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

- anonymous

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

- schrodinger

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

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

- TuringTest

talking about matrix transformations I assume

- anonymous

it also belongs to matrix in linear algebra!

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

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

- TuringTest

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

- anonymous

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

- anonymous

@TuringTest Can you explain it to me?!

- anonymous

@TuringTest it's all yours :)

- 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

I also have class in about 5 min...

- anonymous

Sorry for wasting your time! @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

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

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

- TuringTest

sorry no

- anonymous

ok and if n

- TuringTest

the null space must be zero, so no solutions

- anonymous

I get confused

- 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

has no*

- anonymous

invertible means the det. is not equal zero?

- 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

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

- 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

:(

- slaaibak

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

- 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

right

- TuringTest

ah damn

- AravindG

am i late?

- anonymous

No

- 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

thats nic of u @TuringTest

- 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

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

- slaaibak

non-square matrices aren't invertible.

- anonymous

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

- anonymous

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

- anonymous

how about onto?

- slaaibak

1 2
2 1
2 1

- anonymous

This is not one - to -one
right?!

- slaaibak

Why not? Solve it for Ax=0

- 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

how can I know if it's onto?

- slaaibak

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

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.