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smr_kml

How to know if the matrix is on-to and one-to-one?! Is there an easy way to know that?!

  • one year ago
  • one year ago

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  1. cibychak
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    onto and one to one are not props of a matrix.they belong to functions

    • one year ago
  2. TuringTest
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    talking about matrix transformations I assume

    • one year ago
  3. smr_kml
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    it also belongs to matrix in linear algebra!

    • one year ago
  4. Jemurray3
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    That's not true, you can assign such properties to any transformation.

    • one year ago
  5. TuringTest
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    http://www.mast.queensu.ca/~spencer/apsc174/11onto.pdf

    • one year ago
  6. cibychak
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    oh.thanks guys,i'm just a high school level mathematician :)

    • one year ago
  7. smr_kml
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    @TuringTest Can you explain it to me?!

    • one year ago
  8. Jemurray3
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    @TuringTest it's all yours :)

    • one year ago
  9. TuringTest
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    I would just go down the checklist for a given transformation is n<m ? if so then not onto if the column space is \(\mathbb R^m\) it is onto etc. I was really hoping you'd take it over @Jemurray3; I've never even heard of the term "onto" until recently

    • one year ago
  10. TuringTest
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    I also have class in about 5 min...

    • one year ago
  11. smr_kml
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    Sorry for wasting your time! @TuringTest

    • one year ago
  12. TuringTest
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    not wasted at all, I'll be happy to see other's explanations and save the medal for someone who deserves it !

    • one year ago
  13. slaaibak
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    to check one-to-one, just check if it has only the trivial solution. that is x=0 is the only solution to Ax=0

    • one year ago
  14. smr_kml
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    So when it has only solution Ax = 0, it will be one to one?!

    • one year ago
  15. TuringTest
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    sorry no

    • one year ago
  16. smr_kml
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    ok and if n<m so it's not onto? right?

    • one year ago
  17. TuringTest
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    the null space must be zero, so no solutions

    • one year ago
  18. smr_kml
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    I get confused

    • one year ago
  19. TuringTest
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    that part is right Ax=0 has not solutions but the trivial one if A is invertable if A is invertable the transformation is one-to-one

    • one year ago
  20. TuringTest
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    has no*

    • one year ago
  21. smr_kml
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    invertible means the det. is not equal zero?

    • one year ago
  22. slaaibak
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    the dimension of the nullspace should be zero...I think even if the dimension of the nullspace is zero, it still contains the zero vector and hence the nullspace can't be zero.. or am I wrong?

    • one year ago
  23. TuringTest
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    right it also implies like 12 other equivalent facts about the matrix

    • one year ago
  24. TuringTest
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    Nullspace is the dimension of the solution space for Ax=0 and the zero vec has no dimension, so I think that is wrong slaaibak I make a lot of mistakes with this stuff too though

    • one year ago
  25. smr_kml
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    :(

    • one year ago
  26. slaaibak
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    Nullspace is the solution space to Ax=0. nullity is the dimension of the nullspace

    • one year ago
  27. Jemurray3
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    Then I shall supplement: If we have an nxm matrix, it maps m-dimensional vectors into n-dimensional space. If our transformation is onto, that means that every vector in n-dimensional space is the image of some vector in m-dimensional space under our transformation. Put another way, if we call our matrix A, then we have at least one m-dimensional vector x such that Ax = b for any vector b in R^n. To check for this, try to see if the (m) columns of A span n-dimensional space. I assume you are familiar with the concept of spanning sets, so I'll just say that obviously we must have that m >= n (a necessary, but not sufficient condition). We also require that at least n of the m columns are linearly independent. A one-to-one transformation as stated above means that every vector in m-dimensional space is associated to one and only one vector in n-dimensional space. Therefore, we should be able to go from one vector to another and back with no ambiguity, i.e. the transformation is invertible. For this, we'd need a square matrix whose determinant is not equal to zero.

    • one year ago
  28. smr_kml
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    right

    • one year ago
  29. TuringTest
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    ah damn

    • one year ago
  30. AravindG
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    am i late?

    • one year ago
  31. smr_kml
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    No

    • one year ago
  32. TuringTest
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    Good explanation JM I'm off to class somebody take away my medals and give them to slai and JM, I don't know why I have the most :/ later guys!

    • one year ago
  33. AravindG
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    thats nic of u @TuringTest

    • one year ago
  34. slaaibak
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    correct me if I'm wrong, but a 3 * 2 matrix transformation can also be one-to-one? so checking if its invertible is not the only criteria

    • one year ago
  35. smr_kml
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    if it's invertible so it's one-to-one?

    • one year ago
  36. slaaibak
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    non-square matrices aren't invertible.

    • one year ago
  37. smr_kml
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    I know that and if the det is = 0 so it's not invertible

    • one year ago
  38. Jemurray3
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    Invertible matrices are one-to-one, yes. And @slaaibak can you come up with an example of one?

    • one year ago
  39. smr_kml
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    how about onto?

    • one year ago
  40. slaaibak
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    1 2 2 1 2 1

    • one year ago
  41. smr_kml
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    This is not one - to -one right?!

    • one year ago
  42. slaaibak
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    Why not? Solve it for Ax=0

    • one year ago
  43. Jemurray3
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    It just so happens that invertible matrices are also onto, but a matrix doesn't necessarily HAVE to be invertible in order to be onto.

    • one year ago
  44. smr_kml
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    how can I know if it's onto?

    • one year ago
  45. slaaibak
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    Check if it's consistent for Ax=b. for every b?

    • one year ago
  46. Jemurray3
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    @slaaibak good example. Invertibility is a sufficient, but not necessary condition. I'll revise to say linear independence of columns.

    • one year ago
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