Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook