rank(A)=rank(cA), where c is scalar

Discussion in 'Undergraduate Math' started by Jen, Nov 2, 2005.

  1. Jen

    Jen Guest

    How can I prove that the rank(A)=rank(cA) where c is
    a scalar and A is an mxn matrix?

    I know that when you row reduce a matrix you can divide
    the rows by that common "c" number, but how can I formally
    write it up????
     
    Jen, Nov 2, 2005
    #1
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  2. You can't... unless you require that c be a ->nonzero<- scalar.
    Multiplying a matrix by an invertible matrix does not change the rank:
    that is, if C is an invertible mxm matrix, and D is an invertible nxn
    matrix, then rank(A) = rank(CA) = rank(AD) = rank(CAD).

    So set C to be the diagonal matrix whose diagonal entries are all 1/c
    (here is where you need c is nonzero); this is exactly the same as
    multiplying each row by 1/c.

    --
    ======================================================================
    "It's not denial. I'm just very selective about
    what I accept as reality."
    --- Calvin ("Calvin and Hobbes")
    ======================================================================

    Arturo Magidin
     
    Arturo Magidin, Nov 2, 2005
    #2
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  3. Jen

    Jen Guest

    How can I prove that the rank(A)=rank(cA) where c is
    You can't... unless you require that c be a ->nonzero<- scalar.
    Multiplying a matrix by an invertible matrix does not change the rank:
    that is, if C is an invertible mxm matrix, and D is an invertible nxn
    matrix, then rank(A) = rank(CA) = rank(AD) = rank(CAD).

    So set C to be the diagonal matrix whose diagonal entries are all 1/c
    (here is where you need c is nonzero); this is exactly the same as
    multiplying each row by 1/c.


    If c is a nonzero scalar, and I want rank(A)=rank(cA),
    then why do you have 1/c? Can I have C be the diagonal
    matrix whose diagonal enteries are all c and not 1/c?
    This way rank(A)=rank(cA).
    Is this permitted?
     
    Jen, Nov 3, 2005
    #3
  4. Because that's the matrix that does what you said you wanted: "divide
    each entry by c", starting with cA.
    Yes: if you want to start with A, then you would do that.
    A little bit of thought would have answered your question. If you
    really do not see that what you are asking was exactly the same as I
    said, then you did not understand the answer and it will be worthless
    to you. Try to understand what is being said.

    --
    ======================================================================
    "It's not denial. I'm just very selective about
    what I accept as reality."
    --- Calvin ("Calvin and Hobbes")
    ======================================================================

    Arturo Magidin
     
    Arturo Magidin, Nov 3, 2005
    #4
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