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Linear Algebra Vector Spaces: Prove equivalence

  1. Feb 28, 2010 #1
    1. The problem statement, all variables and given/known data
    Prove that the following are equivalent:
    1. N(A)=0
    2. A is nonsingular
    3. Ax=b has a unique solution for each b that exists in R^n.

    2. Relevant equations

    3. The attempt at a solution
    I think you prove this by showing that 1 implies 2, 2 implies 3, & 3 implies 1.
    But after that I don't know how to prove that.
  2. jcsd
  3. Feb 28, 2010 #2
    Start by trying one of the three implications you must show.
    For example, for 1 implies 2, what does N(A) = 0 tell you about the equation Ax=0? Can you imply this to determine that A is invertible?
  4. Feb 28, 2010 #3
    if N(A)=0 then in the equation Ax=0, x equals 0. But then how do you show that A is invertible from that?
  5. Feb 28, 2010 #4
    What does Ax=0 tell you about the columns of A? Let a1, ...., an be some vector in A, and the components of x are x1,..., xn. Then from Ax=0, you get equations of the form a1x1 + ..... + anxn = 0, and from N(A)=0 you get that all the xi's are 0. What can you conclude from this?
  6. Feb 28, 2010 #5
    Ok I think I figured the first part out:
    The columns of A are linearly independent.
    So, A is row equivalent to the identity matrix, and Ax=0 and Ix=0 have only the solution, x=0. So, the A=E1E2...Ek, which says that the product of invertiable matrices is invertiable and E is invertiable, so A is invertable.

    So, then How do you imply that 2 equals 3, and 3 equals 1?
  7. Feb 28, 2010 #6
    Give them a try, pick one of the two. What have you done so far?

    For example, 2 to 3:
    A is non singular, So A-1 exists. Then multiply Ax=b by A-1 . What does this imply? Is this unique?
    Last edited: Feb 28, 2010
  8. Feb 28, 2010 #7
    And the other directions are probably similar. It's all a matter of shuffling between invertibility, linear dependence, etc.
  9. Feb 28, 2010 #8
    ok, i think that for 2 implies 3. A being nonsingular says that A inverse exists. So, then you can show that x= (inverseA)*B which proves #3. Is this right?

    Now, the one I am stuck on is 3 implies 1. Any hints to get me started?
  10. Feb 28, 2010 #9
    Yes, you are correct and it is easy to understand that this is in fact a unique solution (since for all other solutions you can simply left multiply by A-1 to get the same conclusion.

    I have not yet thought about 3 to 1. Think about it. Maybe there is some trick you can use from what you already proved in this problem.

    You are assuming that Ax=b has a unique solution. Then b=0 also has a unique solution, and you know what it is.
  11. Feb 28, 2010 #10
    I still have no idea on the last one. I have looked all through my book...
  12. Mar 1, 2010 #11


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    That shows that Ax=b has a solution. You should prove that this solution is unique.
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