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How to know if the matrix is onto and onetoone?!
Is there an easy way to know that?!
 one year ago
 one year ago
How to know if the matrix is onto and onetoone?! Is there an easy way to know that?!
 one year ago
 one year ago

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

TuringTestBest ResponseYou've already chosen the best response.1
talking about matrix transformations I assume
 one year ago

smr_kmlBest ResponseYou've already chosen the best response.0
it also belongs to matrix in linear algebra!
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.2
That's not true, you can assign such properties to any transformation.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
http://www.mast.queensu.ca/~spencer/apsc174/11onto.pdf
 one year ago

cibychakBest ResponseYou've already chosen the best response.0
oh.thanks guys,i'm just a high school level mathematician :)
 one year ago

smr_kmlBest ResponseYou've already chosen the best response.0
@TuringTest Can you explain it to me?!
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.2
@TuringTest it's all yours :)
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
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

TuringTestBest ResponseYou've already chosen the best response.1
I also have class in about 5 min...
 one year ago

smr_kmlBest ResponseYou've already chosen the best response.0
Sorry for wasting your time! @TuringTest
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
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

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

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

smr_kmlBest ResponseYou've already chosen the best response.0
ok and if n<m so it's not onto? right?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
the null space must be zero, so no solutions
 one year ago

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

smr_kmlBest ResponseYou've already chosen the best response.0
invertible means the det. is not equal zero?
 one year ago

slaaibakBest ResponseYou've already chosen the best response.2
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

TuringTestBest ResponseYou've already chosen the best response.1
right it also implies like 12 other equivalent facts about the matrix
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
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

slaaibakBest ResponseYou've already chosen the best response.2
Nullspace is the solution space to Ax=0. nullity is the dimension of the nullspace
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.2
Then 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.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
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

AravindGBest ResponseYou've already chosen the best response.0
thats nic of u @TuringTest
 one year ago

slaaibakBest ResponseYou've already chosen the best response.2
correct 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
 one year ago

smr_kmlBest ResponseYou've already chosen the best response.0
if it's invertible so it's onetoone?
 one year ago

slaaibakBest ResponseYou've already chosen the best response.2
nonsquare matrices aren't invertible.
 one year ago

smr_kmlBest ResponseYou've already chosen the best response.0
I know that and if the det is = 0 so it's not invertible
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.2
Invertible matrices are onetoone, yes. And @slaaibak can you come up with an example of one?
 one year ago

smr_kmlBest ResponseYou've already chosen the best response.0
This is not one  to one right?!
 one year ago

slaaibakBest ResponseYou've already chosen the best response.2
Why not? Solve it for Ax=0
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.2
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

smr_kmlBest ResponseYou've already chosen the best response.0
how can I know if it's onto?
 one year ago

slaaibakBest ResponseYou've already chosen the best response.2
Check if it's consistent for Ax=b. for every b?
 one year ago

Jemurray3Best 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.
 one year ago
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