Expert: Sherry Wallin - 10/15/2010

Question
Dear Sherry  Wallin,

Can you clear my doubts on the following questions which I do not understand.

I cannot figure out why when we map f:real number to M(R) given by
f(r)=Matrix [0 0, -r r] are injective but not surjective?

Then why when we map f:R- M(R) given by
f(r)=Matrix [0 0, 0 r] is injective and surjective?

Is there some ways we can see the difference between the two. Seek for your help to enlighten me.

Thanks.

Regards
Raymond


Answer
Raymond~


*I cannot figure out why when we map f:real number to M(R) given by
f(r)=Matrix [0 0, -r r] are injective but not surjective?*

This function is injective because If f(a) = f(b) then a = b. To see this consider this mini proof:

Suppose f(a) = [00,-aa} and f(b) = [00,-bb]. Two matrices are equal if their corresponding entries are equal meaning -a = -b and a = b thus this function is injective.

Notice you have +-r in your matrix, that means in the rule you take r from the domain to r in the range and r in the domain  to -r in the range. If you try to go back to the domain r goes back to r but -r has no where defined to go, this is why this function is not surjective.

On the other hand the 2nd function is injective because if f(a) = f(b) the a = b and it is also surjective because the r in the range will be the image of the r in the domain.

I hope this has helped you understand a bit. You can also for the first function find a counter example that shows that the function isn't surjective, like for instance let r = 2 then r gets mapped to [0 0,-2 2] but what is -2 the image of in the domain?

Math Prof