Question

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

Answer 1

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

Answer 2

talking about matrix transformations I assume

Answer 3

it also belongs to matrix in linear algebra!

Answer 4

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

Answer 5

http://www.mast.queensu.ca/~spencer/apsc174/11onto.pdf

Answer 6

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

Answer 7

@TuringTest Can you explain it to me?!

Answer 8

@TuringTest it's all yours :)

Answer 9

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

Answer 10

I also have class in about 5 min...

Answer 11

Sorry for wasting your time! @TuringTest

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

sorry no

Answer 16

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

Answer 17

the null space must be zero, so no solutions

Answer 18

I get confused

Answer 19

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

Answer 20

has no*

Answer 21

invertible means the det. is not equal zero?

Answer 22

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?

Answer 23

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

Answer 24

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

Answer 25

:(

Answer 26

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

Answer 27

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.

Answer 28

right

Answer 29

ah damn

Answer 30

am i late?

Answer 31

No

Answer 32

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!

Answer 33

thats nic of u @TuringTest

Answer 34

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

Answer 35

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

Answer 36

non-square matrices aren't invertible.

Answer 37

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

Answer 38

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

Answer 39

how about onto?

Answer 40

1 2
2 1
2 1

Answer 41

This is not one - to -one
right?!

Answer 42

Why not? Solve it for Ax=0

Answer 43

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.

Answer 44

how can I know if it's onto?

Answer 45

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

Answer 46

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