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Rank and nullity

  1. Feb 22, 2010 #1
    1. The problem statement, all variables and given/known data
    For any (nxn) matrix A, prove or disprove with a counter example:
    1. Rank(A^2) <= rank(A)
    2. Nullity(A^2) <= nullity(A)


    2. Relevant equations
    Rank = dimension of range
    Nullity = dimension of null space


    3. The attempt at a solution
    I have been trying a few examples with both nonsingular and singular matrices and in all the cases I have found that rank(a^2) = rank(a) and like wise for the nullity. But obviously isn't enough to convince me so I was looking for some help please.
     
  2. jcsd
  3. Feb 22, 2010 #2
    For 1) try to think of the matrices in terms of their associated linear transformations [itex]T : k^n \to k^n[/itex]. Then what 1 says is that the dimension of [itex]T(T(k^n))[/itex] is less than the dimension of [itex]T(k^n)[/itex]. Can you prove this? (Just think of [itex]T(k^n)[/itex] as a vector space, and prove or reference the general statement that [itex]\dim T(U) \leq \dim U[/itex] for subspaces U of [itex]k^n[/itex]).

    For a counterexample try,
    [tex]A = \left[ \begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array} \right][/tex]
     
  4. Feb 22, 2010 #3
    I have no idea what you are trying to say with your thing "k^n" and this thing you are refering to as "T"...
     
  5. Feb 22, 2010 #4
    And that matrix can not be a counter example for the first part because then A^2 equals the zero matrix and the rank of that is 0. The rank of A would be 1. So then the dimension of A would be greater than the dimension of A^2 which fits the property in the question.
     
  6. Feb 22, 2010 #5
    You cannot construct a counterexample for the first part since it's true, but for exactly the reason you mentioned it works as a counterexample for 2 (nullity of A is 1 and nullity of A^2 is 2).

    By k I was merely referring to an arbitrary field, but maybe you only work with real or complex matrices. Or perhaps it was just my notation that threw you off. Anyway if you think a matrix has real entries, then set [itex]k=\mathbb{R}[/itex]. If you think it can have complex entries set [itex]k=\mathbb{C}[/itex]. If you're working over an arbitrary field, then just consider k as an arbitrary field (some linear algebra courses work solely in subfields of [itex]\mathbb{C}[/itex], while other work in general fields).

    I don't know if you have covered linear transformations yet (I sort of assumed since you mentioned the range which usually refers to the range of the linear transformation). If you have then by T I meant the linear transformation determined by the matrix A. What this means is that T is the function:
    [tex]T(X) = AX[/tex]
    where X is a n-dimensional vector. This is a function [itex]\mathbb{R}^n \to \mathbb{R}^n[/itex] if we deal with real numbers only.

    Anyway there is no reason to work with linear transformations if you feel more comfortable working with matrices (my personal preference is just to work with linear transformations). Let me try to restate my hint in the language of matrices.

    The important idea is that you can't possibly increase the rank of A by multiplying it by another matrix. If A has rank r, then we can row-reduce A to a matrix C whose last n-r rows are zero. That is we can write A as A=TC where T is a product of elementary matrices and C has its last n-r rows 0, but then CA has its last n-r rows 0 and therefore CA has rank less than or equal to r. T is invertible so multiplication by it doesn't change the rank and therefore TCA = AA = A^2 has rank less than or equal to r=rank(A).
     
  7. Feb 22, 2010 #6
    I have not learned about linear transformations yet.

    However, what you said in terms of matrices doesn't seem to really prove to how the rank(A^2) <= rank(A). To me you seem to have proved the statement rank(C) <= rank(A) where C is the reduced echelon form of A.
     
  8. Feb 22, 2010 #7
    You say, "That is we can write A as A=TC where T is a product of elementary matrices and C has its last n-r rows 0, but then CA has its last n-r rows 0 and therefore CA has rank less than or equal to r."

    You begin talking about CA so I am assuming that you got CA from the equation A=TC which you gave, but wouldn't this not really work since the equation really is (C^-1)A?
     
  9. Feb 23, 2010 #8
    No I'm actually talking about CA. The point is that TCA = A^2 so we want to gain some knowledge of CA. From the formula for matrix multiplication it should be possible to show that the ith row of C is 0 imply that the ith row of CA will be 0.

    What this proves is merely that CA has at most r rows that are non-zero, so its rank can't possibly be larger than r. Multiplication of a matrix by an invertible matrix doesn't change its rank so TCA also can't have rank larger than r, or in other words [itex]rank(TCA)\leq r[/itex]. This is what we wanted since TC = A so TCA= A^2 and r=rank(A) so substituting these we get,
    [tex]rank(A^2) = rank(TCA) \leq r = rank(A)[/tex]
     
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