Results 1 to 2 of 2

Math Help - linear algebra

  1. #1
    Member
    Joined
    Dec 2008
    Posts
    154

    linear algebra

    Let A =
    [ 3 0 1 ]
    [-4 1 -2 ]
    [ -4 0 -1 ] which is a 3x3 matrix. and let I be a 3x3 identity matrix.
    Find a basis for the intersection of R(A-I) and N(A-I), where R(A-I) is the range of A-I and N(A-I) is the nullspace of A-I.

    the solution says this basis is [1 -2 -2]^T (this T means transpose).
    and the procedure is the following.

    1. Find a basis {x1,x2,..,} for R(A-I)
    2. Let X be the matrix that consists of the vectors in this basis for R(A-I).
    3. Find a basis {v1,v2,....,}for N((A-I)X)
    4. B={Xv1, Xv2,...,..} is the basis of the intersection.

    i ve been trying to do this problem by following the procedure but i cant get the solution and i dont know what i am doing wrong. please help me.can please anyone explain and show me how to find this basis?
    Last edited by Kat-M; January 1st 2009 at 10:34 PM. Reason: added some more stuff
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,201
    Thanks
    1789
    Looks to me like it is just a matter of doing the calculation.

    A- I= \left[\begin{array}{ccc}2 & 0 & 1 \\ -4 & 0 & -2 \\ -4 & 0 & -2\end{array}\right]

    The null space is the set of all (x, y, z) such that
    A- I= \left[\begin{array}{ccc}2 & 0 & 1 \\ -4 & 0 & -2 \\ -4 & 0 & -2\end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right]= \left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right]
    That means we must have 2x+ z= 0, -4x- 2z= 0, and -4x- 2z= 0, all of which reduce to the single equation z= -2x. Of course, y does not appear in that equation and can be anything. That tells us that any vector in the nullity is of the form <x, y, -2x>= x<1, 0, -2>+ y<0, 1, 0> so {<1, 0, -2>, <0, 1, 0>} is a basis for the null space.

    Of course, any vector in the range is of the form
    A- I= \left[\begin{array}{ccc}2 & 0 & 1 \\ -4 & 0 & -2 \\ -4 & 0 & -2\end{array}\right]\left[\begin{array}{c} x \\ y // z \end{array}\right]= \left[\begin{array}{c}2x+ z \\ -4x- 2z \\ -4x- 2z\end{array}\right]
    or <2x+ z, -4x- 2z, -4x- 2z>= (2x+z)<1, -2, -2> for any numbers x, z so <1, -2, -2> is a basis for the range, as you say.

    But <1, -2, -2>= <1, 0, -2>- 2<0, 1, 0> and so is itself in the null space. That means that the range is contained in the null space: the intersection of range and null space is just the range itself. A basis for the intersection is a basis for the range, <1, -2, -2>.

    And, Since Ax is in the range which is in the null space of A, A(Ax)= A^2x= 0 for all x: A^2= 0.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: August 1st 2011, 11:00 PM
  2. Basic Linear Algebra - Linear Transformations Help
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: December 7th 2010, 04:59 PM
  3. Replies: 2
    Last Post: December 6th 2010, 04:03 PM
  4. Replies: 7
    Last Post: August 30th 2009, 11:03 AM
  5. Replies: 3
    Last Post: June 2nd 2007, 11:08 AM

Search Tags


/mathhelpforum @mathhelpforum