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Matrix A is invertible iff A is onto?

  1. Oct 6, 2012 #1
    According to my professor,
    For an nxn matrix A that corresponds to a linear transformation, "A is invertible" is equivalent to "A is onto".
    Also "A is invertible" is equivalent to "A is one-to-one"


    But then "A is onto" should be equivalent to "A is one-to-one", but is this always the case for linear transformations? I mean, if a linear transformation is onto, is it necessarily one-to one? And if a lin transf is one-to one, is it necessarily onto?
     
  2. jcsd
  3. Oct 6, 2012 #2

    Ray Vickson

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    If ##A:F^n \rightarrow F^n## (F = ℝ or ℂ) is linear and onto, is it 1:1? Well, assume there exist ##x_1 \neq x_2 \in F^n ## giving Ax1 = Ax2. Then we have ##Ax = 0,## where x = x1-x2. Since the vector x is not the zero vector, that means that the columns of A are linearly dependent, and that means that the range of A is spanned by fewer than n columns, and that means that A is not onto. That is a contradiction to the assumption that A is onto. Therefore, A is 1:1.

    You should be able to go the other way as well.

    RGV
     
  4. Oct 6, 2012 #3
    This is the case when the spaces involved are of the same dimension. Say a linear transformation maps X onto Y, and dim X = dim Y, prove that by considering the basis in Y.
     
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