smr_kml 3 years ago How to know if the matrix is on-to and one-to-one?! Is there an easy way to know that?!

1. cibychak

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

2. TuringTest

talking about matrix transformations I assume

3. smr_kml

it also belongs to matrix in linear algebra!

4. Jemurray3

That's not true, you can assign such properties to any transformation.

5. TuringTest
6. cibychak

oh.thanks guys,i'm just a high school level mathematician :)

7. smr_kml

@TuringTest Can you explain it to me?!

8. Jemurray3

@TuringTest it's all yours :)

9. TuringTest

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

10. TuringTest

I also have class in about 5 min...

11. smr_kml

Sorry for wasting your time! @TuringTest

12. TuringTest

not wasted at all, I'll be happy to see other's explanations and save the medal for someone who deserves it !

13. slaaibak

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

14. smr_kml

So when it has only solution Ax = 0, it will be one to one?!

15. TuringTest

sorry no

16. smr_kml

ok and if n<m so it's not onto? right?

17. TuringTest

the null space must be zero, so no solutions

18. smr_kml

I get confused

19. TuringTest

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

20. TuringTest

has no*

21. smr_kml

invertible means the det. is not equal zero?

22. slaaibak

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?

23. TuringTest

right it also implies like 12 other equivalent facts about the matrix

24. TuringTest

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

25. smr_kml

:(

26. slaaibak

Nullspace is the solution space to Ax=0. nullity is the dimension of the nullspace

27. Jemurray3

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.

28. smr_kml

right

29. TuringTest

ah damn

30. AravindG

am i late?

31. smr_kml

No

32. TuringTest

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!

33. AravindG

thats nic of u @TuringTest

34. slaaibak

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

35. smr_kml

if it's invertible so it's one-to-one?

36. slaaibak

non-square matrices aren't invertible.

37. smr_kml

I know that and if the det is = 0 so it's not invertible

38. Jemurray3

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

39. smr_kml

40. slaaibak

1 2 2 1 2 1

41. smr_kml

This is not one - to -one right?!

42. slaaibak

Why not? Solve it for Ax=0

43. Jemurray3

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.

44. smr_kml

how can I know if it's onto?

45. slaaibak

Check if it's consistent for Ax=b. for every b?

46. Jemurray3

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