anonymous
  • anonymous
How to know if the matrix is on-to and one-to-one?! Is there an easy way to know that?!
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
onto and one to one are not props of a matrix.they belong to functions
TuringTest
  • TuringTest
talking about matrix transformations I assume
anonymous
  • anonymous
it also belongs to matrix in linear algebra!

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anonymous
  • anonymous
That's not true, you can assign such properties to any transformation.
TuringTest
  • TuringTest
http://www.mast.queensu.ca/~spencer/apsc174/11onto.pdf
anonymous
  • anonymous
oh.thanks guys,i'm just a high school level mathematician :)
anonymous
  • anonymous
@TuringTest Can you explain it to me?!
anonymous
  • anonymous
@TuringTest it's all yours :)
TuringTest
  • TuringTest
I would just go down the checklist for a given transformation is n@Jemurray3; I've never even heard of the term "onto" until recently
TuringTest
  • TuringTest
I also have class in about 5 min...
anonymous
  • anonymous
Sorry for wasting your time! @TuringTest
TuringTest
  • TuringTest
not wasted at all, I'll be happy to see other's explanations and save the medal for someone who deserves it !
slaaibak
  • 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
anonymous
  • anonymous
So when it has only solution Ax = 0, it will be one to one?!
TuringTest
  • TuringTest
sorry no
anonymous
  • anonymous
ok and if n
TuringTest
  • TuringTest
the null space must be zero, so no solutions
anonymous
  • anonymous
I get confused
TuringTest
  • 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
TuringTest
  • TuringTest
has no*
anonymous
  • anonymous
invertible means the det. is not equal zero?
slaaibak
  • 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?
TuringTest
  • TuringTest
right it also implies like 12 other equivalent facts about the matrix
TuringTest
  • 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
anonymous
  • anonymous
:(
slaaibak
  • slaaibak
Nullspace is the solution space to Ax=0. nullity is the dimension of the nullspace
anonymous
  • anonymous
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.
anonymous
  • anonymous
right
TuringTest
  • TuringTest
ah damn
AravindG
  • AravindG
am i late?
anonymous
  • anonymous
No
TuringTest
  • 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!
AravindG
  • AravindG
thats nic of u @TuringTest
slaaibak
  • 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
anonymous
  • anonymous
if it's invertible so it's one-to-one?
slaaibak
  • slaaibak
non-square matrices aren't invertible.
anonymous
  • anonymous
I know that and if the det is = 0 so it's not invertible
anonymous
  • anonymous
Invertible matrices are one-to-one, yes. And @slaaibak can you come up with an example of one?
anonymous
  • anonymous
how about onto?
slaaibak
  • slaaibak
1 2 2 1 2 1
anonymous
  • anonymous
This is not one - to -one right?!
slaaibak
  • slaaibak
Why not? Solve it for Ax=0
anonymous
  • anonymous
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.
anonymous
  • anonymous
how can I know if it's onto?
slaaibak
  • slaaibak
Check if it's consistent for Ax=b. for every b?
anonymous
  • anonymous
@slaaibak good example. Invertibility is a sufficient, but not necessary condition. I'll revise to say linear independence of columns.

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