Results 1 to 2 of 2

Math Help - A few proofs in ranks and null

  1. #1
    Super Member
    Joined
    Mar 2006
    Posts
    705
    Thanks
    2

    A few proofs in ranks and null

    Let U be a linear operator on a finite dimensional vector space V, prove that:

    a) N(U) \subseteq N(U^2) \subseteq N(U^3) \subseteq ... \subseteq N(U^k) \subseteq N(U^k+1) \subseteq . . .

    b) If  rank (U^m) = rank (U^{m+1}) , then  N(U^m) = N(U^k) \ \ \ \forall k \geq m

    c) Let T be linear operator on V whose characteristic polynomial splits, and let  \lambda _1 \lambda _2 , . . . , \lambda _k be distinct eigenvalues of T. Then T is diagonalizable if and only if rank (T- \lambda _i I ) = rank ((T- \lambda _i I )^2 ) for 1 \leq i \leq k
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by tttcomrader View Post
    a) N(U) \subseteq N(U^2) \subseteq N(U^3) \subseteq ... \subseteq N(U^k) \subseteq N(U^k+1) \subseteq . . .
    I start you off. Let x\in N(U) then by definition (I assume it is the nullspace) U(x) = \bold{0}\implies U(U(x)) = U(\bold{0}) = \bold{0}. Thus, x\in N(U^2). Now generalize the argument to prove that N(U^{k}) \subseteq N(U^{k+1}).

    b) If  rank (U^m) = rank (U^{m+1}) , then  N(U^m) = N(U^k) \ \ \ \forall k \geq m
    Using the rank nullity theorem we find that \text{rank}(U^m) + \text{nullity}(U^m) = \text{rank}(U^{m+1})+\text{nullity}(U^{m+1}) thus, \text{nullity}(U^m) = \text{nullity}(U^{m+1}). Thus, \text{dim}N(U^m) = \text{dim}N(U^{m+1}) since N(U^m) is a subspace of N(U^{m+1}) it actually means N(U^m) = N(U^{m+1}) (because their dimensions are the same). Now generalize this argument for m+2,m+3,....
    ---
    Sorry I cannot help with (c), my linear algebra is not that great.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Null spaces, Ranges, Nullities, and Ranks
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: June 9th 2009, 09:06 PM
  2. Eigenvalue with ranks
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: April 26th 2008, 12:54 PM
  3. Replies: 3
    Last Post: October 6th 2007, 02:01 PM
  4. ranks
    Posted in the Statistics Forum
    Replies: 5
    Last Post: July 30th 2007, 09:36 PM
  5. Spearmans rank multiple ranks
    Posted in the Advanced Statistics Forum
    Replies: 7
    Last Post: November 25th 2006, 11:43 AM

Search Tags


/mathhelpforum @mathhelpforum