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

This Question is Closed

cibychak
 3 years ago
Best ResponseYou've already chosen the best response.0onto and one to one are not props of a matrix.they belong to functions

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1talking about matrix transformations I assume

smr_kml
 3 years ago
Best ResponseYou've already chosen the best response.0it also belongs to matrix in linear algebra!

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.2That's not true, you can assign such properties to any transformation.

cibychak
 3 years ago
Best ResponseYou've already chosen the best response.0oh.thanks guys,i'm just a high school level mathematician :)

smr_kml
 3 years ago
Best ResponseYou've already chosen the best response.0@TuringTest Can you explain it to me?!

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.2@TuringTest it's all yours :)

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1I 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

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1I also have class in about 5 min...

smr_kml
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry for wasting your time! @TuringTest

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1not wasted at all, I'll be happy to see other's explanations and save the medal for someone who deserves it !

slaaibak
 3 years ago
Best ResponseYou've already chosen the best response.2to check onetoone, just check if it has only the trivial solution. that is x=0 is the only solution to Ax=0

smr_kml
 3 years ago
Best ResponseYou've already chosen the best response.0So when it has only solution Ax = 0, it will be one to one?!

smr_kml
 3 years ago
Best ResponseYou've already chosen the best response.0ok and if n<m so it's not onto? right?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1the null space must be zero, so no solutions

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1that part is right Ax=0 has not solutions but the trivial one if A is invertable if A is invertable the transformation is onetoone

smr_kml
 3 years ago
Best ResponseYou've already chosen the best response.0invertible means the det. is not equal zero?

slaaibak
 3 years ago
Best ResponseYou've already chosen the best response.2the 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
 3 years ago
Best ResponseYou've already chosen the best response.1right it also implies like 12 other equivalent facts about the matrix

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1Nullspace 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

slaaibak
 3 years ago
Best ResponseYou've already chosen the best response.2Nullspace is the solution space to Ax=0. nullity is the dimension of the nullspace

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.2Then I shall supplement: If we have an nxm matrix, it maps mdimensional vectors into ndimensional space. If our transformation is onto, that means that every vector in ndimensional space is the image of some vector in mdimensional space under our transformation. Put another way, if we call our matrix A, then we have at least one mdimensional 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 ndimensional 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 onetoone transformation as stated above means that every vector in mdimensional space is associated to one and only one vector in ndimensional 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.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1Good 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
 3 years ago
Best ResponseYou've already chosen the best response.0thats nic of u @TuringTest

slaaibak
 3 years ago
Best ResponseYou've already chosen the best response.2correct me if I'm wrong, but a 3 * 2 matrix transformation can also be onetoone? so checking if its invertible is not the only criteria

smr_kml
 3 years ago
Best ResponseYou've already chosen the best response.0if it's invertible so it's onetoone?

slaaibak
 3 years ago
Best ResponseYou've already chosen the best response.2nonsquare matrices aren't invertible.

smr_kml
 3 years ago
Best ResponseYou've already chosen the best response.0I know that and if the det is = 0 so it's not invertible

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.2Invertible matrices are onetoone, yes. And @slaaibak can you come up with an example of one?

smr_kml
 3 years ago
Best ResponseYou've already chosen the best response.0This is not one  to one right?!

slaaibak
 3 years ago
Best ResponseYou've already chosen the best response.2Why not? Solve it for Ax=0

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.2It just so happens that invertible matrices are also onto, but a matrix doesn't necessarily HAVE to be invertible in order to be onto.

smr_kml
 3 years ago
Best ResponseYou've already chosen the best response.0how can I know if it's onto?

slaaibak
 3 years ago
Best ResponseYou've already chosen the best response.2Check if it's consistent for Ax=b. for every b?

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.2@slaaibak good example. Invertibility is a sufficient, but not necessary condition. I'll revise to say linear independence of columns.
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