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Left invertible

  1. Feb 28, 2010 #1
    I found on wikipedia that the following statements are equivalant:
    1. Matrix A is left invertible
    2. Ax=0 => x=0

    I couldn't find the proof so I try to do it myself.

    From 1. to 2. is easy. Assume A is left invertible. If Ax=0, then x=Ix=A-1Ax=A-10 = 0 .

    I can't figure out how to do 2=>1. Any help is appreciated.

    Things that might prove useful:
    - A is injective
    - dim(ker(A))=0
    - rank(A)=n (if A is an m x n matrix)
    - [tex]n\leq m[/tex]
     
  2. jcsd
  3. Mar 1, 2010 #2
    If the solutions to [tex]Ax=0[/tex] are only the trivial one, then its RREF is the identity matrix, thus it is invertible.

    And another:
    [tex]Ax=0[/tex] are only the trivial solution, then [tex]\textrm{det}A\ne 0[/tex] thus it is invertible.
     
  4. Mar 1, 2010 #3
    Why and why?

    Note that A is not (necessarily) a square matrix
     
  5. Mar 1, 2010 #4
    Only square matrices can be inverted, thus the statements only make sense for them.

    All of the statements above I used can be shown to be equivalent, but I need to know what level you are at so I can know which equivalent statements to use.

    [tex]Ax=0 \Rightarrow \textrm{RREF}(A)=I[/tex] follows from the rank-nullity theorem, have you seen this theorem?

    [tex]Ax=0 \Rightarrow \textrm{det}A\ne0[/tex] follows from properties of the determinant and also properties of elementary matrices, are you familiar with these?
     
  6. Mar 1, 2010 #5
    This is simply not true. I have a better source ("Linear algebra done wrong"), but I'll use wikipedia to quote:
    I underlined what I'm trying to proof.

    I know the theorems but I don't see how they are applicable to a non-square matrix.
     
  7. Mar 1, 2010 #6
    Ah I see, just a left-inverse, in that case then you must avoid mentioning determinants although my first statement still holds.
     
  8. Mar 1, 2010 #7
    That statement does hold indeed, and helped me to get:
    rank(A) + dim(ker(A)) = rank(A) = n
    rank(At) + dim(ker(At))= rank(A) + dim(ker(At)) = n + dim(ker(At)) = m
    Hence: [tex]m\geq n[/tex]

    But it doesn't get me any further.
     
  9. Mar 2, 2010 #8
    But [tex]Ax=0[/tex] having only the trivial solution [tex]\Rightarrow \textrm{ker}A={ \emptyset }[/tex] which again seems to imply that its square... I don't know, I'll have to have a think about it.
     
  10. Mar 2, 2010 #9
    It's not true that Ker(A)= empty set, because [tex]0 \in \textrm{ker}(A)[/tex] . And why do you think it implies that it's square?
     
  11. Mar 2, 2010 #10
    Ax = 0 => x = 0
    ==> rank(A) = n = rank(ATA)
    ==> ATA is invertible.

    Using this you should get the ball rolling... Or I may be wrong.
     
  12. Mar 3, 2010 #11
    (ATA)-1ATA = I

    That's it, thanks :)
     
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