I'm pretty sure it's p^n.
I have the proof of dimV=m and dimU=n meaning dimHom(V,U)=mn. How do I transform this proof to the one I want?
I need help with this:
The set Hom(V,W) is the collection of all linear transformations from the F-space V to the F-space W. Suppose that F,V, and W are all finite. Suppose that F=Zp for some prime p, that V is n-dimensional over F, and W is n-dimensional over F. How many elements does Hom(V,W) have?
Thank you for any help.
Okay, I think I understand know.
According to you |Hom(V,W)|=nm.
But V is a n-dimensional vector space on field F=Z_p.
Thus, it has p^n elements.
Because that is the total number of linear combinations.
But W is a n-dimensional vector space on field F=Z_p.
Thus, it has p^n elements.
Thus,
|Hom(V,W)|=p^n * p^n= p^(2n).
But I am not sure. I am not familar with Linear Algebra at a very high level. Thus, I can be wrong. However this answer makes sense.
Since V and W are field themselves and have the same cardinality a famous theorem from field theory states the V and W are isomorphic to each other. Hence the number of different linear transformations (of homomorphism in this case) from V to W is the same the same the number of homomorphisms from V to V because of isomorphism. But then we are asking how many elements are in the endomorphism ring End(V)? According to this Ring of endomorphiss.. The Endomorphism ring is isomorphic to M_nn(V), that is the ring of all n x n matrices having the elements of the field as its members. Since there are n rows and n colomns and for each entry there are p different entries (number of elements in F) thus in total there are p^(2n) many different matrices. Which confirms with the theorem.