by
    -2 1 0
    0 -2 0
    0 0 3

    Where do I even begin this one?

    PS: Is there a way to add matrices on this forum? It's a little messy this way.
     
  2. jcsd
  3. Oct 17, 2011 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    To write matrices or other math formulas, use LaTeX. There are a number of tutorials on the internet. On this board you begin LaTeX code by [ tex ] (without the spaces) and end with [ \tex ] (without the spaces). You can do "in line" LaTeX with [ itex ] and [ /itex ]. To do a matrix use \ begin{bmatrix} and \ end{bmatrix} (without the spaces) for a matrix with braces:
    [tex]\begin{bmatrix}4 & 4 & 1 \\ 12 & 12 & 3\\ 36 & 36 & 9\end{bmatrix}[/tex]
    use "pmatrix" to get the same thing with ( ):
    [tex]\begin{pmatrix}4 & 4 & 1 \\ 12 & 12 & 3\\ 36 & 36 & 9\end{pmatrix}[/tex]

    No, it's not. That is [itex](A+ 2I)^2[/itex]. Isn't [itex]E_1= A(A+ 2I)^2[/itex]? (I don't know what that "a" in the formula was supposed to be.)

    If I have done the calculation correctly,
    [tex]E_1= A(A+ 2I)^2= \begin{bmatrix}12 & 12 & 3 \\ 36 & 36 & 9\\ 108 & 108 & 27\end{bmatrix}[/tex]
    the kernel of [itex]E_1[/itex] would be the set of all (x, y, z) such that
    [tex]\begin{bmatrix}12 & 12 3 \\ 36 & 36 & 9\\ 108 & 108 & 27\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}12x+ 12y+ 3z \\ 36x+ 36y+ 9z \\ 108x+ 108y+ 27z\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0\end{bmatrix}[/tex]
    which gives the three equations 12x+ 12y+ 3z= 0, 36x+ 36y+ 9z= 0, 108x+ 108y+ 27z= 0 which are all multiples of one another. In fact, they all reduce to 4x+ 4y+ z= 0 or z= -4x- 4y. That means that any vector in the kernel can be written as
    [tex]\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}x \\ y \\ -4x- 4y\end{bmatrix}= x\begin{bmatrix}1 \\ 0 \\ -4\end{bmatrix}+ \begin{bmatrix}0 \\ 1 \\ -4\end{bmatrix}[/tex]
    so the kernel is the two-dimensional subspace spanned by those two vectors.

    The Image of [itex]E_1[/itex] is the set of all (x, y, z) such that
    [tex]\begin{bmatrix}12 & 12 3 \\ 36 & 36 & 9\\ 108 & 108 & 27\end{bmatrix}\begin{bmatrix}a \\ b\\ c \end{bmatrix}= \begin{bmatrix}12a+ 12b+ 3c \\ 36a+ 36y+ 9z \\ 108x+ 108y+ 27z\end{bmatrix}= \begin{bmatrix}x \\ y \\ z\end{bmatrix}[/tex]
    for some (a, b, c). That gives the three equations 12a+ 12b+ 3c= x, 36b+ 36c+ 9z= y, and 108a+ 108b+ 27c= z. If we subtract 3 times the first equation from the second we get y- 3x= 0. If we subtract 3 times the second equation from the third, we get z- 3y= 0. That is, z= 3y= 3(3x)= 9x. Any vector in the Image of [itex]E_1[/itex] must be of the form
    [tex]\begin{bmatrix}x \\ y \\ z \end{bmatrix}= \begin{bmatrix} x \\ 3x \\ 9x\end{bmatrix}= x\begin{bmatrix}1 \\ 3 \\ 9\end{bmatrix}[/tex]

    However, that may have nothing to do with your problem because it is not at all clear how you intended to define "[itex]E_1[/itex]".
     
    Last edited: Oct 17, 2011
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