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Matrix rank proof

  1. Aug 27, 2011 #1
    1. The problem statement, all variables and given/known data

    Show that for all nxn matricies A with real entries we have
    rk(A*A) = rk(A) where A* is the transpose of A.

    2. Relevant equations



    3. The attempt at a solution

    I'm working over a vector space V.

    Im(A) = {A(v) | vEV}

    Im(A*A) = {A*(A(v)) | vEV}

    So Im(A*A) is a subset of Im(A)
    So rk(A*A) =< rk(A)

    Ker(A) = {A(w) = 0 | wEV}

    Ker(A*A) = {A*(A(w)) = 0 | wEV}

    so Ker(A*A) is a subset of Ker(A)
    so dimKer(A*A) =< dimKer(A)

    dimKer(A) = dimV - rk(A)

    so -rk(A*A) =< -rk(A)
    so rk(A*A) >= rk(A)

    Combining the results yields rk(A*A) = rk(A)


    Is this correct??
     
  2. jcsd
  3. Aug 27, 2011 #2

    micromass

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    Why are these two things true?? They don't exactly seem trivial to me...
     
  4. Aug 27, 2011 #3

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    Note that you have reversed your definition of Ker.
    It should be:
    Ker(A) = {wEV | A(w) = 0}
     
  5. Aug 27, 2011 #4
    Im(A*A) = {A*(A(v)) | vEV}

    A maps the vector v to another vector v1.
    Then A* maps the vector v1 to another vector v2, but the set of all v1s can either be less than or equal to the set of original vectors, then A* has either the original amount of vectors or a smaller amount to map to v2.
    So Im(A*A) is a subset of Im(A)

    Similar logic can be applied to the Kernel.
     
  6. Aug 27, 2011 #5

    micromass

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    That only means that Im(A*A) has less elements than Im(A), which is true. This doesn't imply that it's a subset. You didn't prove that every element in Im(A*A) is in Im(A).
     
  7. Aug 27, 2011 #6
    Ah I see. Any points in the right direction? :smile:
     
  8. Aug 27, 2011 #7

    micromass

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    Yes. The trick is to prove that

    [tex]Ker(A^*A)=Ker(A)[/tex]

    Now, I claim that [itex]Ker(A)\subseteq Ker(A^*A)[/itex] (why??)
    To see the other inclusion, assume that A*Ax=0. What happens if you multiply both sides with x*?
     
  9. Aug 27, 2011 #8
    A*Ax = 0, then A*Axx* = 0
    But if x is a nx1 matrix and x* a 1xn matrix xx* is a 1x1 matrix?
    But the only way this can be zero is if either A*A is 0?

    As for the kernel fact, I'm a bit confused..
     
  10. Aug 27, 2011 #9

    micromass

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    Multiply the other side by x* :smile:

    And use that in general z*z=|z| for vectors z.

    Let x be in Ker(A). You know that A(x)=0. Can you prove that A*A(x)=0?
     
  11. Aug 27, 2011 #10
    Well if Ax = 0 then A*A(x) = A*(0) = 0, as under a linear operator 0 is always mapped to 0.
     
  12. Aug 27, 2011 #11

    micromass

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    Yes, that's ok!
     
  13. Aug 27, 2011 #12
    I'm afraid I don't see how that shows

    [itex]Ker(A)\subseteq Ker(A^*A)[/itex]
     
  14. Aug 27, 2011 #13

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    In words it says: if x is in the kernel of A, then it is also in the kernel of A*A.

    Rephrasing: for every x in V we have: if Ax = 0, then A*Ax = 0.
     
  15. Aug 27, 2011 #14
    Right I understand now! :smile:

    So if [itex] Ker(A)\subseteq Ker(A^*A) [/itex]

    Then by taking the dimension of each side and applying the rank-nullity theorem and applying my original result then we get the answer?
     
  16. Aug 27, 2011 #15

    I like Serena

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    Nope.
    Previously you had the subset relationship the wrong way around.
    You'll see if you try it again.
    Sorry.
     
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