1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Help with proof

  1. Apr 21, 2006 #1
    Prove: If k !=0, then A and kA have the same rank.
    Given:
    k is in the set of all Reals.

    proof:
    WOLOG let A be an M X N matrix
    Assume m < n
    => the max rank(A) can be is m. (i.e. the row space and the column space)
    => the max the rank(kA) can be is also m, because multiplying a matrix by a constant does not change the dimensions of the matrix
    =>
    # of leading variables stays the same for A and KA
    =>
    Rank(A)=Rank(KA)..
     
    Last edited: Apr 22, 2006
  2. jcsd
  3. Apr 21, 2006 #2

    AKG

    User Avatar
    Science Advisor
    Homework Helper

    How can you say that m < n WOLOG? Also, what are the "leading variables"? Anyways, there are a number of ways to define rank. What definition do you have? Some hints: If you have a set of vectors that are linearly independent, when you multiply them each by the scalar k (non-zero), aren't they still linearly independent? Same for if you have a set of linearly DEpendent vectors? Also, if you have a subspace of some vector space, and you multiply each vector in that subspace by a non-zero scalar, don't you get the same subspace?
     
  4. Apr 22, 2006 #3
    By leading variables i meant the leading 1's. Also i edited, i meant to assume m < n.
    How does stating it is L.I. help?
     
    Last edited: Apr 22, 2006
  5. Apr 13, 2009 #4
    I have a problem.
    Suppose that {u1,u2,...,um} are vectors in R^n. Prove, directly that span{u1,u2,...,um} is a subspace of R^n.
    How do I go about doing this. Thanks.
     
  6. Apr 13, 2009 #5
    To address georgeh's question to AKG, the rank of a matrix is the number of linearly independent vectors it contains. So let's say you have an MxN matrix of rank R. Then it contains R linearly independent columns (or rows), and the rest, if any, are dependent on that set. All you have to show is that multiplying a set of linearly independent vectors by a constant doesn't make them linearly dependent, and that multiplying a vector dependent on that set by the same constant doesn't make it independent.

    To squenshl: you need to show four things:
    - that span {u1, u2,...um} is a subset of R^n
    - that that subset contains the zero vector
    - that that subset is closed under addition (if u and v are elements of the span, then u+v is an element)
    - and that it is closed under scalar multiplication (if u is an element of the span, than k*u is an element).
     
  7. Apr 14, 2009 #6
    Okay.
    I've done the first 2 but how do I show that it is closed under addition & multplication.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Help with proof
  1. Help with a proof. (Replies: 10)

  2. Proofs help (Replies: 7)

  3. Help with a proof! (Replies: 8)

  4. Help with a Proof (Replies: 1)

  5. Proof help! (Replies: 3)

Loading...