Matrix A is invertible iff A is onto?

[Aziza]

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?

 

[Ray Vickson]

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?

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

 

[voko]

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.