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Image and kernel of a matrix

  1. Feb 24, 2015 #1
    1. The problem statement, all variables and given/known data
    I know this stuff isnt complicated but the definitions of my book are very formal and confusing. I have to find the image and the kernel of these two matrices:

    [itex]
    A= \begin{bmatrix}1&2&3\end{bmatrix}
    B= \begin{bmatrix}2&3\\6&9\end{bmatrix} [/itex]


    3. The attempt at a solution
    my book says the image is the span of the column vectors so for A would that be:

    [itex] im(A) = x_1[1] + x_2[2] + x_3[3] [/itex] ???

    and since 3 = 1 + 2 I think that makes 3 redundant so it would be x1[1] + x2[2]?
    and the kernel is when the system = 0 so when
    x1 = -2x2 - 3x3
    if x2 = t and x3 = r then
    [itex] ker(A) = [-2t-3r] ?? [/itex]

    for B
    [itex]im(B) = x_1 \begin{bmatrix}2\\6\end{bmatrix} + x_2 \begin{bmatrix}3\\9\end{bmatrix} [/itex]
    [itex] ker(B) = \begin{bmatrix}2&3&|0\\6&9&|0\end{bmatrix} = \begin{bmatrix}1&3/2&|0\\6&9&|0\end{bmatrix}
    =
    \begin{bmatrix}1&3/2&|0\\0&0&|0\end{bmatrix} [/itex]

    so x1 = -3/2 x2
    if x2 = t then
    [itex] ker(B) = [-3/2t] [/itex]??

    am i doing these right? I would appreciate if someone can help explain these concepts to me without using the formal abstract definitions
     
    Last edited: Feb 24, 2015
  2. jcsd
  3. Feb 24, 2015 #2
    man this is messy. Is there at tutorial for making matrices in LatEx?
     
  4. Feb 24, 2015 #3

    Mark44

    Staff: Mentor

    https://www.physicsforums.com/help/latexhelp/ -- towards the end.
    Here's how I do them.
    \begin{bmatrix} 2 & 3 \\ 6 & 9 \end{bmatrix}

    Remove the spaces and this is how the above is rendered:
    $$\begin{bmatrix} 2 & 3 \\ 6 & 9 \end{bmatrix}$$
     
  5. Feb 24, 2015 #4

    BiGyElLoWhAt

    User Avatar
    Gold Member

    here, this:
    \left [ \begin{array}{c}
    1 & 2 \\
    3 & 4\\
    \end{array}\right ]
    renders as this:
    ## \left [ \begin{array}{c}
    1 & 2 \\
    3 & 4\\
    \end{array}\right ] ##
    You can replace the \left [ with \left ( or \left | (for determinants) or I think...
    ##\left \{ \right \}##
    yup, you can use \left \{ as well, you need that \ in front of the { (as well as for the right) because the symbals { } are special for latex. This tells it that it's just another character.
    Apparently the matrix components render even when you don't have Latex enabled. It's 1 (ampersand) 2 (ampersand) and double slash (\\) to end the line.

    Also, 1 + 2 does in fact equal 3. However, 1*x + 2*y does not in fact equal 3z, so the 3 is not redundant.
     
  6. Feb 24, 2015 #5

    BiGyElLoWhAt

    User Avatar
    Gold Member

    Ahhh, this mark guy, always stealing my thunder. :wink:
     
  7. Feb 24, 2015 #6

    Mark44

    Staff: Mentor

    Is A a row matrix? If so, the columns span a one-dimensional space.
    2 is a multiple of 1, and 3 is a multiple of 1 as well.
    Your matrix, if I understand what you wrote, can be considered to be a mapping from R3 to R1. It turns out that ker(A) is two-dimensional, so you need two vectors for a basis.
     
  8. Feb 24, 2015 #7
    ok thank you guys i cleaned it up a bit. how did i do?
     
  9. Feb 24, 2015 #8

    Mark44

    Staff: Mentor

    Lots better.

    For B, note that the two vectors you show are linearly dependent. The second vector is a scalar multiple of the first (or vice-versa).
     
  10. Feb 24, 2015 #9
    are my image and kernel for A correct? B the the vector
    [itex] \begin{bmatrix}3\\9\end{bmatrix} [/itex] is a mutliple of [itex] \begin{bmatrix}2\\6\end{bmatrix} [/itex]
    if you multiply it by k = 2/3 correct? so do I only include
    [itex]\begin{bmatrix}2\\6\end{bmatrix} [/itex] ??
    i only vaguely understand why I am doing the things I am doing
     
  11. Feb 24, 2015 #10
    oops i mean k = 3/2
     
  12. Feb 24, 2015 #11

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    If A is a linear transformation from vector space U to vector space V, then the "image" of A is the set of vectors, y, in V such that Ax= y for some vector x in U and the "kernel" is the set of all vectors, u, in U such that Au= 0.

    It is still not clear to me what "A" is here. You talk about two matrices, label one of them "A" but do not label the other so it is hard to understand. I think you mean that "A" is the row matrix, [itex]\begin{bmatrix}1 & 2 & 3\end{bmatrix}[/itex] which takes vectors in R3 to vectors in R1. If the vector in R3 is u= <x , y, z> then Au= x+ 2y+ 3z. For u to be in the kernel of A, we would have to have x+ 2y+ 3z= 0. What vectors, <x, y, z>, have that property? The image of A is the set of all "vectors" in R1 (i.e. single numbers), a, such that a= x+ 2y+ 3z for some numbers, x, y, and z.

