Proving Unitary Operators = e^iA for Self Adjoint Matrix

[skyroot]

Homework Statement

Prove or disprove: if U is in the vector space of complex n x n matrices, then U is unitary if and only if U= e^iA, where A is some self adjoint matrix in same vector space, all of whose eigenvalues lie in the interval [0,2pi)

Homework Equations

A is self adjoint; A* = A, and A is unitarily equivalent to a diagonal matrix, A=PDP*, where P is a unitary matrix whose columns consist of eigenvectors of A which form an orthonormal basis for the vector space.
e^iA= P(a diagonal matrix with first entry, D'₁₁= e^it₁, and last entry D'_nn=e^it_n)P* . Where t₁ thru t_n
are the (real) eigenvalues of A, and P is the matrix defined above.
Unitary operators:
UU*= I , the defn of unitary
Unitary operators are normal, all eigenvalues have abs value 1, and ||Uv|| = ||v|| for any vector in the vector space U is acting on.

The Attempt at a Solution

I've proved it one of the directions, starting by assuming U=e^iA, putting it into the form i described above, and showing U*=e^-iA, simply multiplying UU* gives I, which is the defn of unitary operator. I think I need a small hint for how to approach the proof from the other direction,i.e. starting with assuming that U is unitary can one show that it equals e^iA for some self adjoint A. Questioning if only the first direction is true, and if the other direction can be proved at all. Trying to think of a counterexample. Thank you for your help.

 

[skyroot]

Nevermind, i got it