Results 1 to 4 of 4

Math Help - Linear Transformation (Rank and Nullity)

  1. #1
    Newbie
    Joined
    May 2010
    Posts
    15

    Post Linear Transformation (Rank and Nullity)

    Hello,

    I need some help in this problem....

    A = [1 -2 -1, 1 -3 0, 3 -8 -1] and let linear transformation T: R^3--> R^3 be defined by T(x) = Ax, where x = [ x1 x2 x3].

    a) Determine rank(T) and nullity (T)

    b) Determine bases for the range R(T) and the null space N(T).

    (If N(T) = {[0 0 0]} state that N(T) has no basis)...


    I know we have to use the Gauss Jordan to figure that matrix out but I am unable to reduce it properly, perhaps there is another way of doing it??

    please help!

    thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    14,971
    Thanks
    1121
    I can see no difficult in reducing it- it seems easier than most of this kind.
    \begin{bmatrix}1 & -2 & 1 \\ 1 & -3 & 0 \\ 3 & -8 & -1\end{bmatrix}

    Start by subtracting the first row from the second row and subtracting three times the first row from the third row.
    \begin{bmatrix}1 & -2 & 1 \\ 0 & -1 & -1 \\ 0 & -2 & -2\end{bmatrix}
    Can you finish that?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2010
    Posts
    15
    thanks for the response Sir. When I started simplifying using Gauss Jordan I ended up getting---
    x = 2(y) + z
    z = y...If I consider 1st column as x, 2nd as y and 3rd column as z and then set three of these equations equal to 0...
    So the N(A), null space of A = Span {z [3 1 1] }

    but then this is what I get when I simplify A, then how do I implement it on T(x) = Ax...and then find the rank and null space of T?...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,150
    Thanks
    591
    Halls, your first row is incorrect. a13 should be -1.

    vedicmath, how many non-zero rows did you end up with in rref(A)? that is the rank of A, so any two columns of A would be a basis (it is customary to choose the rows of A corresponding to the pivot columns of rref(A)).

    what is the nullity of A, by the rank-nullity theorem? you have a spanning set, does it have the same number of elements as the dimension of the null space (that is one criterion for a basis)?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: January 16th 2012, 10:39 AM
  2. linear algebra - rank-nullity
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: December 24th 2010, 02:30 PM
  3. Rank and Nullity of a linear transformation
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: June 29th 2009, 09:10 AM
  4. A question about rank/image of a linear transformation
    Posted in the Advanced Algebra Forum
    Replies: 7
    Last Post: February 11th 2009, 04:54 PM
  5. Linear Transformation, Nullity and rank...
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: September 28th 2008, 09:24 PM

Search Tags


/mathhelpforum @mathhelpforum