SU_L x SU_R and SU_V x SU_A equivalence

I want to show that the group [itex]SU_L(N)\times SU_R(N)[/itex] is the same as [itex]SU_V(N)\times SU_A(N)[/itex] - i.e. that it is possible to rewrite the transformation:



[tex]

\begin{cases}

\psi_L \to \psi'_L=V_L\,\psi_L\\

\psi_R \to \psi'_R=V_R\,\psi_R

\end{cases}

[/tex]



, where [itex]V_L[/itex] and [itex]V_R[/itex] are [itex]N\times N[/itex] [itex]SU(N)[/itex] matrices and [itex]\psi_L[/itex] and [itex]\psi_R[/itex] are [itex]N[/itex]-component vectors, as:



[tex]

\begin{cases}

\psi_L \to \psi'_L=V A\,\psi_L\\

\psi_R \to \psi'_R=V A^\dagger\, \psi_R

\end{cases}

[/tex]



, where again [itex]V[/itex] and [itex]A[/itex] are [itex]N\times N[/itex] [itex]SU(N)[/itex] matrices. The system:



[tex]

\begin{cases}

V_L=VA &\\

V_R=VA^\dagger

\end{cases}

[/tex]



must be solved for [itex]V[/itex] and [itex]A[/itex]:



[tex]

A^2=V_R^\dagger V_L

[/tex]



So, [itex]A[/itex] is the square root of the [itex]SU(N)[/itex] matrix [itex]V_R^\dagger V_L[/itex].



Problems:



1. The only way I can think of to give meaning to the square root of a matrix is a naive series expansion, but how to prove convergence?



2. Does the series expansion defines an [itex]SU(N)[/itex] matrix?





Random thoughts:



Probably 2. is the less problematic. I believe that it is true that if the square of a matrix equals an [itex]SU(N)[/itex] matrix, that matrix still lives in [itex]SU(N)[/itex]. It is not necessary to find the explicit expression of [itex] A[/itex] and [itex]V[/itex] so it would be enough to prove that the square root of an [itex]SU(N)[/itex] matrix is an [itex]SU(N)[/itex] matrix. Another approach is to consider infinitesimal elements of the group, then the previous system can be easily satisfied in the Lie algebra of the group. Is this enough to conclude that there are finite elements in the group that satisfy the same relationships?



Thank you for your time.