Results 1 to 3 of 3

Math Help - Advance linear transformation exercise

  1. #1
    Junior Member
    Joined
    Jun 2009
    Posts
    58

    Exclamation Advance linear transformation exercise

    Help please with these exercise

    Let V be a finite-dimensional vector space and let T be a linear operator on V.
    Suppose that rank (T^2) = rank (T),

    Prove that the range and null space of T are disjoint therefore have only the zero vector in common.

    THANKS
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by osodud View Post
    Help please with these exercise

    Let V be a finite-dimensional vector space and let T be a linear operator on V.
    Suppose that rank (T^2) = rank (T),

    Prove that the range and null space of T are disjoint therefore have only the zero vector in common.

    THANKS
    (1) by the rank-nullity theorem \text{nullity}(T)=\text{nullity}(T^2). we also know that the null space of T is a subspace of the null space of T^2. so the null space of T is equal to the null space of T^2.

    now if v=T(u) is in the null space of T, then T^2(u)=T(v)=0 and thus v=T(u)=0 by (1).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Apr 2009
    Posts
    678
    Thanks
    1
    Quote Originally Posted by osodud View Post
    Help please with these exercise

    Let V be a finite-dimensional vector space and let T be a linear operator on V.
    Suppose that rank (T^2) = rank (T),

    Prove that the range and null space of T are disjoint therefore have only the zero vector in common.

    THANKS
    1. Prove null-space of T= null space of T^2, under the given conditions. (This can be done by showing they have same dimension).

    2. Using 1 above establish that if x\in Null(T) AND x \in Range(T) implies x=0. As x \in Range(T) there is a  y \in V such that T(y)=x. What can you say about T^2(y) and hence T(y) and hence x from this? That should provide you the prove.
    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. Replies: 0
    Last Post: April 12th 2011, 01:57 PM
  3. Problem with simple linear algebra exercise
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: February 3rd 2010, 11:24 AM
  4. Linear Algebra.Linear Transformation.Help
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: March 5th 2009, 02:14 PM
  5. Linear Transformation
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: February 21st 2007, 09:12 AM

Search Tags


/mathhelpforum @mathhelpforum