Show that there do not exist invertible matrices such that and .
Suppose, on the contrary, that such matrices do exist. Then
Using the fact that we deduce that
The last means that which is impossible because both and are invertible ( and so must be the product ). Hence, the conclusion follows.
Suppose that all eigenvalues of are positive real numbers. Show that
Let the eigenvalues of be , . Consider the Jordan normal form of ; this Jordan form is an upper triangular matrix that has the eigenvalues of in the main diagonal. Let this matrix be called . Furthermore , as a matrix and its Jordan normal form are similar. As is upper triangular, its inverse is given by an upper triangular matrix whose diagonal entries are the inverses of the diagonal entries of . That is,
Let be positive real numbers such that . Prove that
Well if we apply AM-GM to we obtain
and similarly if we apply AM – GM to we obtain
We have successively,
Let be a group homomorphism. We proved in this question that its kernel is infinite. We are proving now that it is also uncountable.
Before we proceed with the proof we are stating that not all homomorphisms are of the form . We can find non trivial homomorphisms. But all continuous are of the above form.
Let be the group of roots of unity. Both groups and are uniquely divisible and thus are vector spaces over .
Since the positive reals are closed under multiplication, it’s easy to see that
Using the axiom of choice , we construct a group homomorphism
Hence the kernel is uncountable.
Note: is a direct summand of . That is because
( since is injective ) and so the exact sequence
Let be a homomorphism. Prove that kernel of is infinite.
If is a finite subgroup then all must have norm , i.e. . ( Otherwise otherwise is an infinite sequence of distinct elements in . )
Suppose , that is finite and is the kernel of . Then let and
So, . But this is a contradiction since .
Hence, no finite subgroup of can be a kernel.
Note: The homomorphisms that can easily be described are for the form .