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Singular matrix theory

  1. Jan 15, 2013 #1

    joshmccraney

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    hey guys


    given [itex]Ax=B[/itex] where A is a square matrix and x and B are vectors, can anyone tell me why a singular matrix (that is, the determinant = 0) implies one of two situations: infinite solutions or zero solutions? a proof would be nice. i read through pauls notes but there was no proof.

    thanks all!
     
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  3. Jan 15, 2013 #2

    Simon Bridge

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    If ##\mathbf{A}\vec{x}=\vec{y}## then ##\mathbf{A}^{-1}\vec{y}=\vec{x}## provided the inverse exists.

    If the matrix ##\mathbf{A}## is singular, it does not have an inverse.
    Another name for it is "degenerate".

    What does that tell you about the solutions?
    (Think about it in terms of solving simultaneous equations.)
     
  4. Jan 15, 2013 #3

    HallsofIvy

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    An n by n square matrix represents a linear transformation, A, from Rn to Rn. If it is "non-singular", then it maps all of Rn to all of Rn. That is, it is a "one to one" mapping- given any y in Rn there exist a unique x in Rn such that Ax= y.

    But we can show that, for any linear transformation, A, from one vector space, U, to another, V, the "image" of A, that is, the set of all vectors y, of the form y= Ax for some x, is a subspace of V and that the "null space" of A, the set of all vectors, x, in U such that Ax= 0, is a subspace of U. Further, we have the "dimension theorem". If "m" is dimension of the image of A (called the "rank" of A) and "n" is the dimension of the nullspace of A (called the "nullity" of A) then m+ n is equal to the dimension of V. In particuar, if U and V have the same dimension, n, and the rank of A is m with m< n, then the nullity of A= m-n> 0.

    It is further true that if A(u)= v and u' is in the nullspace of A then A(u+ u')= A(u)+ A(u')= v+ 0= v.

    The result of all of that is this: If A is a singular linear transformation from vector space U to vector space V, then it maps U into some subspace of V. If y is NOT in that subspace then there is NO x such that Ax= y. If y is in that subspace then there exist x such that Ax= y but also, for any v in the nullity of A (which has non-zero dimension and so contains an infinite number of vectors) A(x+ v)= y also so there exist an infinite number of such vectors.
     
  5. Jan 16, 2013 #4

    joshmccraney

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    thanks this makes tons of sense!
     
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