紧急求教相关矩阵的分解

Yes, you can.
Please refer to the help file for sqrtm() in Matlab, then you will find the function sqrtm() acts as:
For a square matrix A with eigen-decomposition A=V*D*inv(V) where D=diag{a_1,...,a_n} with a_1,...,a_n being the eigenvalues, the function sqrtm() returns B:=sqrtm(A)=V*diag{sqrt(a_1),...,sqrt(a_n)}*inv(V), so that B^2=A.
For a PSD matrix R, it has the eigen-decomposition as R=U'*D*U, where U is unitary (U'*U=I) and the eigenvalues a_1,...,a_n>=0. Then R1:=sqrt(R)=U'*diag{sqrt(a_1),...,sqrt(a_n)}*U which is also PSD (so that Hermitian) with R1'=R1. Then R1*R1'=R1^2=R.
Note that given a PSD matrix R, the solution for R1*R1'=R may not be unique. Neither is for R1^2=R. However, for a PSD R with R1^2=R and R1 also PSD, the solution R1 is unique. And sqrtm(R) will return this unique solution.