    The kernel of the two by two matrix is <x, y> such that [itex]\begin{bmatrix}2 & 3 \\ 6 & 9\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}2x+ 3y \\ 6x+ 9y\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}[/itex]. So we must have 2x+ 3y= 0 and 6x+ 9y= 0. What x and y satisfy that?

    The image is the subspace of all vectors <a, b> such that [itex]\begin{bmatrix}2 & 3 \\ 6 & 9 \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}2x+ 3y \\ 6x+ 9y\end{bmatrix}= \begin{bmatrix}a \\ b\end{bmatrix}[/itex]. What can you say about a and b if 2x+ 3y= a and 6x+ 9y= b for some x and y?
     
  13. Feb 24, 2015 #12

    BiGyElLoWhAt

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    Gold Member

    When in doubt, Wikipedia.
     
  14. Feb 24, 2015 #13
    A is the row matrix [1, 2, 3] and the other matrix is the 2x2 matrix that I labeled B. wikipedia isnt helping because the formal definitions are just confusing me. let me know If I understand correctly so far
    the im(B) is
    [itex] = im(B) = \begin{bmatrix}y_1\\y_2\end{bmatrix} =
    \begin{bmatrix}2&3\\6&9\end{bmatrix} [/itex]
    so
    [itex] \begin{bmatrix}2&3&|y_1\\6&9&|y_2\end{bmatrix} = \begin{bmatrix}1&3/2&|y_1/2\\6&9&|y_2\end{bmatrix} =
    \begin{bmatrix}1&3/2&|y_1/2\\0&0&|y_2-6y_1\end{bmatrix} [/itex]
     
    Last edited: Feb 24, 2015
  15. Feb 24, 2015 #14
    so y1 = 2x1 + 3x2
    and y2 = 6y1 = 12x1 + 18x2
    if x1 = t and x2 = r then
    [itex] = im(B) =
    \begin{bmatrix}2t+3r\\12t+18r\end{bmatrix} [/itex]

    sorry I am pretty confused =[
     
  16. Feb 24, 2015 #15
    is [-3/2t] the correct kernel of B?
     
  17. Feb 24, 2015 #16

    BiGyElLoWhAt

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    Gold Member

    What's the highest dimension of a column vector that can be formed by multiplying a column vector with your matrix?
    You have
    ##\left [ \begin{array}{c}
    2 & 3 \\
    6 & 9 \\
    \end{array} \right ]
    \left [ \begin{array}{c}
    x_1\\
    ...\\
    x_n\\
    \end{array} \right]
    ##
    What's the largest n allowed? This gives you the dimension that you should be looking for.
    Once you find that, your goal is to figure out what vectors can be formed by multiplying B*x. This is the image. As several peole have pointed out, you don't have linlearly independant covectors in your matrix B. (covectors=row vector) To handle this, you can do something similar to what you were trying to do in solving for the kernal of A.

    In this context, you're trying to take vectors on a vector space X and map them to a vector space Y. With this particular transform (B) what kind of vectors can you get in Y if you only take vectors from X and map them to Y using the transform B?
     
  18. Feb 24, 2015 #17

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    I don't know what you mean by "[-3/2t]". I said before that the kernel of B was the subspace of vectors <x, y> such that 2x+ 3y= 0 and 6x+ 9y= 0. Of course the second equation is just 3 times the first (if they were not dependent, (0, 0) would be the only solution). The set of all <x, y> such that 2x+ 3y= 0 form a subspace as the kernel must be. In fact, <x, y>= <3, -2> is a solution and so is any multiple.
     
  19. Feb 24, 2015 #18

    Mark44

    Staff: Mentor

    I don't follow what you're doing to get y2.
    The two equations are
    ##2x_1 + 3x_2 = y_1##
    ##6x_1 + 9x_2 = y_2##
    Solve this system for ##x_1## and ##x_2##, preferably using row reduction in an augmented matrix. The basic idea is to determine whether there is a solution <x1, x2> for any given <y1, y2>. As it turns out, there is not, and this fact leads you to a basis for the image of B.
    No.
     
  20. Feb 24, 2015 #19
    when I got 0 0 y2-6y1 in the bottom row I was gonna say it was inconsistent but then i thought it is consistent
    [itex] y_2= 6y_1 [/itex] and since
    [itex] y_1 = 2x_1 + 3x_2 [/itex]
    i just multiplied it by 6 to get y2. if it is inconsistent does that always mean it is linearly dependent? but now that it is inconsistent im not sure what to do to get the image. there are 2 columns of B so that means there is 2 rows of the x vector and that the y vector will be 2x1 because there is 2 rows of B and 1 column of the x vector. the first row of y (2x1) I can use 2t + 3r, but what do i put for the second row?
     
    Last edited: Feb 25, 2015
  21. Feb 25, 2015 #20
    would the image of B besimply
    [itex] im(B) = \begin{bmatrix}x_1\\3x_1\end{bmatrix} [/itex] because any column vector is of that form for example plugging in 2 into x_1 for the first column in B and 3 into X_1 for the second column in B?
     
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