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Homework Help: Non Invertable matrix

  1. Nov 4, 2008 #1
    1. The problem statement, all variables and given/known dataLet A be an nxn matrix. If A is row equivalent to a matrix B and there is a non-zero column matrix C such that BC=0, prove that A is singular

    2. Relevant equations

    3. The attempt at a solutionIm not quite sure but since B and A are row equivalent than there reduced echelon forms will be the same ? and therefore AC=0 and i was wondering if since A multipliyed by a non zero matrix equals zero does that mean that A in singular?
  2. jcsd
  3. Nov 4, 2008 #2


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    AC=0 yes... try a proof by contradiction; assume that A is non-singular and hence invertible...what happens when you multiply both sides of AC=0 by the inverse of A?
  4. Nov 4, 2008 #3
    Ok so if you Multiply both sides by Ainverse you would get AinverseAC=Ainverse0, which equals IC=0 which is C=0 and since C is not 0 this is a contradiction and therefore proves A is singular?
  5. Nov 4, 2008 #4


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    Technically, it only proves that the inverse of A does not exist, but there is a theorem that tells you any square matrix is singular iff it has no inverse, so assuming you are allowed to use that theorem, then you've shown A is singular.
  6. Nov 4, 2008 #5


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    What definition are you using for "singular"? Gabbagabbahey seems to be interpreting "singular" as meaning the matrix has determinant 0. I would tend to define "singular" as meaning "non-invertible" but, as gabbagabbahey says, they are equivalent.
  7. Nov 4, 2008 #6
    The definition i learned was that singular means non invertable
  8. Nov 4, 2008 #7
    Singular means non-invertible, and non-invertible implies that its determinant and the product of its eigenvalues is zero.
  9. Nov 4, 2008 #8
    The definition that I learned for a singular matrix A is that A's reduced row echelon form is NOT the identity matrix.
  10. Nov 4, 2008 #9
    There are many ways to calculate inverses. Your definition follows with the computation of the inverse of A by doing row operations to [A|I] until it becomes [I|A^-1] (assuming that A is invertible). Another way of calculating inverses is by dividing the cofactor matrix of A transpose by its determinant. Thus, if the determinant is zero, the inverse does not exist.
    Last edited: Nov 4, 2008
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