1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Find the left nullspace of matrix

  1. Apr 30, 2012 #1

    sharks

    User Avatar
    Gold Member

    The problem statement, all variables and given/known data

    [tex]B=
    \displaystyle\left[ {\begin{array}{*{20}{c}}
    1&0&-2&1 \\
    1&2&-2&3 \\
    -2&1&3&0
    \end{array}} \right][/tex]
    Find:
    1. The nullspace of B.
    2. The left nullspace of B.

    The attempt at a solution
    I was able to find the nullspace of B. but i can't figure out why the left nullspace isn't working out, although i'm quite sure that i'm following the right procedure.
    [tex]N(B)=
    \displaystyle\left[ {\begin{array}{*{20}{c}}
    1 \\
    -1 \\
    1 \\
    1
    \end{array}} \right][/tex]
    I'm wondering if there's a shortcut way of finding the nullspace, without working the whole problem over again, starting with the transpose of B?

    So, to find left nullspace of B:
    [tex]B^T=
    \displaystyle\left[ {\begin{array}{*{20}{c}}
    1&1&-2 \\
    0&2&1 \\
    -2&-2&3 \\
    1&3&0
    \end{array}} \right][/tex]
    I then did the same steps as when finding the nullspace, by reducing to row echelon form, which is:
    [tex]B^T=
    \displaystyle\left[ {\begin{array}{*{20}{c}}
    1&1&-2 \\
    0&2&1 \\
    0&0&-1 \\
    0&0&1
    \end{array}} \right][/tex]
    To find the nullspace, as usual, solving: [itex]A.x=0[/itex] where matrix [itex]x[/itex] is a 3x1 column matrix, containing three variables: [itex]x_1,\,x_2,\,x_3[/itex].

    But then i'm stuck, as all the 3 variables are pivot variables! So, there are no free variables. I am unable to express the pivot variables in terms of non-existent free variables. But i still persevered to solve the equation and i got [itex]x_1=x_2=x_3=0[/itex]. This would mean that the left nullspace of B is a zero 4x1 column matrix, which i believe is wrong.

    At this point, i don't know what to do, as i've checked some worked-out examples and they managed to do it with different matrices, but with this particular matrix, i think i need to try a different method or maybe i'm doing something wrong.
     
    Last edited: Apr 30, 2012
  2. jcsd
  3. Apr 30, 2012 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    While knowing how to solve problems by "row-reduction", etc. certainly simplifies calculations and eases the problem, don't lose track of the basic definitions!

    The left null set of B is, by definition, the space of vectors [itex]\begin{bmatrix}a & b & c & d\end{bmatrix}[/itex], such that
    [tex]\begin{bmatrix}a & b & c\end{bmatrix}\begin{bmatrix}1 & 0 & -2 & 1 \\ 1 & 2 & -2 & 3 \\ -2 & 1 & 3 & 0 \end{bmatrix}= \begin{bmatrix}a+ b- 2c \\ 2b+ c \\ -2a- 2b+ 3c \\ a+ 3b\end{bmatrix}= \begin{bmatrix}0 \\ 0\\ 0\\ 0\end{bmatrix}[/tex]

    That gives equations a+ b- 2c= 0, 2b+ c= 0, -2a+ 3b+ 3c= 0, and a+ 3b= 0. The "left null space" consists of all vectors [a, b, c] satisfying those equations.
     
  4. Apr 30, 2012 #3

    sharks

    User Avatar
    Gold Member

    Hi HallsofIvy

    According to the definition you've given, it should be possible to adapt it to: [itex]B.N(B)=0[/itex]. Correct?

    But, when multiplying, i get this vector: [itex][0, 4, 0]^T[/itex]. Does this mean that i've miscalculated the nullspace, N(B)?
     
  5. Apr 30, 2012 #4

    Mark44

    Staff: Mentor

    B times any vector that belongs to its right nullspace has to come out to the zero vector.

    Similarly, if x is in the left nullspace of B, then x*B has to be the zero vector.
     
  6. Apr 30, 2012 #5

    sharks

    User Avatar
    Gold Member

    It seems that i've indeed miscalculated the right nullspace, N(B), as i don't get a zero vector.
     
  7. Apr 30, 2012 #6

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    When you found the reduced echelon form U you essentially carried out row operations on B, so you would have gotten the results by multiplying B on the left by a lower-triangular matrix E; that is, U = EB. (You can find E by doing the same row operations on the 3x3 identity matrix that you did on B; alternatively, you could have used an augmented matrix A = [B|I] and do those row operations on A; the first 4 columns give you U and the last three give you E.)

    Anyway, suppose you do have E (which is invertible). If y is a row vector we have yU = zB, where z = yE. So, if y is in the left-null space of U (and that is easy to find, because U is upper-triangular), then we get z from z = yE. The row vector z is in the left null-space of B. Conversly, if a row vector z is in the left null-space of B, the vector y = z*Inverse(E) is in the left null-space of U, so there is a 1-1 correspondence between these two null spaces.

    RGV
     
  8. Apr 30, 2012 #7

    Mark44

    Staff: Mentor

    No, your work is correct for the right nullspace. B * <1, -1, 1, 1>T = <0, 0, 0>T.

    For the left nullspace, you want a vector x in R>3 such that x*B = <0, 0, 0, 0>T. You have it in post #1, except that it should be a vector in R3, not R4.
     
  9. Apr 30, 2012 #8

    sharks

    User Avatar
    Gold Member

    So, i should aim for this:
    [tex]N(B^T).B=0[/tex]
    These are the corresponding mxn sizes of the matrices: (1x3).(3x4)=(1x4).

    I get these equations:
    a+ b- 2c= 0 .......(1)
    2b+ c= 0 ...........(2)
    -2a+ 3b+ 3c= 0 ...(3)
    a+ 3b= 0 ...........(4)

    Now, to solve them...

    I added (2) and (4) to get: a+5b+c=0 .....(5)
    I then solved the system of 3 linear equations; (1), (3) and (5).
    I get x=y=z=0. What am i doing wrong?
     
    Last edited: Apr 30, 2012
  10. Apr 30, 2012 #9

    Mark44

    Staff: Mentor

    Why do you think this is wrong? It's possible for a nullspace to consist of just a single vector - if so it's the zero vector.
     
  11. Apr 30, 2012 #10

    sharks

    User Avatar
    Gold Member

    OK, so the left nullspace of B is: [itex]\begin{bmatrix}0 & 0 & 0\end{bmatrix}[/itex]
    According to what i learned, the left nullspace is a subspace of Rm, where m is the number of rows. Therefore, the left nullspace should be a vector in R1, not R3.

    Here are the definitions from my book:
    The nullspace and row space are subspaces of Rn.
    The left nullspace and the column space are subspaces of Rm.
    m and n, refer to the number of rows and columns respectively.
     
    Last edited: Apr 30, 2012
  12. May 1, 2012 #11

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Well, the left null space is a subspace of Rm, a collection of vectors, not just a vector.

    Te 0 vector is in any subspace so those are certainly true of the 0 vector.
     
    Last edited by a moderator: May 1, 2012
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Find the left nullspace of matrix
  1. Nullspace matrix (Replies: 7)

Loading...