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Nullspace of a matrix

  1. Mar 13, 2014 #1
    1. The problem statement, all variables and given/known data

    Let U be the subspace of R4 given by:
    U = the nullspace of the matrix

    [0 0 2 4
    0 3 -4 2]

    3. The attempt at a solution

    let v = (v1,...v4) and w = (w1...w4)
    (0,0,2,4) = (λ1v1 + λ2v2 + λ3v3 + λ4v4)
    (0,3,-4,2) = (λ1w1 + λ2w2 + λ3w3 + λ4w4)



    I haven't come across such problem before nor were taught what a nullspace of a matrix is.
    Could someone provide some explanation as to the question?
     
    Last edited: Mar 13, 2014
  2. jcsd
  3. Mar 13, 2014 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Surely, you could look up "nullspace" in the index of your textbook? The "nullspace" of a matrix is the set of all vectors that the matrix maps into the 0 vector (i.e. makes "null"). It is easy to show that the set of all such vectors is a subspace (if Au= 0 and Av= 0 then [itex]A(\alpha u+ \beta v)= \alpha Au+ \beta Av= 0[/itex]).

    So here you are asked to find all [itex]\begin{bmatrix}x \\ y \\ z \\ u\end{bmatrix}[/itex] such that
    [tex]\begin{bmatrix}0 & 0 & 2 & 4 \\ 0 & 3 & -4 & 2\end{bmatrix}\begin{bmatrix}x \\ y \\ z\\ u\end{bmatrix}= \begin{bmatrix}2z+ 4u \\ 3y- 4z+ 2u\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}[/tex]

    Solve the equations 2z+ 4u= 0 and 3y- 4z+ 2u= 0. (There is not a unique answer.)
     
  4. Mar 13, 2014 #3
    It's not in the content of the book. Search engine showed rather dissappointing results too.
     
  5. Mar 13, 2014 #4


    0x + 0y +2z +4u = 0
    0x + 3y -4z +2u = 0

    0 0 2 4 | 0
    0 3 -4 2 | 0

    R1<->R2

    0 3 -4 2 | 0
    0 0 2 4 | 0

    z and u are free variables. There exists infinite solutions to the SOLE.
    (x,y,z,u) = (x,4/3z-2/3u,-2u,u)

    The questions askes for a spanning set for U matrix. So if the null matrix (0,0,0,0) spans(x,y,z,u), then, (0,0,0,0) is a linear combination of (x,y,z,u).

    Edit: λ1(x)+λ2(4/3z-2/3u) +λ3(-2u) + λ4(u) = (0,0,0,0)
     
    Last edited: Mar 13, 2014
  6. Mar 13, 2014 #5

    Mark44

    Staff: Mentor

    The matrix above is not completely reduced. In fact, all you did was switch the two rows.
    (0, 0, 0, 0) is the zero vector in R4. It does NOT span anything other than itself. In your sentence that starts "So if the null matrix ...", there is not one thing that is true.
    You need to pay more attention to definitions, especially those for "nullspace" and "span". Look in the index (at the back of the book) to see where these terms appear in the book. If a term is used on more than one page, it usually is defined on the first page of the list. The table of contents at the front of the book is probably less helpful, as it has a list of the chapters without much detail on what's in the chapter.
     
  7. Mar 13, 2014 #6
    0 0 2 4 | 0
    0 3 -4 2 | 0

    R1<->R2

    0 3 -4 2 | 0
    0 0 2 4 | 0

    RREF....

    0 3 0 -6| 0
    0 0 2 4 | 0

    Leading variables: y, z
    Non-leading: x,u

    2z + 4u = 0
    z = -2u
    3y -6u = 0
    y = 2u

    (x,y,z,u) = (x,2u,-2u,u)
     
    Last edited: Mar 13, 2014
  8. Mar 13, 2014 #7

    Mark44

    Staff: Mentor

    The 4th number in the top row is wrong.
    You should check your work.
    This matrix product ...
    $$ \begin{bmatrix}0 & 0 & 2 & 4 \\ 0 & 3 & -4 & 2 \end{bmatrix} \begin{bmatrix}x \\ 2u \\ -2u \\ u \end{bmatrix}$$
    should result in the column vector <0, 0>. Does it?
     
  9. Mar 13, 2014 #8

    0, 3, 0, 10 | 0
    0, 0, 2, 4| 0

    Leading variables: y, z
    Non-leading variables: x, u

    2z + 4u = 0
    2z = -4u
    z = -2u

    3y + 10u = 0
    y = -10u/3

    (x,y,z,u) = (x,-10u/3,-2u,u)

    [0, 0, 2, 4; 0, 0, 2, 4] [x; -10u/3; -2u;u] = [0;0]
     
  10. Mar 13, 2014 #9

    Mark44

    Staff: Mentor

    It's easy to make mistakes when you're reducing a matrix, so it's important to check your work (as I suggested in my previous post). Also, it's helpful to indicate in your work what you're doing. Off to the left of my matrices, if I'm adding -3 times row 2 to row 1 (for example), I put -3 to the left of row 2, and 1 to the left of row 1, with an arrow going from row 2 to row 1.

    That way, if my answer doesn't check out, I can go back and look at my individual steps. Having an audit trail, so to speak, makes it easier to quickly check my row-reduction work.
     
  11. Mar 13, 2014 #10

    Mark44

    Staff: Mentor

    That's more like it! Congrats!

    Your solution could be written as <x, 0, 0, 0> + <0, -(10/3)u, -2u, u>, or better yet, as x<1, 0, 0, 0> + u<0, -10/3, -2, 1>, where x and u are arbitrary real numbers.

    Since there are two vectors, the nullspace of your matrix is a two-dimensional subspace of R4. In other words, it is a plane in four dimensions. The vectors {<1, 0, 0, 0>, <0, -10/3, -2, 1>} span the nullspace, and since they are linearly independent (which I can tell by inspection), they form a basis for the nullspace.
     
    Last edited: Mar 13, 2014
  12. Mar 13, 2014 #11
    Just to clarify, and ensure I don't get lost in the intermediate step (I like looking at big picture).
    Is the correct question I ought to be asking myself at the onset is:
    Given [0, 0, 2, 4; 0, 0, 2, 4], what values must the vector[x;y;z;u] be such that
    [0, 0, 2, 4; 0, 0, 2, 4][x;y;z;u] = [0;0] ?
    The way I see it, [x;y;z;u] is really just the scalar in the linear combination.
     
    Last edited: Mar 13, 2014
  13. Mar 13, 2014 #12
    Strange, the system rejected my answer.
    It asked what the spanning set of U is.
    (x,-10u/3,-2u,u) is apparently not the right one...
     
    Last edited by a moderator: Mar 13, 2014
  14. Mar 13, 2014 #13

    Mark44

    Staff: Mentor

    Yes.
    No, <x, y, z, u> is a vector in R4. Its coordinates (the x, y, z, and u values) are scalars.
     
  15. Mar 13, 2014 #14

    Mark44

    Staff: Mentor

    The spanning set would be a set of one or more vectors. What I wrote at the tail end of post #10 is probably what they're looking for.
     
  16. Mar 13, 2014 #15
    They're looking for the vectors in format as such, e.g., (1,2,3,4), (5,6,7,8)
     
  17. Mar 13, 2014 #16

    Mark44

    Staff: Mentor

    See post 10. Note that I incorrectly copied one coordinate of one vector, but it's fixed now.
     
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