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Finding x of a mapping in R4

  1. Jan 29, 2016 #1
    1. The problem statement, all variables and given/known data
    Linear algebra:

    Find all x in R4 space that are mapped into the zero vector for the given matrix A

    A=
    | 1 -4 7 -5 |
    | 0 1 -4 3|
    | 2 -6 6 -4|


    2. Relevant equations
    None.... it's linear algebra. Don't ban me.

    3. The attempt at a solution
    I tried RREF on this after setting this Ax=b to zero (b = 0 vector)
    |1 0 -9 7 0 |
    |0 1 -4 3 0 |
    |2 -6 6 -4 0 |

    And I solved...

    using x1=x x2=y x3=z etc..... because 9x1 looks confusing
    (x,y,z,w)

    I have
    x - 9z + 7w =0
    y-4z+3w=0

    The answer in the book is

    z times some column (9;4;1;0) + w times some column (-7 -3 0 1)
    (Nevermind the messy notation this probably isn't necessary for you anyway)

    BUT I just can't get this right!
    I just can't understand this somehow.

    I just completely forgot how to represent this properly.

    I mean technically
    "
    x - 9z + 7w =0
    y-4z+3w=0
    "

    is an answer, right?

    However the book just has it in this certain notation, and also the question asks "find all x" that are mapped....

    But... just please help me figure out why I can't do this.

    At first I was even confused that why X had four rows...... Idk how to represent this. I looked in an earlier section in the book and that doesn't even help. Maybe I need some sleep. :l
     
  2. jcsd
  3. Jan 29, 2016 #2

    fresh_42

    Staff: Mentor

    What happened to your 3rd equation?
    Which techniques are you supposed to apply?
    Have you heard about row reduction?
     
  4. Jan 29, 2016 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Your matrix is 4 by 3 so any vector it can be applied to must have 4 rows.
    Essentially you want to solve Ax= 0 where x is a column matrix with 4 rows.
    You want to solve
    [tex]\begin{bmatrix}1 & -4 & 7 & -5 \\ 0 & 1 & -4 & 3\\ 2 & 6 & 6 & 4\end{bmatrix}\begin{bmatrix}w \\ x \\ y \\ z\end{bmatrix}= \begin{bmatrix} w- 4x+ 7y- 5z \\ x- 4y+ 3z \\ 2w+ 6x+ 6y+ 4z\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}[/tex]
    Personally, I wouldn't "row reduce", that's too advanced for me! I would just solve the three equations w- 4x+ 7y- 5z= 0, x- 4y+ 3z= 0, 2w+ 6x+ 6y+ 4z= 0. For example, if we multiply the first equation by 2 and subtract the third equation from that, the "w" terms cancel leaving -8x+ 8y+ 6z= 0. Now solve that together with x- 4y+ 3z= 0.

    Because there are only three equations but four "unknowns" we cannot solve for specific values of all four unknowns. But we probably can solve for three in terms of the fourth. That would give a one dimensional kernel.
     
  5. Jan 29, 2016 #4
    That's what I used, RREF. With MATLAB, actually.

    But the bottom row is all zeroes once you set it up with the zero column. As well as the rightmost column since you have to solve for the zero vector, which is kind of an unusual matrix. So there's no third equation. I mean you can make one, but yep. I'm also not sure which third equation you might be talking about

    Thanks hallsofivy but i'm just concerned that your answer doesn't seem to match the book's answer. What's a one dimensional kernel?
    And wouldn't you represent x as a linear combination of vectors? At least that's what the book's answer is.
     
    Last edited: Jan 29, 2016
  6. Jan 29, 2016 #5
    Oops I just realized that I forgot that the X I was solving for had to be 4x1. Somehow I didn't totally grasp that.

    But whatever the case, maybe this is my question.

    Forget everything above..
    How do I set this up if I need to solve for b?

    Given a matrix A, and needing to find all x in R4 mapped to the zero vector, do I do what you did above and just add a row of zeros to the right?

    This confuses me because to solve for b in Ax=b, you can just augment the matrix and solve for x. To solve for x in Ax=b you can just add b to the right side (in this case three zeroes).

    Why... does that work? Or does it at all?

    I'm not sure why I can't figure this problem out, but yes, I'm about one month into linear aalgebra. I should know how to do this but i'm just incredibly tired and I can't think properly. From what I can tell I thought my answer is right. Although it just seems that the book is missing the vector x and y.
     
  7. Jan 29, 2016 #6

    Mark44

    Staff: Mentor

    The above is an augmented matrix, one where the last column represents the zero vector in the matrix equation Ax = 0. Although there's nothing wrong with putting that last column of zeroes in, there's really no point in doing so. There is no possibility it will change.
    From your two equations above, you can solve for x, y z, and w like so:
    Code (Text):
    x = 9z - 7w
    y = 4z - 3w
    z = z
    w =       w
    The last two equations are obviously true, and are helpful in getting vectors out of this work.

    If you stare at the four equations above long enough, you should see that what we really have is this:
    ##\begin{bmatrix} x \\ y \\ z \\ w \end{bmatrix} = z\begin{bmatrix}9 \\ 4 \\ 1 \\ 0 \end{bmatrix} + w\begin{bmatrix} -7 \\ -3 \\ 0 \\ 1\end{bmatrix}##
     
  8. Jan 29, 2016 #7
    I think I understand. Thanks a lot. I'm going to try this again.
     
  9. Jan 29, 2016 #8

    fresh_42

    Staff: Mentor

    Try to solve it by hand. Then we see whether the kernel is one- or two-dimensional.
    I did not say you should use row reduction. I asked what you were supposed to use?
     
  10. Jan 29, 2016 #9
    I should use whatever I can use... RREF works for this case. There was no specified method. However we've been taught row reduction.
     
  11. Jan 29, 2016 #10

    Mark44

    Staff: Mentor

    RREF is fine (which I did by hand -- the matrix reduces pretty quickly).

    Do you understand what I did in post #6?
     
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