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Nullspace of non-zero 4x4 matrix

  1. Apr 27, 2005 #1
    Hello everyone,
    it's final's time next week :cry: , so I will be posting here more often than usual :biggrin:
    Here is one problem I came across when doing review:

    The nullspace of non-zero 4x4 matrix cannot contain a set of 4 lin. indep. vectors. (T/F)

    The way I was thinking is that if I solve a homogeneous s-m with this matrix, and if the dimension of nullspace is 4, that means that there have to be 4 free variables in the homogeneous s-m, but matrix is just 4x4.
    And then there is this rank-nullity theorem that n = rank(A) + nullity(A), so in this case rank(A) = 0, is it ever possible? My guess is not.
    Does the same hold for dim of nullspace (nullity): it has to have at least one solution (trivial, where everyting = 0, but that does not mean that the nullity is an empty set!) ?
    Is it correct?

    Thanks in advance!
    Last edited: Apr 27, 2005
  2. jcsd
  3. Apr 27, 2005 #2


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    Have you learnt what bases are? If you have 4 linearly independent (nonzero) vectors, they form a basis for the space. The action of the matrix on any vector in your space is determined by the action on the 4 basis-vectors.
    What does that mean for Av (A = matrix, v=vector), where v is any vector.

    Ofcourse, the rank equation works fine too. What can you conclude from A if its rank is zero?
  4. Apr 27, 2005 #3
    Yeah, I know about bases. And this part I understand.

    this is a bit blurry.

    I can't say it's non-existent, but if it's rank = 0, dimens. of rowspace = 0!
    Even if all the rows/cols are lin. dep. rank would be at least 1.
    But if it's zero I dunno, it cannot be zero-vector, since it's one of the conditions. That's why ask.
    n = dimesion of nullspace? but then what :confused:
  5. Apr 27, 2005 #4
    Precisely. The rank of a non-zero matrix is never zero (the ONLY vector space of dimension zero is the set containing only the zero vector, and its "basis" is the empty set). Thus if you find that null(A)=4 => rank(A)=0, you've answered the question.
  6. Apr 27, 2005 #5


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    Equivalently: a 4 by 4 matrix maps vectors from R4 to R4- which is 4 dimensional. The null space of any linear operator (or matrix) is a subspace of the domain space (here R4). If it contains 4 linearly independent vectors the its dimension is at least 4. In fact, the only 4 dimensional subspace of R4 is R4 itself. Saying the null space contains 4 independent vectors is simply saying the null space is R4 itself: the matrix maps every vector into the 0 vector and so is the 0 matrix.
  7. Apr 27, 2005 #6
    Then a follow-up question: does nullity has to be at least 1? since it always has a solution as zero-vector? Which I assume to be non-empty, so it's 1.

    On one hand it makes sense as I described it above, on the other hand, it doesn't: by rank-nullity thm that would mean that matrix contains at least one lin. dep. row/col: n = rank(A) + nullity(A).
    But looking at a matrix A with all cols/rows lin. indep: rank(A) = n, which would mean nullity(A) = 0.

    Is it a contradiction? Or am I missing something?

    Thanks again.
  8. Apr 27, 2005 #7
    The nullity of a matrix is the dimension of the nullspace of the matrix. The nullspace of any matrix contains the zero vector. In the case that the nullspace is the zero vector (and only in this case), the nullity of the matrix is 0, because (as I mentioned in my last post), the dimension of the vector space containing only the zero vector is zero (a basis for the vector space has zero elements - ie. the basis is the empty set).
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