Let 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
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The Attempt at a Solution
Im 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?
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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?
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?
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.
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.
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.