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Matrices: Technical Questions

  1. Sep 9, 2004 #1
    Hey Guys,
    I was solving some practice questions on linear algebra and I came accross this one. I've solved many of its kind, but this one has a variation i'm not so sure about.. Here's the question:

    Let A =
    | 0 3 -6 3 |
    | 0 -3 6 -3 |

    Find a basis of nullspace (A). (a set of basic solutions for the homogeneous system)

    (That's a matrix, sorry about not using latex, I'm kinda new here)

    Now, I know how to do a normal question, by doing elementary row operations and then factoring the parameters.. But what do I do when the first column is all zeroes?? do I make the first basic solution a column of 4 zeroes? or what?
    Your help would be greatly appreciated, thx..
  2. jcsd
  3. Sep 10, 2004 #2


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    Staff Emeritus
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    The problem asks you to find a basis for the null space of A. A itself, in this example, is from R4 to R2. If we represent a vector in R4 as < x, y, z, u> then your two equations are 0x+ 3y- 6z+ 3u= 0 and 0x- 3y+ 6z- 3u= 0.

    The first column being 0 means that x disappears from the equations! Okay, that's easy- that just means that x can be anything and that any vector that is a multiple of <1, 0, 0, 0> is in the null space: one basis vector is <1, 0, 0, 0>.

    The two equations are 3y- 6z+ 3u= 0 and -3y+ 6z- 3u= 0. It's easy to see that the second equation is just -1 times the first: you really dont have 2 independent equation. (Normally, you would try to "solve" the equations by eliminating unknowns until you don't have enough equations left. IF you had enough independent equations to completely solve for each unknown, the null space would be just {0}.)
    Since you really have just the one equation in 3 unknowns, solve for one of them in terms of the other 2: 3y= 6z- 3u so y= 2z- u. Now, to be sure you get independent vectors, take one of the remaining unknowns to be 1 and the other 0:
    If z= 1 and u= 0, then y= 2(1)- 0= 2. One basis vector for the null space is <0,2,1,0>.
    If z= 0 and u= 1, then y= 2(0)- 1= -1. Another basis vector for the null space is <0,-1,0, 1>.

    The null space of A is 3 dimensional and has basis <1, 0, 0, 0>, <0, 2, 1, 0> and
    <0, -1, 0, 1>.

    You understand, of course, that the basis of the null space is not unique: we could have solved for z or u instead of y as well as choosing numbers other than 0 and 1 to try and arrived at a different answer.
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