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Showing that the 0 matrix is the only one with rank = 0

  1. Apr 2, 2017 #1
    1. The problem statement, all variables and given/known data
    Prove that if rank(A) = 0, then A = 0.

    2. Relevant equations


    3. The attempt at a solution
    This seems like a very easy problem, but I just want to make sure I am not missing anything.

    rank(A) = dim(Im(A)) = 0, so Im(A) = {0}. Thus, A is by definition the zero matrix.

    My only concern is whether I can conclude from Im(A) = {0} that A is the zero matrix. That is, that there is no other matrix that satisfies this property.
     
  2. jcsd
  3. Apr 2, 2017 #2

    fresh_42

    Staff: Mentor

    What does ##\operatorname{Im}A = 0## mean?
     
  4. Apr 2, 2017 #3
    I meant to write something like ##\text{Im}(A)= \{\vec{0} \}##
     
  5. Apr 2, 2017 #4

    fresh_42

    Staff: Mentor

    Yes, of course, but what does it mean? What is the definition of ##\operatorname{Im} A=0##? Or ##= \{\vec{0}\}##, that isn't the point. Then compare it to what ##A=0## means.
     
  6. Apr 2, 2017 #5
    Well, that means that the matrix ##A## maps everything to ##\vec{0}##. So of course the matrix matrix satisfies this. I just don't really know how I would show that the zero matrix is the only matrix that satisfies this.
     
  7. Apr 2, 2017 #6

    fresh_42

    Staff: Mentor

    ##\operatorname{Im}A =0 \,\Longleftrightarrow\,A(v)=0 \,\forall\, v\in V \,\Longleftrightarrow\,A=0##
    It's simply the definitions, i.e. it works in both directions.
     
  8. Apr 2, 2017 #7

    vela

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    Staff Emeritus
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    You could find the columns of A by multiplying by the appropriate vector, e.g., (1,0,0,...,0), (0,1,0,...,0).
     
  9. Apr 3, 2017 #8
    A corollary of Rank-nullity says that the rank of a matrix is the same as the order of the highest order nonzero minor in that matrix. Therefore, if a matrix has rank zero, it means that its every entry must be the zero element.

    Alternatively, you could invoke the Rank-Nullity theorem. You could view the matrix ##A ## as a linear operator (there's another theorem that allows you to do this) ##A : U\to V ##.
    The rank of ##A ## is then defined to be the dimension of the subspace ##A(U) =:\mbox{ran}A = \mbox{im}A##. R-N gives you
    [tex]
    \mbox{dim}(U) = \mbox{dim}(\mbox{ker}A ) + \mbox{dim}(\mbox{ran}A)
    [/tex]
    What is the only linear operator whose nullity is ##\mbox{dim}(U) ##? By nullity I mean the dimension of the kernel of ##A##.
     
    Last edited: Apr 3, 2017
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